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The derivative becomes a toolkit

A weather app on a phone shows one day’s temperature. How fast is it changing at 10 am?
hour°C
One day’s temperature, midnight to midnight. It falls until early morning, rises until the afternoon, then falls through the evening.

Open the weather app on a phone, or the weather box in a newspaper. It often shows a graph of one day’s temperature. The line rises through the morning. It peaks in the afternoon. Then it falls through the evening and the night.

hour°C
Four questions of one graph: how fast at 10 am, when rising, when falling, when hottest, when coldest?

You can ask four questions of that one graph. How fast is the temperature changing at 10 am? When is it rising, and when is it falling? When is the day hottest? When is it coldest?

hour°C
hour°C an hour
The day’s temperature, and under it its slope from Chapter 5. What does the lower curve tell you about the upper one?

Chapter 5 gave you the slope of every function you meet in Class 12. The slope is a number at each point of a graph. Here is the question to carry to the last page. What does that slope tell you about the quantity itself?

One cubic, three marks: its slope 5.17 at x equals 0.35, an arrow where it falls, and dots where it is flat.

It has four parts, like the questions about the temperature. How fast does the quantity change when something else changes? Where does it rise, and where does it fall? Where does it reach a peak or a lowest point? And which choice makes it as large, or as small, as it can be?

You will answer all four with the same slope from Chapter 5. You use the existing derivative formulas you have, in new ways.

Brahmagupta
A SCHOLAR INDIA REMEMBERS

Brahmagupta was an astronomer at Bhillamala, now Bhinmal in Rajasthan, around 628 CE. His tradition predicted eclipses from the geometry of shadows; when an eclipse came at a different time than predicted, the tables were corrected. He also treated zero as a number in its own right: he asked what a zero means. Ask that of the temperature graph. At the hottest moment, how fast is the temperature changing? At a rate of $0$. Has nothing happened? No. That $0$ is how you find the peak.

Brahmagupta · the brahminy kite
Check yourself
  1. Chapter 5 built the derivative $f'(x)$ as a limit. What does Chapter 6 do with it?
    1. redefines $f'(x)$ using areas instead of limits
    2. applies that same $f'(x)$ to practical questions
    3. restricts $f'(x)$ to functions already known to increase
    4. replaces $f'(x)$ with a faster computation rule
    Check your answer
    1. redefines $f'(x)$ using areas instead of limits — Areas and integration are a separate tool from a different chapter, not a redefinition used here.
    2. ✓ applies that same $f'(x)$ to practical questions — (B) Chapter 6 takes the derivative already built in Chapter 5 and puts it to work on practical questions.
    3. restricts $f'(x)$ to functions already known to increase — The toolkit applies to any function with a derivative, not only ones already known to increase.
    4. replaces $f'(x)$ with a faster computation rule — Chapter 6 never introduces a faster or different derivative rule, it reuses the one Chapter 5 already built.
  2. Why can this chapter answer “is this quantity growing right now” without re-deriving the derivative from a limit?
    1. a new growth-rate formula is introduced, independent of $f'(x)$
    2. $f'(x)$ was built in Chapter 5; this chapter reuses it
    3. growth is read straight off a graph, never from $f'(x)$
    4. the chapter only ever handles constant functions
    Check your answer
    1. a new growth-rate formula is introduced, independent of $f'(x)$ — No independent growth-rate formula appears here, every tool in this chapter routes through $f'(x)$.
    2. ✓ $f'(x)$ was built in Chapter 5; this chapter reuses it — (B) $f'(x)$ was built once, in Chapter 5, and this chapter reuses it without rebuilding it from a limit.
    3. growth is read straight off a graph, never from $f'(x)$ — This chapter’s tools all read the sign or value of $f'(x)$, not a graph, to answer growth questions.
    4. the chapter only ever handles constant functions — A chapter built only for constant functions, where $f'(x)=0$ always, would have nothing left to ask.
  3. Rates of change, maxima and minima, and optimisation sit in one chapter because
    1. each topic is solved without ever computing $f'(x)$
    2. each topic needs its own separate derivative rule
    3. each topic reads the same $f'(x)$ in a different way
    4. each topic replaces $f'(x)$ with a numerical estimate
    Check your answer
    1. each topic is solved without ever computing $f'(x)$ — Every topic in this chapter computes $f'(x)$ at some point; none of them skip it.
    2. each topic needs its own separate derivative rule — One derivative, $f'(x)$, is reused across every topic, no topic needs a rule built just for it.
    3. ✓ each topic reads the same $f'(x)$ in a different way — (C) Rates, extrema, and optimisation are three different questions asked of the same underlying $f'(x)$.
    4. each topic replaces $f'(x)$ with a numerical estimate — No topic here trades the exact derivative for a numerical estimate, all the work stays exact.

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Before you start

You have already used a tool like each one below, long before now. Try each one below.

If any of these felt new, read the page named before going on.

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Rate of change via the chain rule

▲CONCEPT
A pump pushes in 900 cm³ of air a second. The air goes in steadily, but the radius does not grow steadily.
r=5r=10r=15
The balloon at radius 5, 10 and 15 cm. It holds about 524, 4,189 and 14,137 cm³ of air.

Think of blowing up a balloon with a pump at a birthday party. Each stroke of the pump pushes in about the same amount of air. At first the balloon swells fast. Later it seems to grow more and more slowly, though you pump just as hard. The air goes in at a steady rate, but the size does not grow at a steady rate.

rVV = (4/3)πr³
One more centimetre of radius: from 5 to 6 cm takes 381 cm³ of air, from 15 to 16 cm takes 3,020.

Two amounts are changing here, the air inside and the radius. They are linked, because a bigger radius holds more air. So how fast one changes is linked to how fast the other changes.

t
→
r(t)
slope dr/dt
→
V(r)
slope dV/dr = 4πr²
→
V
dV/dt = dV/dr × dr/dt
Two functions in a row: time gives the radius, the radius gives the volume. Their slopes multiply, dV/dt = dV/dr × dr/dt.

If $y = f(x)$, the slope $d y/d x$ is the RATE OF CHANGE of $y$ with respect to $x$. Its value at $x = x_0$ is the rate at that point. Now let $y$ and $x$ both change with time $t$. Chapter 5’s chain rule links their rates:

$d y/d t = d y/d x \cdot d x/d t$.

This needs $y$ to have a slope in $x$, and $x$ to have a slope in $t$.

dr/dt = 2.865 cm/s
→
× dV/dr
100π ≈ 314.16 at r = 5
→
dV/dt = 900 cm³/s
At r = 5 the volume grows 314.16 cm³ per centimetre of radius. So 900 cm³ a second grows the radius 2.865 cm a second.
rdr/dt
The same 900 cm³ a second: the radius grows 2.865 cm a second at r = 5, 0.716 at 10, only 0.318 at 15.

Let us check it on the balloon. Say the pump puts in 900 cubic centimetres of air a second. How fast does the radius grow when $r = 5$?

Worked example

Find $d r/d t$ at $r = 5$ when $d V/d t = 900$

  1. $V = (4/3) \pi r^3$
    The volume of a sphere of radius $r$.
  2. $d V/d r = 4 \pi r^2$
    Differentiate with respect to $r$: the 3 comes down, and $(4/3) \cdot 3 = 4$.
  3. $d V/d r = 4 \pi \cdot 5^2 = 100 \pi$
    Put in $r = 5$.
  4. $d V/d t = d V/d r \cdot d r/d t$
    The chain rule: the volume depends on the radius, and the radius depends on time.
  5. $900 = 100 \pi \cdot d r/d t$
    Put in the two rates you know, $d V/d t = 900$ and $d V/d r = 100 \pi$.
  6. *$d r/d t = 9/\pi$, about 2.9 cm a second*
    Divide both sides by $100 \pi$.
  7. $100 \pi \cdot 9/\pi = 900$
    Put the answer back: it gives the 900 the pump puts in.

The worked example below does the same at $r = 15$. There the radius grows at only about 0.32 cm a second.

A ripple’s radius grows from 3 cm to 4 cm at 0.5 cm a second. The ring, about 22.0 square centimetres, is the area added.

So a rate you cannot measure easily comes from one you can. The formula that links the two quantities does the work.

Your turn: the side of a square grows at 2 cm a second. How fast does its area grow when the side is 5 cm?

Brahmagupta
A SCHOLAR INDIA REMEMBERS

The air goes in at a steady rate. Predict: does the radius grow at a steady rate too? At $r = 5$ the radius grows about $2.9$ cm a second. At $r = 15$, only about $0.32$. The guess was wrong: the radius does not grow steadily. Each extra centimetre of radius needs a shell of air over the whole surface, $4 \pi r^2$, and that is why. At $r = 15$ that surface is $9$ times bigger, so the radius grows $9$ times more slowly.

Brahmagupta · the brahminy kite
Check yourself
  1. In the chain rule $d y/d t = ((d y)/(d x)) \cdot ((d x)/(d t))$, what does $d y/d t$ measure?
    1. how fast $x$ changes with $y$
    2. how fast $t$ changes with $x$
    3. how fast $y$ changes with $x$ alone
    4. how fast $y$ changes with time
    Check your answer
    1. how fast $x$ changes with $y$ — That inverts the roles, $x$ changing with $y$ is the separate rate $d x/d y$, not $d y/d t$.
    2. how fast $t$ changes with $x$ — $d t/d x$ is a different rate again, not the one the chain rule combines the factors to find.
    3. how fast $y$ changes with $x$ alone — $d y/d x$ alone leaves time out entirely; it is only one factor multiplied in the chain rule.
    4. ✓ how fast $y$ changes with time — (D) $d y/d t$ is the rate at which $y$ changes as time passes, the physical rate the chain rule builds.
  2. A sphere’s volume $V$ depends on its radius $r$, and $r$ changes with time $t$. Which product gives $d V/d t$?
    1. $((d V)/(d r)) \cdot ((d r)/(d t))$
    2. $((d r)/(d t)) \cdot ((d r)/(d t))$
    3. $((d V)/(d t)) \cdot ((d r)/(d t))$
    4. $((d V)/(d r)) \cdot ((d t)/(d r))$
    Check your answer
    1. ✓ $((d V)/(d r)) \cdot ((d r)/(d t))$ — (A) $d V/d t = ((d V)/(d r)) \cdot ((d r)/(d t))$ multiplies the geometric rate by the time rate, exactly as the chain rule requires.
    2. $((d r)/(d t)) \cdot ((d r)/(d t))$ — Squaring $d r/d t$ never brings $V$ into the product at all, so it cannot give a volume rate.
    3. $((d V)/(d t)) \cdot ((d r)/(d t))$ — This reuses $d V/d t$ inside its own definition, the chain rule needs two OTHER rates multiplied.
    4. $((d V)/(d r)) \cdot ((d t)/(d r))$ — $d t/d r$ inverts the time rate; the correct factor is $d r/d t$, not its reciprocal.
  3. A cube’s volume $V$ depends on its edge length $s$, and $s$ grows with time. Which quantity does $d V/d s$ and $d s/d t$ together give directly?
    1. $d^2 V/d t^2$
    2. $d V/d s$ alone, with no time term
    3. $d s/d V$
    4. $d V/d t$
    Check your answer
    1. $d^2 V/d t^2$ — A second time derivative needs more than one multiplication of the two given first-order rates.
    2. $d V/d s$ alone, with no time term — Dropping the time factor leaves only the geometric rate $d V/d s$, never the physical rate asked for.
    3. $d s/d V$ — $d s/d V$ is the reciprocal pairing of the two rates, not the product they form together.
    4. ✓ $d V/d t$ — (D) Multiplying $d V/d s$ by $d s/d t$ gives $d V/d t$, the same chain-rule pattern as the sphere example.
  4. A student writes $d V/d t = ((d V)/(d r)) + ((d r)/(d t))$ for a sphere’s volume. What is wrong?
    1. the formula should use $d t/d r$ instead of $d r/d t$
    2. the two rates must be divided, not added
    3. the two rates must be multiplied, not added
    4. the formula needs a third rate multiplied in
    Check your answer
    1. the formula should use $d t/d r$ instead of $d r/d t$ — Swapping in $d t/d r$ inverts the time rate, which is a different mistake from the addition shown.
    2. the two rates must be divided, not added — Dividing the two rates is not the chain rule’s operation either, multiplication is what is required.
    3. ✓ the two rates must be multiplied, not added — (C) The chain rule multiplies the two rates together; addition is the wrong operation entirely.
    4. the formula needs a third rate multiplied in — Only two rates combine here; adding a third would double-count the single chain already in play.
The same balloon at radius 12 cm and 15 cm. Air goes in at 900 cubic centimetres a second, and the step takes 7.7 seconds.
dr/dt = 1/π ≈ 0.318
→
× 4π(15)²
900π
→
900 cm³/s
The check: 1/π cm a second times 900π gives 900 cm³ a second, the rate the pump puts in.
Worked example

Find $d r/d t$ for a balloon’s radius when its volume grows at $900$ $\text{cm}^3$/s

