Chapter 6 · Application of Derivatives

Exercise 6.3 · Q5: Absolute maximum and minimum of $x^3$ ⁠, $\sin x+\cos x$ ⁠, $4x-\tfrac12x^2$ ⁠, $(x-1)^2+3$ on closed intervals

Exercise 6.3Absolute extrema on a closed interval · Section 6.4.1

The question

5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:

(i) $f(x)=x^3,\ x\in[-2,2]$ (ii) $f(x)=\sin x+\cos x,\ x\in[0,\pi]$

(iii) $f(x)=4x-\dfrac12x^2,\ x\in\left[-2,\dfrac92\right]$ (iv) $f(x)=(x-1)^2+3,\ x\in[-3,1]$

Exam answer

(i) $f'(x) = 3x^2 = 0$ gives $x = 0$ ⁠, which lies in $[-2, 2]$ ⁠.

\begin{aligned} f(-2) &= -8 \\ f(0) &= 0 \\ f(2) &= 8 \end{aligned}

Absolute maximum value $f(2) = 8$ ⁠; absolute minimum value $f(-2) = -8$ ⁠.

(ii) $f'(x) = \cos x - \sin x = 0$ gives $\tan x = 1$ ⁠, so $x = \frac{\pi}{4}$ in $[0, \pi]$ ⁠.

\begin{aligned} f(0) &= 1 \\ f\left(\tfrac{\pi}{4}\right) &= \sqrt2 \\ f(\pi) &= -1 \end{aligned}

Absolute maximum value $f\left(\frac{\pi}{4}\right) = \sqrt2$ ⁠; absolute minimum value $f(\pi) = -1$ ⁠.

(iii) $f'(x) = 4 - x = 0$ gives $x = 4$ ⁠, which lies in $\left[-2, \frac92\right]$ ⁠.

\begin{aligned} f(-2) &= -10 \\ f(4) &= 8 \\ f\left(\tfrac92\right) &= \tfrac{63}{8} \end{aligned}

Absolute maximum value $f(4) = 8$ ⁠; absolute minimum value $f(-2) = -10$ ⁠.

(iv) $f'(x) = 2(x - 1) = 0$ gives $x = 1$ ⁠, the right end point of $[-3, 1]$ ⁠.

\begin{aligned} f(-3) &= 19 \\ f(1) &= 3 \end{aligned}

Absolute maximum value $f(-3) = 19$ ⁠; absolute minimum value $f(1) = 3$ ⁠.

For (iv) the answer key prints "Absolute minimum value = 19, absolute maximum value = 3", with maximum and minimum interchanged.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

On a closed interval a continuous function always has a greatest and a least value (Theorem 5). Each one occurs either at an end point or at an interior point; at an interior one the derivative must be zero (Theorem 6). So the only candidates are the end points and the critical points inside. Evaluate $f$ at all of them: the largest value is the absolute maximum, the smallest is the absolute minimum. No second derivative test is needed.

Given, to find, and the plan

Given: Four functions, each differentiable (hence continuous) on the stated closed interval.
To find: The absolute maximum value and the absolute minimum value of each on its interval.
Plan: The Working Rule of Section 6.4.1: step 1, critical points in the interval; step 2, the end points; step 3, the value of $f$ at every one of these; step 4, pick the largest and the smallest.

Basics you need

1Theorems 5 and 6 (Section 6.4.1)

Theorem 5: a continuous function on $[a,b]$ has an absolute maximum value and an absolute minimum value, each attained at least once. Theorem 6: if $f$ is differentiable on the closed interval and attains its absolute maximum or minimum at an interior point $c$ ⁠, then $f'(c)=0$ ⁠. Lesson: absolute extrema on a closed interval

2The Working Rule

Rule step 1: critical points of $f$ in the interval. Rule step 2: the end points. Rule step 3: values of $f$ at all these points. Rule step 4: the largest is the absolute maximum value, the smallest the absolute minimum value.

3Closed intervals

$[a,b]$ includes both end points, so they are genuine candidates. A critical point outside the interval is thrown away. Basics: interval notation

4Exact values

$\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}$ ⁠, $\sin\pi=0$ ⁠, $\cos\pi=-1$ ⁠, $\sqrt2\approx1.414$ ⁠. Basics: sin and cos on the unit circle

Part (i): $f(x)=x^3$ on $[-2,2]$

Critical points in the interval

$f'(x)=3x^2=0$ only at $x=0$ ⁠, which lies in $[-2,2]$ ⁠.

$x^3$ is a polynomial, so it is differentiable (and continuous) everywhere: Theorems 5 and 6 apply, and there are no points of non-differentiability to add.