  1. A spherical balloon’s volume increases at $900$ $\text{cm}^3$ per second; find $d r/d t$ when $r = 15$ cm
    The air goes in at 900 cubic centimetres a second. You want how fast the radius grows when it is 15 cm.
  2. $V = (4/3) \pi r^3$
    The volume of a sphere of radius $r$.
  3. $d V/d t = 4 \pi r^2 \cdot d r/d t$
    Both $V$ and $r$ change with time. So differentiate with respect to $t$, and use the chain rule on $r^3$.
  4. $900 = 4 \pi (15)^2 \cdot d r/d t = 900 \pi \cdot d r/d t$
    Put in the rate you know, $d V/d t = 900$, and $r = 15$.
  5. *$d r/d t = 1/\pi$ cm/s, about $0.32$ cm/s*
    Divide both sides by $900 \pi$. At this size the radius grows only about a third of a centimetre a second.
  6. $4 \pi (15)^2 \cdot 1/\pi = 4 \cdot 225 = 900$
    Check by putting the answer back. It gives 900 cubic centimetres a second, the rate you were given.
Check yourself
  1. For the balloon problem ($d V/d t = 900$, $r = 15$), the chain rule gives $900 = 4 \pi (15)^2 \cdot ((d r)/(d t))$. What does $4 \pi (15)^2$ equal?
    1. $225 \pi$
    2. $900 \pi$
    3. $60 \pi$
    4. $1800 \pi$
    Check your answer
    1. $225 \pi$ — Dropping the coefficient $4$ gives $\pi \cdot 15^2 = 225 \pi$ instead of multiplying it in.
    2. ✓ $900 \pi$ — (B) $4 \pi (15)^2 = 4 \pi \cdot 225 = 900 \pi$, squaring the radius first and keeping the coefficient $4$.
    3. $60 \pi$ — Dropping the square on $15$ gives $4 \pi \cdot 15 = 60 \pi$ instead of squaring it first.
    4. $1800 \pi$ — Doubling the correctly-squared value gives $1800 \pi$, an arithmetic slip after the right setup.
  2. Using $d r/d t = 1/\pi$ cm/s, about how fast is the balloon’s radius growing?
    1. about $3.14$ cm/s
    2. about $0.16$ cm/s
    3. about $0.32$ cm/s
    4. about $900$ cm/s
    Check your answer
    1. about $3.14$ cm/s — $1/\pi$ is roughly $0.32$, not $\pi$ itself, the two are reciprocals, not the same number.
    2. about $0.16$ cm/s — Halving the correct value gives about $0.16$, as if the rate were $1/(2 \pi)$ instead of $1/\pi$.
    3. ✓ about $0.32$ cm/s — (C) $1/\pi$ is roughly $0.32$, so the radius grows about $0.32$ cm each second.
    4. about $900$ cm/s — $900$ cm³/s is the given volume rate, not the radius rate the problem asks to be solved for.
  3. If the same balloon’s volume increased at $1800$ cm³/s instead of $900$, with $r = 15$ cm still, what would $d r/d t$ become?
    1. $1/\pi$ cm/s
    2. $2/\pi$ cm/s
    3. $1/(2 \pi)$ cm/s
    4. $4/\pi$ cm/s
    Check your answer
    1. $1/\pi$ cm/s — $1/\pi$ is the original rate, doubling $d V/d t$ at the same radius must double $d r/d t$ as well.
    2. ✓ $2/\pi$ cm/s — (B) With $r$ fixed at $15$, $d r/d t$ scales directly with $d V/d t$, so doubling $900$ to $1800$ doubles $1/\pi$ to $2/\pi$.
    3. $1/(2 \pi)$ cm/s — Halving $1/\pi$ moves the wrong direction, doubling $d V/d t$ makes the radius grow faster, not slower.
    4. $4/\pi$ cm/s — Quadrupling $1/\pi$ overshoots the effect of doubling $d V/d t$, which only doubles the result.

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Reading the sign of the derivative

▲CONCEPT
hour°C
Morning: slope 1.36 °C an hour, so it is warming. Evening: slope −1.36, so it is cooling. The sign gives the direction.

Look at the day’s temperature graph again. Through the morning the line rises, so its slope is positive. Through the evening the line falls, so its slope is negative. The sign of the slope tells you the direction. You can read it without plotting a single reading.

strictly increasing
y = x³
increasing
Left: a bigger input always gives a bigger output. Right: on the flat stretch two outputs are equal.
hour°C
From 4 pm to midnight, later always means cooler: 32.2 °C at 6 pm, 30 at 8 pm, 27 at 10 pm. Strictly decreasing.

Each direction has an exact name.

rises
falls
level
slope 0 all along
Slope above 0 all along: the curve rises. Below 0 all along: it falls. Exactly 0 all along: it stays level.

The test reads the direction from the sign of $f'$. Let $f$ be continuous on $[a, b]$ and differentiable on $(a, b)$.

y = x²
enrichment
Enrichment. Slope positive all the way from 1 to 3 makes the chord (the straight line from (1, 1) to (3, 9)) have a positive slope too: 4. So f(3) = 9 is above f(1) = 1.

With $f'(x) > 0$ the rise is in fact strict. Chapter 5’s mean value theorem is the reason the test works. Between any two inputs there is a point whose slope equals the slope of the chord. That theorem stays enrichment, so the board does not ask you to prove the test.

Left: the curve is flat for one instant at x equals 2, then climbs on. Right: it stays level from 1.5 to 2.5.

A slope of 0 at one single point does not stop a rise. $f(x) = x^3$ has $f'(0) = 0$, yet it is strictly increasing. Only a slope of 0 along a whole stretch makes $f$ constant there.

f′ < 0
f′ > 0
1
3
Solve 2x − 4 = 0: x = 2 cuts the line in two. Test one value per piece: f′(1) = −2, f′(3) = 2.

To use the test, solve $f'(x) = 0$ first. Those points split the line into intervals. Then test one value in each interval and read the sign.

y = x² − 4x
xy′y′ = 2x − 4
x² − 4x and its slope 2x − 4. The slope is negative before 2, positive after: the curve falls to −4, then rises.

Let us try $f(x) = x^2 - 4x$. Where does it fall, and where does it rise?

Worked example

Find where $f(x) = x^2 - 4x$ falls and where it rises

  1. $f'(x) = 2x - 4$
    Differentiate term by term: $x^2$ gives $2x$, and $4x$ gives 4.
  2. $2x - 4 = 0$, so $x = 2$
    Set the slope to 0 and solve: add 4 to both sides, then divide by 2.
  3. $f'(1) = -2 < 0$
    Test one value left of 2: the slope is negative, so $f$ falls on $x < 2$.
  4. $f'(3) = 2 > 0$
    Test one value right of 2: the slope is positive, so $f$ rises on $x > 2$.
  5. *$f$ falls until $x = 2$ and rises after it*
    Read the two signs in order: negative, then positive.
  6. $f(1) = -3$, $f(2) = -4$, $f(3) = -3$
    Check with numbers: the values fall to $-4$ at 2, then rise again.

So one sign on each interval gives you the shape of the whole graph.

Your turn: find where $f(x) = x^3 - 3x$ is increasing and where it is decreasing.

Check yourself
  1. To tell whether $f$ is increasing on an interval, the monotonicity test looks at
    1. the value of $f(x)$ at the interval’s midpoint
    2. the sign of $f'(x)$ throughout the interval
    3. the sign of $f''(x)$ throughout the interval
    4. the size of $f'(x)$ compared to $f(x)$
    Check your answer
    1. the value of $f(x)$ at the interval’s midpoint — A single midpoint value cannot certify behaviour across the WHOLE interval on its own.
    2. ✓ the sign of $f'(x)$ throughout the interval — (B) The monotonicity test reads only the sign of $f'(x)$ across the whole interval.
    3. the sign of $f''(x)$ throughout the interval — $f''(x)$’s sign belongs to the second derivative test, used for extrema, not for monotonicity.
    4. the size of $f'(x)$ compared to $f(x)$ — Comparing $f'(x)$ to $f(x)$ mixes two unrelated quantities; only the sign of $f'(x)$ matters.
  2. Suppose $f'(x) < 0$ for every $x$ in $(2, 5)$. What can you conclude about $f$ on that interval?
    1. $f$ is constant throughout $(2, 5)$
    2. $f$ is decreasing throughout $(2, 5)$
    3. $f$ has a local minimum at $x = 2$
    4. $f$ is increasing throughout $(2, 5)$
    Check your answer
    1. $f$ is constant throughout $(2, 5)$ — Constant behaviour requires $f'(x) = 0$ throughout, not a derivative that stays negative.
    2. ✓ $f$ is decreasing throughout $(2, 5)$ — (B) $f'(x) < 0$ throughout $(2,5)$ means $f$ decreases across the whole interval, by the monotonicity test.
    3. $f$ has a local minimum at $x = 2$ — A local minimum needs a sign CHANGE around the point, not a constant negative sign throughout.
    4. $f$ is increasing throughout $(2, 5)$ — A negative $f'(x)$ signals decreasing behaviour, the opposite conclusion from increasing.
  3. A student claims $f$ must be “more strongly” increasing wherever $f'(x)$ is a large positive number than where it is a small positive number. What is wrong with treating the test this way?
    1. the test needs $f''(x)$ to confirm the size claim
    2. the test cannot distinguish increasing from decreasing at all
    3. the test only needs the size of $f'(x)$, not its sign
    4. the test only needs $f'(x)$ positive in sign, not in size
    Check your answer
    1. the test needs $f''(x)$ to confirm the size claim — Bringing in $f''(x)$ answers a different question, about extrema, not this monotonicity claim.
    2. the test cannot distinguish increasing from decreasing at all — The test distinguishes increasing from decreasing perfectly well, using the sign of $f'(x)$ alone.
    3. the test only needs the size of $f'(x)$, not its sign — This reverses the actual rule, the test uses the SIGN of $f'(x)$; its size plays no role at all.
    4. ✓ the test only needs $f'(x)$ positive in sign, not in size — (D) Monotonicity only asks whether $f'(x)$ is positive or negative, a larger positive value is not “more” increasing.

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Local maxima and minima

▲CONCEPT
hour°C
Near 4 pm: an hour before, rising 0.41 °C an hour; at 4 pm, slope 0; an hour after, falling 0.41 °C an hour.
hour°C
Near 4 am, the coldest moment, it runs the other way: falling 0.41 °C an hour, then slope 0, then rising.

Go back to the hottest moment of the day. Just before it, the temperature is still rising, so the slope is positive. Just after it, the temperature is falling, so the slope is negative. At the top itself the slope is zero. The coldest moment, early in the morning, is the other way round. The temperature falls until then and rises after it.

hour°Cnamename
hourslope
Slope from + to −: the day peaks, 33 °C at 4 pm. From − to +: its lowest, 21 °C at 4 am.

So a peak is where the slope changes from positive to negative. A lowest point is where it changes from negative to positive.

y = 3 − |x − 2|
A peak can be a corner, with no slope at all: 3 − |x − 2| peaks at x = 2.
y = 4x − x²
Near x = 2 no value of 4x − x² rises above f(2) = 4. That makes 4 a local maximum.

Such a peak is a LOCAL MAXIMUM. $f$ has a local maximum at $c$ when no value of $f$ near $c$ is bigger than $f(c)$. $f$ has a LOCAL MINIMUM at $c$ when no value of $f$ near $c$ is smaller than $f(c)$. A point $c$ where $f'(c) = 0$, or where $f$ has no slope at all, is a CRITICAL POINT. Every local maximum or minimum inside an open interval is at a critical point.

Each curve is flat at x equals 2. Rising then falling gives a peak, falling then rising a dip, rising on both sides neither.

The FIRST DERIVATIVE TEST decides what happens at a critical point. Let $f$ be continuous at a critical point $c$.

The arch peaks at x equals 2, height 4. Its slope is positive on the left and negative on the right.
f′ > 0
f′ < 0
1.9
2.1
Test just either side of 2: f′(1.9) = 0.2 and f′(2.1) = −0.2. Positive, then negative: a local maximum.

Let us check it on $f(x) = 4x - x^2$, the arch in the figure.

Worked example

Test the critical point of $f(x) = 4x - x^2$

  1. $f'(x) = 4 - 2x$
    Differentiate term by term.
  2. $4 - 2x = 0$, so $x = 2$
    Set the slope to 0 and solve: $x = 2$ is the one critical point.
  3. $f'(1.9) = 0.2 > 0$
    Test just left of 2: the slope is positive.
  4. $f'(2.1) = -0.2 < 0$
    Test just right of 2: the slope is negative.
  5. *A local maximum at $x = 2$, and $f(2) = 4$*
    Positive, then negative: the first derivative test gives a peak, of height $8 - 4 = 4$.
  6. $f(1.9) = 3.99$ and $f(2.1) = 3.99$, both below 4
    Check either side: both values are a little below the peak.

With this test you read every peak and every lowest point off the signs alone.

Your turn: find the critical points of $f(x) = x^4 - 4x^3$. Use the test to decide what happens at each one.

Check yourself
  1. A point $x=a$ is called a critical point of $f$ when
    1. $f'(a) = 0$ or $f'(a)$ fails to exist
    2. $f'(a)$ is the largest value of $f'$ anywhere
    3. $f(a) = 0$ or $f(a)$ fails to exist
    4. $f''(a) = 0$ or $f''(a)$ fails to exist
    Check your answer
    1. ✓ $f'(a) = 0$ or $f'(a)$ fails to exist — (A) A critical point is where $f'(a) = 0$ or $f'(a)$ does not exist.
    2. $f'(a)$ is the largest value of $f'$ anywhere — A critical point needs $f'(a)$ to be zero or undefined, not to be the largest value $f'$ ever reaches.
    3. $f(a) = 0$ or $f(a)$ fails to exist — The function’s own value, $f(a)$, does not define a critical point, its DERIVATIVE does.
    4. $f''(a) = 0$ or $f''(a)$ fails to exist — The second derivative belongs to a different test; a critical point is defined by the FIRST derivative.
  2. The first derivative test classifies a critical point $a$ as a local maximum when $f'$
    1. stays positive on both sides of $a$
    2. equals zero on both sides of $a$
    3. changes from negative to positive around $a$
    4. changes from positive to negative around $a$
    Check your answer
    1. stays positive on both sides of $a$ — Staying positive on both sides means no sign change at all, so $a$ is neither a max nor a min.
    2. equals zero on both sides of $a$ — $f'$ equal to zero on both sides describes a flat stretch, not a single classified critical point.
    3. changes from negative to positive around $a$ — Negative-to-positive is the OPPOSITE case, that pattern signals a local minimum, not a maximum.
    4. ✓ changes from positive to negative around $a$ — (D) Positive-to-negative around $a$ is exactly the first derivative test’s condition for a local maximum.
  3. The first derivative test classifies a critical point $a$ as a local minimum when $f'$
    1. changes from positive to negative around $a$
    2. changes from negative to positive around $a$
    3. stays negative on both sides of $a$
    4. equals zero only exactly at $a$
    Check your answer
    1. changes from positive to negative around $a$ — Positive-to-negative is the OPPOSITE case, that pattern signals a local maximum, not a minimum.
    2. ✓ changes from negative to positive around $a$ — (B) Negative-to-positive around $a$ is exactly the first derivative test’s condition for a local minimum.
    3. stays negative on both sides of $a$ — Staying negative on both sides means no sign change at all, so $a$ is neither a max nor a min.
    4. equals zero only exactly at $a$ — $f'(a)=0$ alone, without checking the sign on either side, does not classify the point by itself.
  4. At a critical point, $f'$ is positive just before $a$ and positive just after $a$ too: no sign change at all. What does the first derivative test conclude?
    1. the test cannot be applied to this point at all
    2. $a$ must be a local minimum regardless
    3. $a$ is neither a local maximum nor a minimum
    4. $a$ must be a local maximum regardless
    Check your answer
    1. the test cannot be applied to this point at all — The test still applies here, it simply reports that no sign change means no local extremum at all.
    2. $a$ must be a local minimum regardless — A local minimum needs the pattern negative-to-positive; staying positive throughout is different.
    3. ✓ $a$ is neither a local maximum nor a minimum — (C) No sign change around $a$ means the first derivative test reports neither a max nor a min.
    4. $a$ must be a local maximum regardless — A local maximum needs the pattern positive-to-negative; staying positive throughout is different.
▲CONCEPT
cup
y = x²
cap
y = −x²
A cup, y = x², and a cap, y = −x². The cup holds a lowest point at 0; the cap holds a peak.

The SECOND DERIVATIVE TEST reads the bend of the curve at a critical point. Chapter 5 showed that $f''(x) > 0$ bends the curve up like a cup. And $f''(x) < 0$ bends it down like a cap. A cup holds a lowest point, and a cap holds a peak.

Each curve is flat at x equals 2. Cup: a local minimum. Cap: a local maximum. Third: second derivative 0, so no verdict.

Let $f'(c) = 0$, and let $f$ be twice differentiable at $c$.

Here is why the test works when $f''(c) > 0$:

Why it works

Why $f''(c) > 0$ gives a local minimum

  1. $f''(c) = \lim_{x \to c} (f'(x) - f'(c))/(x - c)$
    The second derivative is the derivative of $f'$, written as a limit.
  2. $f''(c) = \lim_{x \to c} (f'(x))/(x - c)$
    Put in $f'(c) = 0$.
  3. $(f'(x))/(x - c) > 0$ for $x$ near $c$
    A positive limit means the fraction is positive for every $x$ close enough to $c$.
  4. $x < c$: $x - c < 0$, so $f'(x) < 0$
    Left of $c$ the bottom is negative, so the top is negative too.
  5. $x > c$: $x - c > 0$, so $f'(x) > 0$
    Right of $c$ the bottom is positive, so the top is positive too.
  6. *A local minimum at $c$*
    The slope goes from negative to positive, so the first derivative test gives a minimum.

The signs swap when $f''(c) < 0$, and the same steps give a local maximum.

y = x²
xy′y′ = 2x
xy″y″ = 2
y″ > 0 all along: a cup
x², its slope 2x and its second derivative 2. The slope climbs through 0 from negative to positive: a lowest point.
y = −x²
xy′y′ = −2x
xy″y″ = −2
y″ < 0 all along: a cap
−x², its slope −2x and its second derivative −2. The slope falls through 0 from positive to negative: a peak.

Check it on two curves you know, $y = x^2$ and $y = -x^2$.

Worked example

Use the second derivative test on $y = x^2$ and $y = -x^2$

  1. $f(x) = x^2$: $f'(x) = 2x$, so $f'(0) = 0$
    Differentiate: the slope is 0 at $x = 0$, so 0 is a critical point.
  2. $f''(x) = 2$, so $f''(0) = 2 > 0$
    Differentiate once more to get the second derivative, and put in $x = 0$.
  3. *$f''(0) > 0$: a local minimum at 0*
    $f''(0) > 0$ bends the curve up, so the second derivative test gives a minimum.
  4. $f(x) = -x^2$: $f'(0) = 0$ and $f''(0) = -2 < 0$
    The same two steps, with $f'(x) = -2x$ and $f''(x) = -2$.
  5. *$f''(0) < 0$: a local maximum at 0*
    $f''(0) < 0$ bends the curve down, so the test gives a maximum.
x⁴
minimum
−x⁴
maximum
x³
neither
x⁴, −x⁴ and x³ all have f′(0) = 0 and f″(0) = 0. Yet one has a minimum, one a maximum, one neither.

Let us see why $f''(c) = 0$ says nothing. Take three curves:

Worked example

Find $f'(0)$ and $f''(0)$ for $x^4$, $-x^4$ and $x^3$

  1. $x^4$: $f'(x) = 4x^3$ and $f''(x) = 12x^2$, both 0 at $x = 0$
    Differentiate twice, then put in $x = 0$.
  2. $-x^4$: $f'(x) = -4x^3$ and $f''(x) = -12x^2$, both 0 at $x = 0$
    The same, with every sign changed.
  3. $x^3$: $f'(x) = 3x^2$ and $f''(x) = 6x$, both 0 at $x = 0$
    The same again: both derivatives are 0 at 0.
  4. *Yet $x^4$ has a minimum at 0, $-x^4$ a maximum, and $x^3$ neither*
    The signs of $f'$ on each side decide, as the figure above shows.

One value of $f''$ cannot tell these three apart.

When $f''$ is easy to find, this test is quicker than checking the signs on each side. When it gives 0, the sign check still decides.

Your turn: use the second derivative test on $f(x) = x^3 - 3x$ at $x = 1$ and at $x = -1$.

Brahmagupta
A SCHOLAR INDIA REMEMBERS

The second derivative test gives $0$ for $x^4$ at $0$. Has the test told you nothing? It has told you one thing: this test cannot decide here. That is not the same as no peak and no lowest point. So take a second reading, the signs. The slope $4x^3$ is $-0.004$ at $x = -0.1$ and $0.004$ at $x = 0.1$. Negative, then positive. So $x^4$ has a lowest point at $0$.

Brahmagupta · the brahminy kite
Check yourself
  1. The second derivative test requires which condition to already hold at $x=a$ before it can be applied?
    1. $f''(a) = 0$
    2. $f'(a) = 0$
    3. $f(a) = 0$
    4. $f'(a)$ fails to exist
    Check your answer
    1. $f''(a) = 0$ — $f''(a)=0$ is the test’s INCONCLUSIVE outcome, not the precondition needed to run it at all.
    2. ✓ $f'(a) = 0$ — (B) The second derivative test starts from a critical point where $f'(a) = 0$, then checks the sign of $f''(a)$.
    3. $f(a) = 0$ — The function’s own value being zero has no bearing on whether $a$ is a critical point in the first place.
    4. $f'(a)$ fails to exist — The second derivative test specifically needs $f'(a)=0$, a derivative that fails to exist rules it out.
  2. At a critical point where $f''(a) > 0$, the second derivative test concludes
    1. $a$ is a local maximum
    2. the test is inconclusive at $a$
    3. $a$ is a local minimum
    4. $a$ is neither: a point of inflection
    Check your answer
    1. $a$ is a local maximum — A positive $f''(a)$ gives a MINIMUM, not a maximum, the sign and the label here are mismatched.
    2. the test is inconclusive at $a$ — The test is conclusive here, “inconclusive” only applies when $f''(a) = 0$, not when it is positive.
    3. ✓ $a$ is a local minimum — (C) $f''(a) > 0$ with $f'(a) = 0$ gives a local minimum.
    4. $a$ is neither: a point of inflection — A positive $f''(a)$ is a definite sign, not zero, so the point is a minimum, not a point of inflection.
  3. At a critical point where $f''(a) < 0$, the second derivative test concludes
    1. the test is inconclusive at $a$
    2. $a$ is a local minimum
    3. $a$ is a local maximum
    4. $a$ is neither: a point of inflection
    Check your answer
    1. the test is inconclusive at $a$ — The test is conclusive here, “inconclusive” only applies when $f''(a) = 0$, not when it is negative.
    2. $a$ is a local minimum — A negative $f''(a)$ gives a MAXIMUM, not a minimum, the sign and the label here are mismatched.
    3. ✓ $a$ is a local maximum — (C) $f''(a) < 0$ with $f'(a) = 0$ gives a local maximum.
    4. $a$ is neither: a point of inflection — A negative $f''(a)$ is a definite sign, not zero, so the point is a maximum, not a point of inflection.
  4. At a critical point where $f''(a) = 0$, what must be done to classify $a$?
    1. conclude $a$ is neither a max nor a min, since $f''(a) = 0$ rules both out
    2. fall back to the first derivative test: it is inconclusive here
    3. conclude $a$ is both a local max and a local min
    4. conclude the function has no critical point at $a$ after all
    Check your answer
    1. conclude $a$ is neither a max nor a min, since $f''(a) = 0$ rules both out — $f''(a)=0$ does not RULE OUT an extremum, it leaves the classification open, not settled as “neither”.
    2. ✓ fall back to the first derivative test: it is inconclusive here — (B) When $f''(a) = 0$, the second derivative test gives no answer, so the first derivative test decides instead.
    3. conclude $a$ is both a local max and a local min — Nothing about $f''(a) = 0$ makes $a$ both a local maximum and a local minimum. The test is silent, so the first derivative test decides.
    4. conclude the function has no critical point at $a$ after all — $f'(a)=0$ already established $a$ as a critical point; a zero second derivative does not undo that.
Worked example

Find the launch angle that gives a projectile’s maximum range, using both tests

  1. $R (\theta) = (v_0^2 \sin (2 \theta))/g$, with $v_0$ and $g$ fixed and $\theta$ the launch angle
    Physics 11, Chapter 3, Motion in a Plane, gives this as the horizontal range of a projectile.
  2. $d R/(d \theta) = (2 v_0^2 \cos (2 \theta))/g$
    differentiate with respect to $\theta$; $v_0^2/g$ is a constant factor, chain rule on $\sin (2 \theta)$.
  3. $d R/(d \theta) = 0$ when $\cos (2 \theta) = 0$
    Set the slope to 0: $2 v_0^2 / g$ is not 0, so the cosine must be.
  4. $2 \theta = \pi/2$, so $\theta = \pi/4$
    For $0 < \theta < \pi/2$, $2 \theta$ lies between 0 and $\pi$, where only $\pi/2$ has cosine 0.
  5. $(d^2 R)/(d \theta^2) = -(4 v_0^2 \sin (2 \theta))/g$
    Differentiate again: the chain rule turns $\cos (2 \theta)$ into $-2 \sin (2 \theta)$.
  6. $\sin (2 \cdot \pi/4) = \sin (\pi/2) = 1$
    Put in $\theta = \pi/4$.
  7. $(d^2 R)/(d \theta^2) = -(4 v_0^2)/g < 0$
    The second derivative is negative at $\pi/4$, so the second derivative test gives a maximum.
  8. for $\theta$ slightly less than $\pi/4$, $\cos (2 \theta) > 0$ so $R$ is increasing; for $\theta$ slightly more, $\cos (2 \theta) < 0$ so $R$ is decreasing
    the first derivative test, a sign change from positive to negative, agrees with the second derivative test.
  9. *$R$ is greatest at $\theta = \pi/4$ (45 degrees), matching the launch-angle fact Physics 11 already states*
    both tests, run on the same function, give the same 45-degree answer.
Peak 19 at x equals 1, dip 15 at x equals 3. Left: the slope’s sign on each side. Right: the second derivative’s sign.
y = x³ − 6x² + 9x + 15
xy′
The slope is 9 at 0, −3 at 2, 9 at 4: rising, falling, rising. A peak at 1, a lowest point at 3.
lowest point to peak
17
The check: f(2) = 17 lies between the lowest point, 15, and the peak, 19, as it must.
Worked example

Classify the critical points of $f(x) = x^3 - 6x^2 + 9x + 15$ using both tests

  1. $f(x) = x^3 - 6x^2 + 9x + 15$; find and classify its critical points
    A cubic. First find where its slope is zero.
  2. $f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$
    Differentiate, then factor. The slope is zero where a bracket is zero.
  3. Critical points at $x = 1$ and $x = 3$
    These are the two critical points.
  4. First derivative test: $f'$ is positive before $x=1$, negative between $1$ and $3$, positive after $3$
    Test one value in each stretch, such as $x = 0$, $2$ and $4$. The slopes there are 9, $-3$ and 9.
  5. $x = 1$ is a local max (positive to negative); $x = 3$ is a local min (negative to positive)
    Rising then falling at 1 gives a peak. Falling then rising at 3 gives a lowest point.
  6. Second derivative test: $f''(x) = 6x - 12$
    Now the second test. Differentiate $f'(x)$ once more.
  7. $f''(1) = -6 < 0$ confirms a max; $f''(3) = 6 > 0$ confirms a min
    At $x = 1$ the curve is a cap, so a peak. At $x = 3$ it is a cup, so a lowest point.
  8. *Both tests agree: local max value $f(1) = 19$, local min value $f(3) = 15$*
    Put each $x$ back into $f$ to get the heights.
  9. $f(2) = 8 - 24 + 18 + 15 = 17$, between $15$ and $19$
    Check with a point between the two. The curve falls from 19 to 15, so $f(2)$ must lie between them.
Check yourself
  1. For $f(x) = x^3 - 6x^2 + 9x + 15$, factoring $f'(x) = 3x^2 - 12x + 9$ gives $3(x-1)(x-3)$. What are the critical points?
    1. $x = 0$ and $x = 3$
    2. $x = 1$ and $x = 3$
    3. $x = -1$ and $x = -3$
    4. $x = 3$ and $x = 9$
    Check your answer
    1. $x = 0$ and $x = 3$ — $x=0$ never solves $(x-1)(x-3)=0$; only $x=1$ and $x=3$ do.
    2. ✓ $x = 1$ and $x = 3$ — (B) $3(x-1)(x-3) = 0$ gives the critical points $x=1$ and $x=3$ directly from the factored form.
    3. $x = -1$ and $x = -3$ — The signs are flipped, $(x-1)(x-3)=0$ gives $x=1$ and $x=3$, not their negatives.
    4. $x = 3$ and $x = 9$ — This mismatches one correct root with $9$, an unrelated coefficient from the original expression.
  2. At $x=3$, $f''(3) = 6$. Combined with $f(3) = 15$, what does this tell us?
    1. $x=3$ is an inflection point with value $15$
    2. $x=3$ is a local maximum with value $15$
    3. $x=3$ is a local minimum with value $19$
    4. $x=3$ is a local minimum with value $15$
    Check your answer
    1. $x=3$ is an inflection point with value $15$ — A positive second derivative confirms a minimum here, it rules out an inflection point, not in.
    2. $x=3$ is a local maximum with value $15$ — A positive second derivative means a cup, which the test reads as a MINIMUM, not a maximum.
    3. $x=3$ is a local minimum with value $19$ — $19$ is the local maximum value at the OTHER critical point, $x=1$, not the value here at $x=3$.
    4. ✓ $x=3$ is a local minimum with value $15$ — (D) $f''(3) = 6 > 0$ confirms a cup at $x=3$, a local minimum, matching the given value $f(3) = 15$.
  3. Both tests are applied to the same critical points $x=1,3$ in this worked example. What does their agreement demonstrate?
    1. the first derivative test becomes unnecessary once the second is available
    2. the two tests can classify the same point independently and agree
    3. the second derivative test is only valid once the first derivative test has run
    4. agreement between the tests proves the function has no other critical points
    Check your answer
    1. the first derivative test becomes unnecessary once the second is available — The second derivative test is inconclusive exactly when $f''(a)=0$, and then the first test is still needed.
    2. ✓ the two tests can classify the same point independently and agree — (B) This worked example shows the two tests agreeing is not a coincidence: each independently reads the same critical point correctly.
    3. the second derivative test is only valid once the first derivative test has run — Each test can classify a critical point on its own; neither is a required first step for the other.
    4. agreement between the tests proves the function has no other critical points — The two tests only classify the critical points already found, they say nothing about points not yet located.
✕MISCONCEPTION
Step 1
f′(x) = 3x², so f′(0)
0
Step 2
so x = 0 is
a critical point
Step 3
so it is a max or a min
?
The working on x³ at 0, step by step. Which step is wrong? Keep your answer: the signs test it.

Here is a working on $f(x) = x^3$. Which step is wrong?

y = x³
Close up at 0: f(−0.1) = −0.001, f(0) = 0, f(0.1) = 0.001. The values keep rising through 0.

Keep your answer, then check the signs.

Worked example

Check the sign of $f'(x) = 3x^2$ on each side of 0

  1. $f'(x) = 3x^2 > 0$ for $x < 0$
    A square is positive for every $x \neq 0$, so the slope is positive left of 0.
  2. $f'(x) = 3x^2 > 0$ for $x > 0$
    The same square on the right: the slope is positive there too.
  3. The sign does not change at 0
    Positive on both sides, so $f$ rises straight through 0.
  4. $f(-0.1) = -0.001$, $f(0) = 0$, $f(0.1) = 0.001$
    Check with numbers: the values keep rising through 0.

The sign check settles it. There is no peak and no lowest point.

local maximum
4x − x² at 23 − |x − 2| at 2
local minimum
x² at 0x⁴ at 0
neither
x³ at 0
Every critical point is a point to test. Some are peaks, some are lowest points, and some, like x³ at 0, are neither.
The curve y equals x cubed has slope 0 at the origin. Its slope is never negative, so it climbs on both sides.

Finishing the test first

A zero slope makes a point one to test, not yet an answer. Every local maximum or minimum where $f$ has a slope sits at $f'(c) = 0$. The reverse is false. Here $f''(0) = 0$ as well, so the second derivative test is silent. Only the sign check can decide, and it says neither.

So when you find a critical point, always finish the test before you name it.

Check yourself
  1. A student claims: “if $f'(a) = 0$, then $x=a$ must be a local max or min.” Is this claim correct?
    1. no: a zero of $f'$ is only a point to test, not a guarantee
    2. no: a zero of $f'$ never produces a local max or min
    3. yes: a zero of $f'$ is always enough to guarantee a local max or min
    4. yes, but only when $f$ is a polynomial of degree three or higher
    Check your answer
    1. ✓ no: a zero of $f'$ is only a point to test, not a guarantee — (A) The claim overreaches: a zero of $f'$ is only a candidate, needing the sign-change or second-derivative test to confirm it.
    2. no: a zero of $f'$ never produces a local max or min — A zero of $f'$ CAN still produce a local extremum, the claim only fails as a blanket guarantee.
    3. yes: a zero of $f'$ is always enough to guarantee a local max or min — The counterexample $f(x)=x^3$ at $x=0$ already shows a zero of $f'$ that is neither a max nor a min.
    4. yes, but only when $f$ is a polynomial of degree three or higher — The counterexample used against the claim, $f(x)=x^3$, is itself a degree-three polynomial.
  2. For $f(x) = x^3$ at $x=0$, $f'(0) = 0$ but $f'(x) = 3x^2 \geq 0$ everywhere. What does this tell us about $x=0$?
    1. $f$ decreases just before $x=0$, so $x=0$ is a local minimum
    2. $f$ never decreases anywhere, so $x=0$ is neither a local max nor a local min
    3. $f$ decreases just after $x=0$, so $x=0$ is a local maximum
    4. $f$ never increases anywhere, so $x=0$ is neither a local max nor a local min
    Check your answer
    1. $f$ decreases just before $x=0$, so $x=0$ is a local minimum — $f'(x)=3x^2$ stays non-negative right up to $x=0$, so $f$ does not actually decrease just before it.
    2. ✓ $f$ never decreases anywhere, so $x=0$ is neither a local max nor a local min — (B) $f'(x) = 3x^2$ never goes negative, so $f(x)=x^3$ never decreases anywhere, $x=0$ is neither a max nor a min.
    3. $f$ decreases just after $x=0$, so $x=0$ is a local maximum — $f'(x)=3x^2$ stays non-negative right after $x=0$ too, so $f$ does not decrease just after it either.
    4. $f$ never increases anywhere, so $x=0$ is neither a local max nor a local min — $3x^2 \geq 0$ means $f'$ is never NEGATIVE, so $f$ never decreases, this option has the direction reversed.
  3. Since $f'(x) = 3x^2$ never changes sign around $x=0$ for $f(x)=x^3$, what kind of point is $x=0$?
    1. an inflection point, not an extremum
    2. a local maximum, since $f'(0) = 0$ there
    3. a local minimum, since $f'(0) = 0$ there
    4. a point outside the domain of $f$
    Check your answer
    1. ✓ an inflection point, not an extremum — (A) With no sign change in $f'$ around $x=0$, the point is classified as an inflection point rather than an extremum.
    2. a local maximum, since $f'(0) = 0$ there — $f'(0)=0$ alone does not make $x=0$ a maximum, with no sign change, no extremum type applies here.
    3. a local minimum, since $f'(0) = 0$ there — $f'(0)=0$ alone does not make $x=0$ a minimum either, for the same reason: no sign change occurred.
    4. a point outside the domain of $f$ — $f(x)=x^3$ is perfectly well-defined at $x=0$, the point sits inside the domain, not outside it.
  4. What extra step does the $f(x)=x^3$ example show is always needed after finding $f'(a) = 0$?
    1. no extra step is needed once $f'(a) = 0$ is confirmed
    2. checking that $a$ lies inside a closed interval before calling it an extremum
    3. checking that $f(a) = 0$ as well before calling $a$ an extremum
    4. checking the sign change of $f'$, or the sign of $f''(a)$
    Check your answer
    1. no extra step is needed once $f'(a) = 0$ is confirmed — This restates exactly the misconception the example refutes, $f'(a)=0$ alone is never enough.
    2. checking that $a$ lies inside a closed interval before calling it an extremum — The closed-interval procedure belongs to a separate topic, absolute extrema, not this local classification.
    3. checking that $f(a) = 0$ as well before calling $a$ an extremum — The function’s own value at $a$ plays no role in classifying it, the DERIVATIVE’s behaviour does.
    4. ✓ checking the sign change of $f'$, or the sign of $f''(a)$ — (D) The example shows a sign-change check (or a look at $f''(a)$) is always the required next step after $f'(a)=0$.

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Absolute max/min on a closed interval

▲CONCEPT
hour°Cnamenamename
From 6 am to 6 pm: hottest 33 °C at 4 pm, slope 0. Coldest 21.8 °C at 6 am, already rising 0.785 °C hourly.

Now look only at the readings from 6 am to 6 pm. The hottest reading may still be in the afternoon, where the slope is zero. But the coldest reading may be at 6 am itself. The temperature is already rising then, so the slope there is not zero. It is the lowest reading only because the clock starts there.

So on a stretch with two ends, the largest and smallest values can sit at a flat point inside it. Or they can sit at an end.

[0, 1]
largest 1
(0, 1)
no largest value
On [0, 1], y = x reaches its largest value, 1. On (0, 1) it never gets there: no largest value.

The largest value of $f$ on the whole interval is its ABSOLUTE MAXIMUM. The smallest is its ABSOLUTE MINIMUM. If $f$ is continuous on a closed interval $[a, b]$, it has both: this guarantee is the EXTREME VALUE THEOREM. On an open interval there may be neither: $f(x) = x$ on $(0, 1)$ has no largest value.

From x equals 0.5 to 4.2: flat points at heights 6 and 2, ends at 5.1 and 8.0. The highest is an end.

To find them on $[a, b]$, follow four steps.

y = 4x − x²
4x − x² on [0, 3]: the list is 0, 4, 3. Largest 4 at the flat point, smallest 0 at the end.
values on the list
0
3
4
The points to compare, on one line: 0, 3 and 4. Read off the smallest and the largest; no test is needed.

Let us try $f(x) = 4x - x^2$ on $[0, 3]$.

Worked example

Find the absolute maximum and minimum of $f(x) = 4x - x^2$ on $[0, 3]$

  1. $f'(x) = 4 - 2x = 0$ at $x = 2$
    Find the critical points: the slope is 0 only at $x = 2$, inside $(0, 3)$.
  2. $f(2) = 8 - 4 = 4$
    Work out $f$ at the critical point.
  3. $f(0) = 0$ and $f(3) = 12 - 9 = 3$
    Add the two ends to the list, and work out $f$ there too.
  4. *Absolute maximum 4 at $x = 2$, absolute minimum 0 at $x = 0$*
    Compare the three values: 4 is the largest, and 0, at an end, is the smallest.
  5. $f'(0) = 4$, not 0
    At the end $x = 0$ the slope is not 0, yet the minimum sits there.
  6. $f(1) = 3$, between 0 and 4
    Check one more input: its value lies between the minimum and the maximum.

So you never need to test which of these points is a peak. You compare the numbers on the list and read off the answer.

Your turn: find the absolute maximum and minimum of $f(x) = x^3 - 3x$ on $[0, 2]$.

Check yourself
  1. The Extreme Value Theorem guarantees an absolute maximum and minimum for $f$ on $[a,b]$ when $f$ is
    1. continuous on the closed interval $[a,b]$
    2. equal to zero at both endpoints
    3. differentiable on the open interval $(a,b)$ only
    4. increasing throughout $[a,b]$
    Check your answer
    1. ✓ continuous on the closed interval $[a,b]$ — (A) Continuity on the closed interval $[a,b]$ is exactly the condition the Extreme Value Theorem needs.
    2. equal to zero at both endpoints — The theorem places no requirement at all on what value $f$ takes at the endpoints.
    3. differentiable on the open interval $(a,b)$ only — Differentiability alone, without continuity on the closed interval, is not the theorem’s actual condition.
    4. increasing throughout $[a,b]$ — A function need not be increasing anywhere for the theorem to still guarantee both extrema.
  2. To find the absolute max and min of $f$ on $[a,b]$, which values must be compared?
    1. $f'$ at every interior critical point and at both endpoints
    2. $f$ at both endpoints only, ignoring interior critical points
    3. $f$ at every interior critical point only, ignoring the endpoints
    4. $f$ at every interior critical point and $f$ at both endpoints
    Check your answer
    1. $f'$ at every interior critical point and at both endpoints — It is the FUNCTION’s values, $f$, that get compared here, not the derivative’s values.
    2. $f$ at both endpoints only, ignoring interior critical points — Ignoring the interior critical points misses candidate values the theorem requires checking.
    3. $f$ at every interior critical point only, ignoring the endpoints — Ignoring the endpoints misses candidate values the Extreme Value Theorem requires checking.
    4. ✓ $f$ at every interior critical point and $f$ at both endpoints — (D) The absolute max/min procedure compares $f$ at every interior critical point AND at both endpoints, leaving nothing out.
  3. After listing $f$’s values at the interior critical points and at both endpoints, how is the absolute maximum identified?
    1. it is the average of all the listed values
    2. it is the value at whichever endpoint is larger
    3. it is the largest value in that whole list
    4. it is the value at the interior critical point closest to the midpoint
    Check your answer
    1. it is the average of all the listed values — An average blends every listed value together, it is not the same thing as the largest one.
    2. it is the value at whichever endpoint is larger — Restricting the comparison to the two endpoints ignores the interior critical values entirely.
    3. ✓ it is the largest value in that whole list — (C) The absolute maximum is simply the largest value found once every candidate on the list has been compared.
    4. it is the value at the interior critical point closest to the midpoint — How close a point sits to the interval’s midpoint has no bearing on the SIZE of $f$ there.
  4. If $f$ is continuous on $[a, b]$ and has no critical points inside $(a, b)$, where must its absolute max and min occur?
    1. nowhere: the Extreme Value Theorem fails without an interior critical point
    2. at the two endpoints, $a$ and $b$
    3. only at whichever endpoint gives a positive value
    4. at the interval’s midpoint by default
    Check your answer
    1. nowhere: the Extreme Value Theorem fails without an interior critical point — The theorem’s guarantee of an absolute max and min does not depend on having an interior critical point.
    2. ✓ at the two endpoints, $a$ and $b$ — (B) With no interior critical points, the comparison list is just the two endpoints, so the extrema sit there.
    3. only at whichever endpoint gives a positive value — The sign of the value plays no role in the procedure, both endpoints are compared regardless.
    4. at the interval’s midpoint by default — The interval’s midpoint plays no role in the comparison unless it happens to be an endpoint itself.
✕MISCONCEPTION
Step 1
f′(x) = 2x, 0 only at
x = 0
Step 2
test x = 0 only: f(0)
0
Step 3
so max = min =
0 ?
The working on x² from −1 to 2. Which step is wrong? Keep your answer, then test the ends.

Here is a working for $f(x) = x^2$ on $[-1, 2]$. Which step is wrong?

The curve y equals x squared from minus 1 to 2. Minimum 0 at the flat point. Maximum 4 at the end, x equals 2.

Keep your answer, then test the ends.

Worked example

Test the ends of $[-1, 2]$ for $f(x) = x^2$

  1. $f(-1) = 1$ and $f(2) = 4$
    Work out $f$ at the two ends of the interval.
  2. The list: $f(0) = 0$, $f(-1) = 1$, $f(2) = 4$
    Put the two ends next to the critical point $x = 0$.
  3. *Absolute maximum 4 at $x = 2$, an end*
    Compare the three values: the largest sits at an end, not at the critical point.
  4. $f'(2) = 2 \cdot 2 = 4$, not 0
    The slope at that end is 4, so it is not a flat point.

So the maximum sits where the slope is not 0. A search for zero slopes would never find it.

values on [−1, 2]
0
1
4
The full list: 0 at x = 0, 1 at x = −1, 4 at x = 2. The maximum, 4, sits at an end.

So the second step is wrong. It left the two ends off the list. On a closed interval the ends always go on the list, next to the critical points.

Check yourself
  1. A student finds the absolute max/min on $[a,b]$ by testing only the critical points, the same procedure used for local max/min. What is missing?
    1. nothing is missing: the procedure is identical to the local case
    2. the interval must first be split into smaller sub-intervals
    3. the second derivative must also be checked at each critical point
    4. the two endpoints must also be evaluated and compared
    Check your answer
    1. nothing is missing: the procedure is identical to the local case — This restates exactly the misconception being described; the procedure is not identical to the local case.
    2. the interval must first be split into smaller sub-intervals — Splitting the interval into smaller pieces is not part of the absolute-extrema procedure at all.
    3. the second derivative must also be checked at each critical point — The second derivative test classifies a local extremum, it does not supply the missing endpoint values.
    4. ✓ the two endpoints must also be evaluated and compared — (D) Testing critical points alone finds LOCAL extrema; the absolute case still needs both endpoints compared in too.
  2. For $f(x) = x^2$ on $[-1, 2]$, the only critical point is $x=0$, giving $f(0) = 0$. What is the absolute maximum on this interval?
    1. $4$, at the endpoint $x=2$
    2. $1$, at the endpoint $x=-1$
    3. $0$, at the critical point $x=0$
    4. $2$, at the endpoint $x=2$
    Check your answer
    1. ✓ $4$, at the endpoint $x=2$ — (A) Comparing $f(-1)=1$, $f(0)=0$, and $f(2)=4$, the absolute maximum is $4$, at the endpoint $x=2$.
    2. $1$, at the endpoint $x=-1$ — $f(-1)=1$ is correctly computed, but it is not the larger of the two endpoint values compared.
    3. $0$, at the critical point $x=0$ — $f(0)=0$ is the absolute MINIMUM on this interval, not the maximum, the two are being swapped.
    4. $2$, at the endpoint $x=2$ — $x=2$ is the right endpoint, but $f(2) = 2^2 = 4$, not $2$, the squaring here is done wrong.
  3. For $f(x)=x^2$ on $[-1,2]$, why does the absolute maximum occur at an endpoint that is never a critical point at all?
    1. $f(x)=x^2$ has no true critical point on this interval
    2. the endpoint $x=2$ is secretly also a critical point of $f$
    3. the critical point test only finds where $f'=0$, not the boundary
    4. every absolute maximum must occur at a point where $f'$ fails to exist
    Check your answer
    1. $f(x)=x^2$ has no true critical point on this interval — $x=0$ IS a genuine critical point here, since $f'(0)=0$, the interval does have one.
    2. the endpoint $x=2$ is secretly also a critical point of $f$ — $f'(2) = 4 \neq 0$, so $x=2$ is a genuine boundary point, never a critical point of $f$ at all.
    3. ✓ the critical point test only finds where $f'=0$, not the boundary — (C) The critical-point test only finds where $f'=0$ inside the interval; the boundary is a separate category the procedure adds back in.
    4. every absolute maximum must occur at a point where $f'$ fails to exist — This is not a requirement of the theorem, plenty of absolute maxima occur where $f'$ exists just fine.
  4. $x=0$ is both a local minimum and the absolute minimum for $f(x)=x^2$ on $[-1,2]$. Does every local extremum inside an interval have to be the absolute extremum too?
    1. yes: every local extremum inside a closed interval is automatically the absolute extremum
    2. no: a local extremum can never also be an absolute extremum
    3. no: a local extremum is only a candidate; it must still be compared against the endpoints
    4. yes, but only when the interval is symmetric about zero
    Check your answer
    1. yes: every local extremum inside a closed interval is automatically the absolute extremum — This example happens to agree, but nothing guarantees it, the endpoints could always outrank the local extremum.
    2. no: a local extremum can never also be an absolute extremum — This very example is a counterexample to that claim: $x=0$ is a local AND an absolute minimum at once.
    3. ✓ no: a local extremum is only a candidate; it must still be compared against the endpoints — (C) A local extremum is only a candidate for the absolute one; it still has to survive comparison against the endpoints.
    4. yes, but only when the interval is symmetric about zero — Whether the interval is symmetric about zero plays no role in this comparison at all.
Points to compare on 1 to 5: ends at 24 and 56, flat points at 29 and 28. Both the maximum and the minimum are ends.
xy′y′ = 6x² − 30x + 36
The slope 6(x − 2)(x − 3) is 0 at 2 and 3 and negative only between them. So 29 and 28 are only local.
least to greatest
24
28
29
56
33
The list 24, 28, 29, 56. The check, f(4) = 33, lies between the least and the greatest.
Worked example

Find the absolute max and min of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$

  1. $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$; find the absolute max and min
    A cubic on a closed interval. The list holds the critical points inside it and both ends.
  2. $f'(x) = 6x^2 - 30x + 36 = 6(x-2)(x-3)$
    Differentiate and factor to find where the slope is zero.
  3. Critical points at $x = 2$ and $x = 3$, both inside $(1, 5)$
    Both roots lie inside $(1, 5)$, so both go on the list.
  4. $f(1) = 24$, $f(2) = 29$, $f(3) = 28$, $f(5) = 56$
    Work out $f$ at all four points, the two ends included.
  5. *Absolute maximum $56$ at $x=5$; absolute minimum $24$ at $x=1$, both at the endpoints*
    The largest and smallest values on the list are the answers. Both are at the ends. The values 29 and 28 are only a local maximum and a local minimum.
  6. $f(4) = 128 - 240 + 144 + 1 = 33$, between $24$ and $56$
    Check one more input inside the interval. Its value lies between the minimum and the maximum, as it must.
Check yourself
  1. For $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1,5]$, factoring $f'(x) = 6x^2 - 30x + 36$ gives $6(x-2)(x-3)$. What are the critical points inside $(1,5)$?
    1. $x=6$ and $x=30$
    2. $x=2$ and $x=3$
    3. $x=1$ and $x=5$
    4. $x=0$ and $x=6$
    Check your answer
    1. $x=6$ and $x=30$ — $6$ and $30$ are coefficients taken from $f'(x)$ itself, not the roots of its factored form.
    2. ✓ $x=2$ and $x=3$ — (B) $6(x-2)(x-3) = 0$ gives the critical points $x=2$ and $x=3$, both inside $(1,5)$.
    3. $x=1$ and $x=5$ — $1$ and $5$ are the interval’s endpoints, not the roots of the factored derivative $(x-2)(x-3)$.
    4. $x=0$ and $x=6$ — Neither $0$ nor $6$ solves $(x-2)(x-3)=0$, only $x=2$ and $x=3$ actually do.
  2. Comparing $f(1)=24$, $f(2)=29$, $f(3)=28$, $f(5)=56$, what is the absolute maximum on $[1,5]$?
    1. $29$, at the critical point $x=2$
    2. $28$, at the critical point $x=3$
    3. $24$, at the endpoint $x=1$
    4. $56$, at the endpoint $x=5$
    Check your answer
    1. $29$, at the critical point $x=2$ — $29$ is only the larger of the two interior critical values, it is not the largest of all four.
    2. $28$, at the critical point $x=3$ — $28$ is smaller than both $29$ and $56$ in the same list, so it cannot be the maximum.
    3. $24$, at the endpoint $x=1$ — $24$ is actually the smallest value in the list, the absolute minimum, not the maximum.
    4. ✓ $56$, at the endpoint $x=5$ — (D) Comparing all four values, $56$ at the endpoint $x=5$ is the largest, so it is the absolute maximum.
  3. The interior values $f(2)=29$ and $f(3)=28$ are the largest and smallest among the CRITICAL points, yet neither is the absolute max or min on $[1,5]$. Why?
    1. the function is not continuous at $x=2$ and $x=3$
    2. both endpoint values, $24$ and $56$, beat them once compared
    3. the endpoints must always outrank the interior critical points by definition
    4. critical-point values are never eligible to be absolute extrema on a closed interval
    Check your answer
    1. the function is not continuous at $x=2$ and $x=3$ — $f(x)=2x^3-15x^2+36x+1$ is a polynomial, so it is continuous everywhere, including at $x=2$ and $x=3$.
    2. ✓ both endpoint values, $24$ and $56$, beat them once compared — (B) Once $24$ and $56$ join the comparison, they fall outside $[28, 29]$ on either side and take both extremes.
    3. the endpoints must always outrank the interior critical points by definition — No such rule exists, whichever value is largest or smallest wins the comparison, endpoint or interior.
    4. critical-point values are never eligible to be absolute extrema on a closed interval — Critical-point values CAN be the absolute extrema in other problems, they simply lost the comparison here.

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Optimisation

▲CONCEPT
Two tins that hold the same oil, 16π cm³. Which one needs less metal?
r=1h=16
r=4h=1
volume = 16π
Two cans that hold the same oil, 16π cm³: one 2 cm across and 16 cm tall, one 8 cm across and 1 cm tall.

Think of a tin can, like the ones cooking oil is sold in. A company must fit a fixed volume of oil into each can. It wants to use as little metal as it can. A tall thin can and a short wide can may hold the same oil. Yet they need different amounts of metal. Somewhere between them is the can that needs the least.

r = 1
→
16/r²
fixed volume
→
h = 16
r = 2
→
16/r²
fixed volume
→
h = 4
r = 4
→
16/r²
fixed volume
→
h = 1
Choose the radius; the fixed volume decides the height. r = 1 gives h = 16, r = 2 gives 4, 4 gives 1.

The metal depends on two lengths, the radius and the height. But the two are tied together. Once you choose the radius, the fixed volume decides the height. So the metal really depends on one length only.

rhh = 16/r²
The constraint as a curve, h = 16/r². Every can on it holds 16π cm³: (1, 16), (2, 4), (4, 1).

Making a quantity as large or as small as it can be is OPTIMISATION. The rule that ties the lengths together is the CONSTRAINT. Solve the problem in four steps.

r=1h=16
r=2h=4
r=4h=1
volume = 16π &middot; best: least surface
Three cans, each holding 16π cm³. Metal: 34π cm² for the tall one, 40π for the wide one, 24π for the middle one.

Let us check the steps on the can. You know the formulas for a cylinder from earlier classes. The volume is $\pi r^2 h$, and the total surface area is $S = 2 \pi r^2 + 2 \pi r h$. Take the volume as $16 \pi$ cubic centimetres.

Worked example

Work out the metal for two cans that each hold $16 \pi$

  1. $\pi r^2 h = 16 \pi$, so $h = 16/r^2$
    Divide both sides by $\pi r^2$: the radius decides the height.
  2. $r = 1$: $h = 16$, and $S = 2 \pi (1)^2 + 2 \pi (1)(16) = 34 \pi$
    Put $r = 1$ into $h$, then both lengths into $S$.
  3. $r = 4$: $h = 1$, and $S = 2 \pi (4)^2 + 2 \pi (4)(1) = 40 \pi$
    The same for $r = 4$: the short wide can needs more metal.

The worked example below finds the best radius, 2 cm, which needs only $24 \pi$.

So a question about the best choice becomes a question about one function and its slope.

Your turn: a rectangle has a perimeter of 20 cm. Write its area in terms of one side, $x$. Which $x$ gives the largest area?

Brahmagupta
A SCHOLAR INDIA REMEMBERS

Back to the can: the constraint gives $h = 16/r^2$. Put in $r = -2$ and you get $h = 4$, the same as for $r = 2$. The metal comes out as $2 \pi (-2)^2 + 32 \pi/(-2) = -8 \pi$. Is a negative number allowed here? Not as a radius, and not as an amount of metal. A length is never below $0$. So write $r > 0$ before you solve. Then every value you test is a can you could make.

Brahmagupta · the brahminy kite
Check yourself
  1. Once the quantity to optimise is written down, the next step is to
    1. differentiate the constraint equation directly
    2. set the quantity to optimise equal to zero
    3. express the quantity as a function of one variable
    4. guess a value and check whether it satisfies the constraint
    Check your answer
    1. differentiate the constraint equation directly — The constraint is used to ELIMINATE a variable, it is not differentiated on its own as a first move.
    2. set the quantity to optimise equal to zero — It is the DERIVATIVE of the quantity that gets set to zero later, not the quantity itself now.
    3. ✓ express the quantity as a function of one variable — (C) Using the constraint to write the quantity as a function of one variable comes straight after writing the quantity down.
    4. guess a value and check whether it satisfies the constraint — Guessing and checking is not the systematic method this chapter’s optimisation procedure follows.
  2. After expressing the quantity as a function of one variable, the next step is to
    1. integrate the function over its whole domain
    2. evaluate the function at $x=0$ and stop there
    3. find its critical points by setting the derivative equal to zero
    4. find its critical points by setting the function itself equal to zero
    Check your answer
    1. integrate the function over its whole domain — Integration answers a different kind of question, it plays no role in this optimisation procedure.
    2. evaluate the function at $x=0$ and stop there — Stopping at the single arbitrary point $x=0$ skips the actual critical-point search entirely.
    3. ✓ find its critical points by setting the derivative equal to zero — (C) Setting the derivative equal to zero is exactly how critical points are found once the single-variable function exists.
    4. find its critical points by setting the function itself equal to zero — Critical points come from setting the DERIVATIVE to zero, not the function’s own value.
  3. Once critical points are found in an optimisation problem, how are they classified as a max or min?
    1. the first or second derivative test, or a boundary comparison
    2. by checking whether the critical point is a whole number
    3. by re-solving the constraint for a second variable
    4. by comparing the critical point only to the constraint equation
    Check your answer
    1. ✓ the first or second derivative test, or a boundary comparison — (A) The same first or second derivative test used earlier classifies an optimisation problem’s critical points, with boundary comparison added on a closed domain.
    2. by checking whether the critical point is a whole number — Whether a critical point happens to be a whole number has nothing to do with classifying it.
    3. by re-solving the constraint for a second variable — Re-solving for a second variable answers a different question, not the classification step at all.
    4. by comparing the critical point only to the constraint equation — The constraint already did its job eliminating a variable, it plays no role in classification.
  4. An optimisation problem’s domain is a closed interval, and a critical point inside it gives a local minimum. Why might this not be the answer to the original problem?
    1. on a closed domain, the boundary must still be compared before the answer is final
    2. a local minimum inside a closed interval is always the answer, so no further check is needed
    3. a closed domain guarantees no boundary comparison is ever required
    4. the constraint must be reapplied after classifying the critical point
    Check your answer
    1. ✓ on a closed domain, the boundary must still be compared before the answer is final — (A) A local minimum from a critical point is still only a candidate; the boundary values must be compared before the true answer is known.
    2. a local minimum inside a closed interval is always the answer, so no further check is needed — This denies exactly the boundary check that a closed domain requires as part of the procedure.
    3. a closed domain guarantees no boundary comparison is ever required — A closed domain is precisely the case where boundary comparison IS required, this reverses the rule.
    4. the constraint must be reapplied after classifying the critical point — The constraint already did its job earlier, reducing the problem to one variable, it is not reapplied here.
Surface area over pi against radius, for a can of fixed volume. The lowest point is at radius 2, where the area is 24 pi.
rS/π2r² + 32/r
rslope4r − 32/r²
Metal over π, 2r² + 32/r, and its slope 4r − 32/r². The slope is negative below r = 2 and positive above.
r=2h=4surface=24πequal: h, w
The best can: radius 2 cm, height 4 cm, as tall as it is wide. It needs 24π ≈ 75.4 cm² of metal.
Worked example

Minimise a closed cylinder’s surface area for a fixed volume of $16 \pi$ $\text{cm}^3$

  1. A closed cylindrical can must hold volume $16 \pi$ $\text{cm}^3$; minimise its total surface area
    The volume is fixed. You want the least surface area, which uses two letters, $r$ and $h$.
  2. $\pi r^2 h = 16 \pi$, so $h = 16/r^2$
    Use the fixed volume to write $h$ in terms of $r$.
  3. $S(r) = 2 \pi r^2 + 2 \pi r h = 2 \pi r^2 + 32 \pi/r$
    Put that $h$ into the surface area. Now $S$ depends on $r$ alone, for $r > 0$.
  4. $S'(r) = 4 \pi r - 32 \pi/r^2$
    Differentiate with respect to $r$.
  5. $4 \pi r - 32 \pi/r^2 = 0$
    Set the slope to 0.
  6. $4 \pi r^3 - 32 \pi = 0$
    Multiply both sides by $r^2$, which is not 0 since $r > 0$.
  7. $r^3 = 8$, so $r = 2$
    Add $32 \pi$ to both sides, divide by $4 \pi$, then take the cube root.
  8. $h = 16/2^2 = 4$
    Find $h$ from the fixed volume, $h = 16/r^2$.
  9. $S''(r) = 4 \pi + 64 \pi/r^3 > 0$ for every $r > 0$
    The second derivative is positive for every $r > 0$, so the curve is a cup all along. Its one flat point, $r = 2$, is the lowest.
  10. *Minimum surface area $S = 24 \pi$ $\text{cm}^2$ at $r=2$, $h=4$*
    Put $r = 2$ back into $S(r)$. The height, 4 cm, is twice the radius, so the can is as tall as it is wide.
  11. $\pi (2)^2 \cdot 4 = 16 \pi$, and $S(1) = 2 \pi + 32 \pi = 34 \pi > 24 \pi$
    Check two things. The can still holds $16 \pi$, and a can of radius 1 needs more metal.
The same 40 m of fence round three plots: 75, 100 and 64 m². The square, 10 m by 10 m, holds the most.
Check yourself
  1. For the closed cylinder with volume $16 \pi$, the constraint $\pi r^2 h = 16 \pi$ gives $h$ in terms of $r$ as
    1. $h = 8/r^2$
    2. $h = 16/r^2$
    3. $h = 16/r$
    4. $h = 16 \pi/r^2$
    Check your answer
    1. $h = 8/r^2$ — Halving the numerator to $8$ does not follow from solving $\pi r^2 h = 16 \pi$ for $h$.
    2. ✓ $h = 16/r^2$ — (B) Dividing both sides of $\pi r^2 h = 16 \pi$ by $\pi r^2$ gives $h = 16/r^2$.
    3. $h = 16/r$ — Dropping the square on $r$ uses $r$ alone in the denominator, instead of the required $r^2$.
    4. $h = 16 \pi/r^2$ — The $\pi$ on both sides of $\pi r^2 h = 16 \pi$ cancels, it should not remain in the final formula for $h$.
  2. Solving $S'(r) = 4 \pi r - 32 \pi/r^2 = 0$ gives $r^3 = 8$. What radius minimises the surface area?
    1. $r = 2$
    2. $r = 2 \pi$
    3. $r = 4$
    4. $r = 8$
    Check your answer
    1. ✓ $r = 2$ — (A) $r^3 = 8$ gives $r = 2$, the cube root of $8$.
    2. $r = 2 \pi$ — Attaching a factor of $\pi$ to the numerical answer does not follow from solving $r^3=8$.
    3. $r = 4$ — Doubling the correct cube root of $8$ gives $4$, not the actual value of $r$.
    4. $r = 8$ — $8$ is the cube itself, $r^3$, not its cube root, which is what $r$ actually equals.
  3. With $r=2$ and $h = 16/r^2 = 4$, what is the minimum total surface area?
    1. $48 \pi$ cm²
    2. $32 \pi$ cm²
    3. $16 \pi$ cm²
    4. $24 \pi$ cm²
    Check your answer
    1. $48 \pi$ cm² — Doubling the correct total is an arithmetic slip after the surface-area formula was set up correctly.
    2. $32 \pi$ cm² — Using only $32 \pi/r$ at $r=2$ drops the two end-cap terms, $2 \pi r^2$, from the total.
    3. $16 \pi$ cm² — $16 \pi$ restates the given volume’s own coefficient, it is not the computed surface area at all.
    4. ✓ $24 \pi$ cm² — (D) $S(2) = 2 \pi (2)^2 + 32 \pi/2 = 8 \pi + 16 \pi = 24 \pi$ cm², the minimum total surface area.

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One derivative, many readings

■RECAP
Applications of Derivatives Rate of change via the chain rule + − Reading the sign of the derivative + − + Local maxima and minima Absolute max/min on a closed interval r h Optimisation
Five idea cards. Arrows run from the slope’s sign to local maxima and minima, then on to absolute values and to optimisation.

The question at the start asked what the slope tells you about the quantity itself.

One cubic, three readings: slope 4.21 at x equals 0.45, arrows for rising and falling, flat points at x equals 1 and 3.
how fast at 10 am
hour°C
rising and falling
hour°C
hottest
hour°C
coldest
hour°C
Answered: 1.57 °C an hour at 10 am; rising from 4 am to 4 pm; hottest 33 °C, coldest 21 °C.

Here is the answer. The slope is a rate. When two quantities are linked, the chain rule links their rates. The sign of the slope tells you where the quantity rises and where it falls. Where the slope is zero, or does not exist, and changes sign, there is a peak or a lowest point. On a closed interval, a continuous $f$ takes its largest and smallest values among those points and the two ends. To make a quantity as large or as small as it can be, first write it in one variable. Then the same tests find the answer.

Check yourself
  1. In this chapter’s recap, the derivative’s VALUE, its SIGN, and its ZERO-CROSSINGS each answer a different question. Which question does its SIGN answer?
    1. what the function’s absolute maximum equals
    2. where a turning point might be
    3. how fast the function is changing right now
    4. whether $f$ increases or decreases
    Check your answer
    1. what the function’s absolute maximum equals — The absolute maximum’s value needs the full closed-interval comparison, not the sign by itself.
    2. where a turning point might be — Flagging a candidate turning point is the job of the derivative’s ZERO-CROSSINGS, not its sign alone.
    3. how fast the function is changing right now — How fast the function is changing right now is read from the derivative’s VALUE, not its sign.
    4. ✓ whether $f$ increases or decreases — (D) The SIGN of the derivative is the one that answers whether $f$ increases or decreases.
  2. A zero slope makes a point one to test, not yet an answer. What must always follow before calling it an extremum?
    1. checking that the function is defined on a closed interval
    2. nothing further: a zero of $f'$ is already proof enough
    3. checking that the zero also satisfies a given constraint
    4. checking the sign change of $f'$, or the sign of $f''$
    Check your answer
    1. checking that the function is defined on a closed interval — The sign-change or second-derivative check applies whether the interval is open or closed, it is not tied to that.
    2. nothing further: a zero of $f'$ is already proof enough — This restates exactly the overreach being warned against, a zero alone is never proof enough.
    3. checking that the zero also satisfies a given constraint — A constraint belongs specifically to optimisation problems, not to classifying every zero of $f'$ in general.
    4. ✓ checking the sign change of $f'$, or the sign of $f''$ — (D) The required follow-up is the sign-change test or the sign of $f''$, exactly the check this chapter builds twice.
  3. On a closed interval, the largest and smallest values are found among the critical points and the two ends. Which earlier idea does this recall?
    1. the chain rule, which multiplies two rates together
    2. the absolute max/min procedure, comparing values against both endpoints
    3. the first derivative test, which only compares values on either side of a critical point
    4. the second derivative test, which only reads the sign of $f''$ at one point
    Check your answer
    1. the chain rule, which multiplies two rates together — The chain rule combines two rates together and involves no endpoints or comparison list at all.
    2. ✓ the absolute max/min procedure, comparing values against both endpoints — (B) This is the same absolute max/min procedure from earlier: interior critical values are compared against both endpoints, kept in the running to the end.
    3. the first derivative test, which only compares values on either side of a critical point — The first derivative test never involves an interval’s endpoints, it compares either side of one point.
    4. the second derivative test, which only reads the sign of $f''$ at one point — The second derivative test reads the sign of $f''$ at one point alone, with no notion of an interval’s endpoints.
Check yourself: the whole chapter
  1. A cube’s volume is $V = x^3$, and its side $x$ grows with time. To find $d V/d t$, what extra fact do you need, beyond the formula for $V$?
    1. $d x/d t$: how fast the side itself is changing.
    2. Nothing else: $d V/d x$ alone already gives $d V/d t$.
    3. The exact value of $t$ at that instant.
    4. $d^2 x/d t^2$: how fast the side’s own rate is changing.
    Check your answer
    1. ✓ $d x/d t$: how fast the side itself is changing. — (A) By the chain rule, $d V/d t = ((d V)/(d x)) \cdot ((d x)/(d t)) = 3x^2 \cdot ((d x)/(d t))$. The missing ingredient is exactly $d x/d t$.
    2. Nothing else: $d V/d x$ alone already gives $d V/d t$. — $d V/d x$ only measures how $V$ changes as $x$ itself changes. It says nothing about time until it is multiplied by $d x/d t$.
    3. The exact value of $t$ at that instant. — The instant matters only to fix where $x$ sits. What turns the formula into a rate is $d x/d t$, not the clock reading.
    4. $d^2 x/d t^2$: how fast the side’s own rate is changing. — Only the first rate, $d x/d t$, is needed here. A second derivative would measure how that rate itself changes, which this question does not ask.
  2. On some interval, the derivative of $f$ is positive everywhere except at exactly one point, where it equals zero. What can you say about $f$ there?
    1. $f$ is increasing up to that point, then decreasing after it.
    2. $f$ is constant at that one point, so it cannot be called strictly increasing overall.
    3. $f$ is still strictly increasing throughout: one isolated zero does not break it.
    4. $f$ is strictly increasing only if that zero point is excluded from the interval.
    Check your answer
    1. $f$ is increasing up to that point, then decreasing after it. — The derivative never turns negative here, so the curve never turns around. It only touches zero for an instant and keeps climbing.
    2. $f$ is constant at that one point, so it cannot be called strictly increasing overall. — Constant needs a whole stretch where the derivative is zero, not a single instant. One isolated zero does not flatten $f$ anywhere nearby.
    3. ✓ $f$ is still strictly increasing throughout: one isolated zero does not break it. — (C) A lone zero of the derivative does not interrupt the climb. Only a whole flat stretch, not a single point, would stop $f$ from being strictly increasing.
    4. $f$ is strictly increasing only if that zero point is excluded from the interval. — Nothing requires removing that point. $f$ climbs straight through it, so the whole interval still counts as strictly increasing.
  3. To find the absolute maximum of a continuous $f$ on $[a, b]$, which values do you compare?
    1. Only the values at the critical points inside $(a, b)$.
    2. Only the two end values, $f(a)$ and $f(b)$.
    3. Only the values where $f''$ is negative, since those are the peaks.
    4. The values at the critical points inside $(a, b)$ and at both ends.
    Check your answer
    1. Only the values at the critical points inside $(a, b)$. — An absolute maximum can sit at an end. For $x^2$ on $[-1, 2]$ it is $4$, at $x = 2$, and the critical point gives only $0$.
    2. Only the two end values, $f(a)$ and $f(b)$. — The ends are not enough. For $4x - x^2$ on $[0, 3]$ the ends give $0$ and $3$, but the critical point gives $f(2) = 4$.
    3. Only the values where $f''$ is negative, since those are the peaks. — A negative $f''$ marks a local peak, not the largest value. An end can be higher, so the ends go on the list too.
    4. ✓ The values at the critical points inside $(a, b)$ and at both ends. — (D) On a closed interval the candidates are the critical points inside and both ends. Compare their values and read off the largest.
  4. A plot fenced with $40$ m has width $w$ and area $A(w) = w (20 - w)$. What is the next step to find the largest area?
    1. Set $A(w) = 0$ and solve for $w$.
    2. Set $A'(w) = 20 - 2w$ equal to $0$ and solve for $w$.
    3. Take the widest plot the fence allows, since a wider plot holds more.
    4. Differentiate the fence length $2w + 2l = 40$ and set the result to $0$.
    Check your answer
    1. Set $A(w) = 0$ and solve for $w$. — $A(w) = 0$ at $w = 0$ and $w = 20$, where the plot has no area. The largest area is where the slope $A'(w)$ is $0$.
    2. ✓ Set $A'(w) = 20 - 2w$ equal to $0$ and solve for $w$. — (B) $A'(w) = 20 - 2w$ is $0$ at $w = 10$, and $A''(w) = -2 < 0$ makes it a maximum: the $10$ m square, $100$ m².
    3. Take the widest plot the fence allows, since a wider plot holds more. — Width and length trade off. At $w = 16$ the plot is $16$ m by $4$ m, only $64$ m², less than the square’s $100$ m².
    4. Differentiate the fence length $2w + 2l = 40$ and set the result to $0$. — The fence length is fixed; it gave the link $l = 20 - w$, already used. The slope to set to $0$ is the area’s.
  5. At $x = a$, the derivative of $f$ changes sign from positive just before $a$ to negative just after $a$. What does the first derivative test conclude?
    1. $x = a$ is a local maximum: the curve peaks there.
    2. $x = a$ is a local minimum: the curve dips there.
    3. $x = a$ is neither, since a zero derivative only marks a critical point.
    4. The test cannot decide without also checking the second derivative there.
    Check your answer
    1. ✓ $x = a$ is a local maximum: the curve peaks there. — (A) Positive before $a$ and negative after means $f$ climbs into $a$ and falls away from it, exactly the shape of a local maximum.
    2. $x = a$ is a local minimum: the curve dips there. — Positive-to-negative means the curve climbs then falls, a peak, which is a maximum, not a minimum.
    3. $x = a$ is neither, since a zero derivative only marks a critical point. — A critical point is only the starting clue. The sign change around it is what the first derivative test uses to classify it as a maximum here.
    4. The test cannot decide without also checking the second derivative there. — The first derivative test already settles it from the sign change alone. The second derivative test is a separate, optional route.

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Where you will meet this

You have already made a choice like this, long before you met the word derivative.

Your turn: A food stall’s profit is $P(x) = 60x - x^2$ for $x$ snacks sold, where $0 \leq x \leq 50$. Is the best profit at a critical point inside the interval, or at an endpoint?

LATER

Where maxima and minima go next

First-year engineering and economics courses extend this chapter’s single-variable optimisation to many variables, choosing a best combination under more than one constraint at once using Lagrange multipliers. This chapter’s one-variable maximum or minimum, found by setting one derivative to zero, is the base case those methods build on.

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Practice set: Exercise 6.1

Exercise 6.1 practises rates of change. You write the formula that links two quantities, then differentiate it with respect to time. Circles, spheres, cubes and ladders all come up.

Exercise 6.1: Rate of change
  1. practice The radius of a circle is increasing at the rate of $4$ cm/s. Find the rate at which the area of the circle is increasing when the radius is $6$ cm.
  2. practice An edge of a variable cube is increasing at the rate of $2$ cm/s. How fast is the volume of the cube increasing when the edge is $9$ cm long?
  3. practice The radius of a circle is increasing at the rate of $0.5$ cm/s. What is the rate of increase of its circumference?
  4. practice The radius of a spherical soap bubble is increasing at the rate of $2$ cm/s. Find the rate at which its surface area is increasing when the radius is $6$ cm.
  5. practice The side of a variable square is increasing at the rate of $6$ cm/s. Find the rate of increase of its area when the side is $10$ cm.
  6. practice A spherical balloon has a variable radius. Find the rate of change of its volume with respect to the radius when the radius is $7$ cm.
  7. practice The length $x$ of a rectangle is decreasing at the rate of $4$ cm/minute and the width $y$ is increasing at the rate of $3$ cm/minute. When $x = 12$ cm and $y = 8$ cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.
  8. practice The radius of an air bubble is increasing at the rate of $1/3$ cm/s. At what rate is the volume of the bubble increasing when the radius is $2$ cm?
  9. practice A spherical balloon is being inflated by pumping in $850$ cubic centimetres of gas per second. Find the rate at which its radius is increasing when the radius is $17$ cm.
  10. practice Sand is pouring from a pipe at the rate of $20$ cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-fourth of the radius of the base. How fast is the height of the sand cone increasing when the height is $5$ cm?
  11. practice A ladder $13$ m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $3$ m/s. How fast is its height on the wall decreasing when the foot of the ladder is $5$ m away from the wall?
  12. practice A spherical balloon has a variable diameter $3 (2 x + 1)$. Find the rate of change of its volume with respect to $x$, when $x = 2$.
  13. practice Which of these is the rate of change of $x^3 - 2 x$ with respect to $x$ at $x = 3$?
    1. $23$
    2. $25$
    3. $27$
    4. $29$
  14. practice The rate of change of the area of a circle with respect to its radius $r$ at $r = 4$ cm is
    1. $6 \pi$
    2. $8 \pi$
    3. $10 \pi$
    4. $4 \pi$
  15. practice A cube’s edge is increasing at the rate of $2$ cm/s. The rate of increase of its surface area, when the edge is $5$ cm, is
    1. $100$
    2. $110$
    3. $120$
    4. $130$
  16. practice Two quantities $x$ and $y$ satisfy $x^2 - y^2 = 16$, and both change with time. If $d x/d t = 3$ units/s, find $d y/d t$ at the instant when $x = 5$ and $y = 3$.
  17. practice A particle moves along the curve $4 y = x^3 + 8$. Find the points on the curve at which the $y$-coordinate changes $12$ times as fast as the $x$-coordinate.
  18. practice A cube’s volume is increasing at the rate of $64$ cm³/s. Find the rate at which its surface area is increasing when the edge is $8$ cm.
  19. practice A kite is flying at a constant height of $90$ m above the ground and moving horizontally away from the person holding the string at the rate of $6$ m/s. Find the rate at which the string is being paid out when $150$ m of string is already out.
Answers
  1. $48 \pi$ cm²/s
  2. $486$ cm³/s
  3. $\pi$ cm/s
  4. $96 \pi$ cm²/s
  5. $120$ cm²/s
  6. $196 \pi$ cm²
  7. (a) perimeter decreasing at $2$ cm/min; (b) area increasing at $4$ cm²/min
  8. $(16 \pi)/3$ cm³/s
  9. $25/(34 \pi)$ cm/s
  10. $1/(20 \pi)$ cm/s
  11. $d y/d t = -5/4$ m/s; the height is decreasing.
  12. $675 \pi$
  13. (B) $25$.
  14. (B) $8 \pi$.
  15. (C) $120$.
  16. $5$ units/s
  17. $(4, 18)$ and $(-4, -14)$
  18. $32$ cm²/s
  19. $4.8$ m/s

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Practice set: Exercise 6.2

In Exercise 6.2 you find where a function is increasing and where it is decreasing. Solve $f'(x) = 0$, then test the sign of $f'$ on each interval.

Exercise 6.2: Increasing and decreasing functions
  1. practice Show that the function given by $f (x) = 4 x - 5$ is increasing on $\mathbb{R}$.
  2. practice Show that the function given by $f (x) = e^{3 x}$ is increasing on $\mathbb{R}$.
  3. practice Show that the function given by $f (x) = \log x$ is increasing on $(0, \infty)$.
  4. practice Find the intervals in which $f (x) = 3 x^2 - 12 x + 7$ is increasing or decreasing.
  5. practice Find the intervals in which $f (x) = 4 x^2 + 5 x - 3$ is increasing or decreasing.
  6. practice Find the intervals in which $f (x) = x^2 - 8 x + 12$ is increasing or decreasing.
  7. practice Show that $f (x) = \sin x$ is (a) increasing on $(0, \pi/2)$ and (b) decreasing on $(\pi/2, \pi)$.
  8. practice Find the intervals in which each of the following functions is strictly increasing or strictly decreasing: (a) $x^2 + 4 x - 5$ (b) $10 - 7 x - 2 x^2$.
  9. practice Prove that the function $f (x) = x^2 + x - 1$ is neither increasing nor decreasing on $(-1, 1)$.
  10. practice Which of the following functions is decreasing on $(0, \pi/2)$?
    1. $\sin x$
    2. $\tan x$
    3. $\cos x$
    4. $\sin 2 x$
  11. practice On which of the following intervals is the function $f (x) = x^{60} + \sin x - 1$ decreasing?
    1. $(0, 1)$
    2. $(\pi/2, \pi)$
    3. $(0, \pi/2)$
    4. None of these
  12. practice Find the intervals in which $f (x) = 2 x^3 - 9 x^2 + 12 x + 5$ is increasing or decreasing.
  13. practice Find the intervals in which $f (x) = 2 x^3 - 3 x^2 - 36 x + 10$ is strictly increasing or strictly decreasing.
  14. practice Find the intervals in which $f (x) = -2 x^3 - 3 x^2 + 12 x + 5$ is strictly increasing or strictly decreasing.
  15. practice For what values of $a$ is the function $f (x) = x^2 + a x + 3$ increasing on the interval $[2, 4]$?
  16. practice Find the values of $x$ for which $f (x) = 2 x^3 - 15 x^2 + 24 x + 7$ is strictly decreasing.
  17. practice Find the values of $x$ for which $y = [x (x - 3)]^2$ is an increasing function.
  18. practice Find the intervals in which $f (x) = 3 x^4 - 4 x^3 - 12 x^2 + 5$ is increasing or decreasing.
  19. practice Find the intervals in which $f (x) = \sin 4 x$, $x \in (0, \pi/2)$, is increasing and decreasing.
  20. practice Find the values of $x$ for which $f (x) = (x + 1)^3 (x - 2)^3$ is an increasing function.
Answers
  1. Proved: $f'(x) = 4 > 0$ for every $x \in \mathbb{R}$, so $f$ is strictly increasing on $\mathbb{R}$.
  2. Proved: $f'(x) = 3 e^{3 x} > 0$ for every $x \in \mathbb{R}$, so $f$ is increasing on $\mathbb{R}$.
  3. Proved: $f'(x) = 1/x > 0$ for every $x > 0$, so $f$ is increasing on $(0, \infty)$.
  4. Decreasing on $(-\infty, 2)$; increasing on $(2, \infty)$.
  5. Decreasing on $(-\infty, -5/8)$; increasing on $(-5/8, \infty)$.
  6. Decreasing on $(-\infty, 4)$; increasing on $(4, \infty)$.
  7. Proved: $f'(x) = \cos x$ is positive on $(0, \pi/2)$ and negative on $(\pi/2, \pi)$.
  8. (a) decreasing on $(-\infty, -2)$, increasing on $(-2, \infty)$; (b) increasing on $(-\infty, -7/4)$, decreasing on $(-7/4, \infty)$.
  9. Proved: $f'(x) = 2 x + 1$ is negative on $(-1, -1/2)$ and positive on $(-1/2, 1)$, so $f$ is neither increasing nor decreasing throughout $(-1, 1)$.
  10. (C) $\cos x$.
  11. (D) None of these.
  12. Increasing on $(-\infty, 1) \cup (2, \infty)$; decreasing on $(1, 2)$.
  13. Increasing on $(-\infty, -2) \cup (3, \infty)$; decreasing on $(-2, 3)$.
  14. Decreasing on $(-\infty, -2) \cup (1, \infty)$; increasing on $(-2, 1)$.
  15. $a \geq -4$.
  16. $1 < x < 4$.
  17. Increasing on $(0, 3/2) \cup (3, \infty)$.
  18. Decreasing on $(-\infty, -1) \cup (0, 2)$; increasing on $(-1, 0) \cup (2, \infty)$.
  19. Increasing on $(0, \pi/8) \cup (3 \pi/8, \pi/2)$; decreasing on $(\pi/8, 3 \pi/8)$.
  20. Increasing on $x > 1/2$; decreasing on $x < 1/2$.

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Practice set: Exercise 6.3

Exercise 6.3 practises maxima and minima. You use the first and second derivative tests and find absolute values on a closed interval. The last questions are optimisation problems.

Exercise 6.3: Maxima and minima
  1. practice Find the maximum or minimum value, if any, of $f (x) = (3 x - 2)^2 + 5$.
  2. practice Find the maximum or minimum value, if any, of $g (x) = -(x + 3)^2 + 8$.
  3. practice Find the maximum or minimum value, if any, of $h (x) = 4 x^2 + 4 x + 7$.
  4. practice Find the maximum or minimum value, if any, of $f (x) = |x - 3| + 2$.
  5. practice Find the local maxima and local minima, if any, of $f (x) = x^3 - 3 x + 5$, and find the corresponding values.
  6. practice Find the local maxima and local minima of $g (x) = x^3 - 6 x^2 + 9 x + 4$, and find the corresponding values.
  7. practice Find the local maxima and local minima of $f (x) = 2 x^3 - 3 x^2 - 12 x + 8$, and find the corresponding values.
  8. practice Find the local maximum and local minimum values, if any, of $f (x) = x^4 - 4 x^3$.
  9. practice Determine whether $x = 1$ is a point of local maximum, local minimum, or neither, for $f (x) = x^3 - 3 x^2 + 3 x + 7$.
  10. practice Prove that $f (x) = e^x$ has neither a local maximum nor a local minimum.
  11. practice Prove that $g (x) = \log x$ has neither a local maximum nor a local minimum.
  12. practice Find the absolute maximum and minimum values of $f (x) = x^3$ on $[-3, 3]$.
  13. practice Find the absolute maximum and minimum values of $f (x) = 6 x - x^2$ on $[-1, 4]$.
  14. practice Find the absolute maximum and minimum values of $f (x) = (x - 2)^2 + 5$ on $[-1, 4]$.
  15. practice Find the maximum profit that a company can make, if the profit function is given by $p (x) = 42 + 60 x - 15 x^2$.
  16. practice Find the absolute maximum and minimum values of $f (x) = x^3 - 6 x^2 + 9 x + 2$ on $[0, 5]$.
  17. practice At what points in $[0, 2 \pi]$ does the function $\sin 2 x$ attain its maximum value?
  18. practice What is the maximum value of $f (x) = \sin x + \cos x$?
  19. practice Find the maximum value of $2 x^3 - 24 x + 50$ on $[1, 3]$. Find the maximum value of the same function on $[-3, -1]$.
  20. practice It is given that at $x = 2$, the function $f (x) = x^4 - 70 x^2 + a x + 9$ attains its maximum value on the interval $[0, 3]$. Find the value of $a$.
  21. practice Find the absolute maximum and minimum values of $f (x) = x + \sin 2 x$ on $[0, \pi]$.
  22. practice Find two positive numbers whose sum is $20$ and whose product is maximum.
  23. practice Find two positive numbers $x$ and $y$ such that $x + y = 16$ and $x y^3$ is maximum.
  24. practice Find two positive numbers whose sum is $18$ and the sum of whose cubes is minimum.
  25. practice A square piece of tin of side $24$ cm is to be made into a box without a top, by cutting a square of side $x$ cm from each corner and folding up the flaps. What value of $x$ makes the volume of the box maximum, and what is that maximum volume?
  26. practice A rectangular sheet of tin $16$ cm by $10$ cm is to be made into a box without a top, by cutting off a square of side $x$ cm from each corner and folding up the flaps. What value of $x$ makes the volume of the box maximum?
  27. practice A closed cylindrical can is to have a total surface area of $96 \pi$ cm². Find the radius and height that maximise its volume.
  28. practice A wire of length $20$ m is cut into two pieces. One piece is bent into a square and the other into a circle. What should be the length of each piece so that the combined area is minimum?
  29. practice Which of these is a point on the curve $x^2 = 4 y$ nearest to the point $(0, 5)$?
    1. $(2 \sqrt{3}, 3)$
    2. $(0, 0)$
    3. $(2, 1)$
    4. $(4, 4)$
Answers
  1. Minimum value $5$ at $x = 2/3$; no maximum value.
  2. Maximum value $8$ at $x = -3$; no minimum value.
  3. Minimum value $6$ at $x = -1/2$; no maximum value.
  4. Minimum value $2$ at $x = 3$; no maximum value.
  5. Local maximum $7$ at $x = -1$; local minimum $3$ at $x = 1$.
  6. Local maximum $8$ at $x = 1$; local minimum $4$ at $x = 3$.
  7. Local maximum $15$ at $x = -1$; local minimum $-12$ at $x = 2$.
  8. Local minimum $-27$ at $x = 3$; $x = 0$ is a critical point but not a local extremum.
  9. Neither: $x = 1$ is a point of inflection, since $f'(x) = 3 (x - 1)^2$ never changes sign.
  10. Proved: $f'(x) = e^x > 0$ for every $x$, so $f$ has no critical point and hence no local extremum.
  11. Proved: $g'(x) = 1/x \neq 0$ anywhere in the domain, so $g$ has no critical point.
  12. Absolute maximum $27$ at $x = 3$; absolute minimum $-27$ at $x = -3$.
  13. Absolute maximum $9$ at $x = 3$; absolute minimum $-7$ at $x = -1$.
  14. Absolute maximum $14$ at $x = -1$; absolute minimum $5$ at $x = 2$.
  15. Maximum profit ₹⁠$102$, at $x = 2$.
  16. Absolute maximum $22$ at $x = 5$; absolute minimum $2$ (attained at both $x = 0$ and $x = 3$).
  17. Maximum value $1$, attained at $x = \pi/4$ and $x = 5 \pi/4$.
  18. $\sqrt{2}$.
  19. Maximum on $[1,3]$ is $32$ (at $x=3$); maximum on $[-3,-1]$ is $82$ (at $x=-2$).
  20. $a = 248$.
  21. Absolute maximum $\pi$ at $x = \pi$; absolute minimum $0$ at $x = 0$, both at the endpoints.
  22. The numbers are $10$ and $10$; maximum product $100$.
  23. $x = 4$, $y = 12$; maximum value of $x y^3$ is $6912$.
  24. $x = 9$, $y = 9$; minimum sum of cubes $1458$.
  25. $x = 4$ cm; maximum volume $1024$ cm³.
  26. $x = 2$ cm; maximum volume $144$ cm³.
  27. $r = 4$ cm, $h = 8$ cm; maximum volume $128 \pi$ cm³.
  28. Square piece $= 80/(\pi + 4)$ m; circle piece $= 20 \pi/(\pi + 4)$ m.
  29. (A) $(2 \sqrt{3}, 3)$.

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Practice set: miscellaneous

The Miscellaneous Exercise mixes all the methods you have met. Each question needs one of them, and part of the work is to spot which.

Miscellaneous Exercise: Chapter 6
  1. practice Find the maximum value of $f (x) = (\log x)/x$.
  2. practice The two equal sides of an isosceles triangle with fixed base $10$ cm are decreasing at the rate of $2$ cm per second. How fast is the area decreasing when the two equal sides are also $10$ cm each?
  3. practice Find the intervals in which $f (x) = x^3 - 3 x^2 - 9 x + 5$ is increasing or decreasing.
  4. practice Find the intervals in which $f (x) = x/(x^2 + 1)$ is increasing or decreasing.
  5. practice Find the absolute maximum and minimum values of $f (x) = \sin^2 x - \cos x$ on $[0, \pi]$.
  6. practice Find, in terms of $R$, the height of the right circular cylinder of greatest volume that can be inscribed in a sphere of radius $R$, and find that maximum volume.
  7. practice An open rectangular tank with a square base is to have a volume of $64$ m³. The base costs ₹⁠$40$ per m² and the sides cost ₹⁠$20$ per m². Find the dimensions for which the cost is minimum, and find that minimum cost.
  8. practice A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $12$ m. Find the width and height of the rectangle so that the window admits the maximum possible light.
  9. practice The sum of the perimeters of a circle and a square is $k$, a constant. Prove that the sum of their areas is least when the side of the square is double the radius of the circle.
  10. practice Prove that $f (x) = x^3 + x^2 + x + 1$ has neither a local maximum nor a local minimum.
  11. practice Find the points on the curve $y = x^2 + 3$ nearest to the point $(0, 6)$.
  12. practice A particle’s displacement, in metres, at time $t$ seconds is given by $s (t) = t^3 - 6 t^2 + 9 t$, $t \geq 0$. Find the interval(s) of time during which the particle moves in the positive direction, and find its velocity at the instant its acceleration is zero.
  13. practice Find the local maximum value, if any, of $f (x) = x \sqrt{1 - x}$, $x < 1$.
  14. practice A cylindrical tank of radius $7$ m is being filled with wheat at the rate of $154$ cubic metres per hour. The depth of the wheat is then increasing at the rate of
    1. $1$ m/h
    2. $0.5$ m/h
    3. $2$ m/h
    4. $1.5$ m/h
  15. practice Find the absolute maximum and minimum values of $f (x) = x^3 - 3 x$ on $[-2, 3]$, and state whether $x = 1$ is a point of local maximum or local minimum.
  16. practice For a cone of fixed slant height $l$, show that the semi-vertical angle at which the volume is maximum satisfies $\tan \theta = \sqrt{2}$.
Answers
  1. Maximum value $1/e$ at $x = e$.
  2. $d A/d t = -(20 \sqrt{3})/3$ cm²/s; the area is decreasing.
  3. Increasing on $(-\infty, -1) \cup (3, \infty)$; decreasing on $(-1, 3)$.
  4. Increasing on $(-1, 1)$; decreasing on $(-\infty, -1) \cup (1, \infty)$.
  5. Absolute maximum $5/4$ at $x = 2 \pi/3$; absolute minimum $-1$ at $x = 0$.
  6. $h = (2 R)/\sqrt{3}$; maximum volume $(4 \sqrt{3} \pi R^3)/9$.
  7. Base side $4$ m, height $4$ m; minimum cost ₹⁠$1920$.
  8. Width $= 24/(4 + \pi)$ m, height $= 12/(4 + \pi)$ m.
  9. Proved: at the critical point $s = 2 r$ (with $r = k/(2 (4 + \pi))$, $s = k/(4 + \pi) = 2 r$), and the second derivative there is positive, confirming a minimum.
  10. Proved: $f'(x) = 3 x^2 + 2 x + 1$ has negative discriminant and positive leading coefficient, so $f'(x) > 0$ for every $x$; $f$ has no critical point at all.
  11. $(\sqrt{10}/2, 11/2)$ and $(-\sqrt{10}/2, 11/2)$.
  12. Positive direction for $0 \leq t < 1$ and $t > 3$; velocity when acceleration is zero ($t = 2$) is $-3$ m/s.
  13. Local maximum $(2 \sqrt{3})/9$ at $x = 2/3$; no local minimum.
  14. (A) $1$ m/h.
  15. Absolute maximum $18$ at $x = 3$; absolute minimum $-2$ (at $x = -2$ and $x = 1$); $x = 1$ is a point of local minimum.
  16. Proved: with $r = l \sin \theta$ and $h = l \cos \theta$, $V (\theta) = (\pi l^3/3) \sin^2 \theta \cos \theta$; setting $V'(\theta) = 0$ gives $2 \cos^2 \theta = \sin^2 \theta$, i.e. $\tan^2 \theta = 2$, so $\tan \theta = \sqrt{2}$ at the maximum.
Board paper practice
  1. practice On which interval is the function $f (x) = x^2 - 6 x + 7$ strictly increasing? (1 mark)
    1. $(-\infty, 3)$
    2. $(3, \infty)$
    3. $(-\infty, \infty)$
    4. nowhere
  2. practice Assertion (A). $x = 0$ is a point of local minimum for $f (x) = x^4$. Reason (R). $f''(0) = 0$, but $f'(x)$ changes sign from negative to positive at $x = 0$. (1 mark)
    1. Both A and R are true and R is the correct explanation of A.
    2. Both A and R are true but R is not the correct explanation of A.
    3. A is true but R is false.
    4. A is false but R is true.
  3. practice Find the interval in which $f (x) = x^3 - 12 x$ is strictly decreasing. (2 marks)
  4. practice Find the values of $k$ for which $f (x) = k x^3 - 9 x^2 + 9 x + 3$ is strictly increasing for every $x \in \mathbb{R}$. (3 marks)
  5. practice Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius $R$ is $4 R/3$, and find the maximum volume as a fraction of the sphere’s own volume. (5 marks)
  6. practice An open cylindrical water tank of radius $r$ metres and height $h$ metres is to hold a fixed volume of $16 \pi$ m³. The base costs ₹⁠$100$ per m² and the curved side costs ₹⁠$50$ per m². (i) Express $h$ in terms of $r$. (1 mark) (ii) Express the total cost $C (r)$ as a function of $r$ alone. (1 mark) (iii) Find the radius that makes the cost minimum, and the minimum cost. (2 marks)
  7. practice Find the absolute maximum and absolute minimum values of $f (x) = x^3 - 3 x^2 + 6$ on $[-1, 2]$. (3 marks)
Answers
  1. (B) $(3, \infty)$.
  2. (A) Both A and R are true and R is the correct explanation of A.
  3. $f$ is strictly decreasing on $(-2, 2)$.
  4. $k \geq 3$.
  5. Height $4 R/3$; maximum volume $(32 \pi R^3)/81$, which is $8/27$ of the sphere’s volume.
  6. (i) $h = 16/r^2$. (ii) $C (r) = 100 \pi r^2 + 1600 \pi/r$. (iii) $r = 2$ m; minimum cost ₹⁠$1200 \pi$.
  7. Absolute maximum $6$ at $x = 0$; absolute minimum $2$, at both $x = -1$ and $x = 2$.

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