Working Rule, step 1. Lesson: absolute extrema
Tabulate and compare

$x$ $-2$ (end) $0$ (critical) $2$ (end)

$f(x)=x^3$ $-8$ $0$ $8$

Absolute maximum value $8$ (at $x=2$ ⁠); absolute minimum value $-8$ (at $x=-2$ ⁠).

The critical point $x=0$ gives neither: $x^3$ only pauses there (Q1(iv) and the Remark after Theorem 2). That is fine; the rule just needs it in the list.

Working Rule, steps 2–4.

$x$	$-2$ (end)	$0$ (critical)	$2$ (end)
$f(x)=x^3$	$-8$	$0$	$8$

y=x^3

[-2,2]

⁠: the extremes are at the two end points; the flat point

(0,0)

is neither.

Part (ii): $f(x)=\sin x+\cos x$ on $[0,\pi]$

Critical points in $[0,\pi]$

$f'(x)=\cos x-\sin x$ ⁠. If $\cos x=0$ (at $x=\frac{\pi}{2}$ ⁠) then $f'=-1\ne0$ ⁠, so at any zero of $f'$ we may divide by $\cos x$ ⁠: $\tan x=1$ ⁠. In $[0,\pi]$ this gives only $x=\frac{\pi}{4}$ (the next solution, $\frac{5\pi}{4}$ ⁠, is beyond $\pi$ ⁠).

$\tan x=1$ needs $\sin x$ and $\cos x$ equal and non-zero: first or third quadrant. Only the first-quadrant angle lies in $[0,\pi]$ ⁠.

$\tan\frac{\pi}{4}=1$ ⁠. Basics: signs of sin and cos

Tabulate and compare

$x$	$0$ (end)	$\frac{\pi}{4}$ (critical)	$\pi$ (end)
$\sin x+\cos x$	$0+1=1$	$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2$	$0+(-1)=-1$

Since $\sqrt2\approx1.414\gt1\gt-1$ ⁠: absolute maximum value $\sqrt2$ (at $x=\frac{\pi}{4}$ ⁠); absolute minimum value $-1$ (at $x=\pi$ ⁠).

The minimum is at an end point. Without the end points in the list we would miss it, which is why step 2 of the rule matters.

Working Rule, steps 2–4.

\sin x+\cos x

[0,\pi]

⁠: starts at

1

⁠, peaks at

\sqrt2

when

x=\frac{\pi}{4}

⁠, ends lowest at

-1

when

x=\pi

⁠.

Part (iii): $f(x)=4x-\frac12x^2$ on $\left[-2,\frac92\right]$

Critical points in the interval

$f'(x)=4-x$ ⁠, zero at $x=4$ ⁠. Since $-2\le4\le\frac92=4.5$ ⁠, it is in the interval.

Always check that the critical point lies in the given interval before using it.

Working Rule, step 1.

Tabulate and compare

$x$	$-2$ (end)	$4$ (critical)	$\frac92$ (end)
$4x-\frac12x^2$	$-8-2=-10$	$16-8=8$	$18-\frac{81}{8}=\frac{63}{8}$

$\frac{63}{8}=7.875$ ⁠, so the values are $-10$ ⁠, $8$ ⁠, $7.875$ ⁠: absolute maximum value $8$ (at $x=4$ ⁠); absolute minimum value $-10$ (at $x=-2$ ⁠).

x=\frac92

⁠:

4\cdot\frac92=18

and

\frac12\cdot\frac{81}{4}=\frac{81}{8}

⁠, so

18-\frac{81}{8}=\frac{144-81}{8}=\frac{63}{8}

⁠. It is close to

8

but smaller, so the peak is at the critical point.

Working Rule, steps 2–4.

4x-\frac12x^2

\left[-2,\frac92\right]

⁠: from

(-2,-10)

up to the peak

(4,8)

⁠, then down a little to

\left(\frac92,\frac{63}{8}\right)

⁠.

Part (iv): $f(x)=(x-1)^2+3$ on $[-3,1]$

Critical points

$f'(x)=2(x-1)$ ⁠, zero only at $x=1$ ⁠, which is the right end point of $[-3,1]$ ⁠. So there is no critical point strictly inside; the candidates are just the end points $-3$ and $1$ ⁠.

Inside $(-3,1)$ ⁠, $x-1\lt0$ ⁠, so $f'(x)\lt0$ ⁠: $f$ is decreasing all the way along the interval. That already tells us the left end is highest and the right end lowest.

Theorem 1, Section 6.3. Lesson: increasing and decreasing
Tabulate and compare

$x$ $-3$ (end) $1$ (end, critical)

$(x-1)^2+3$ $16+3=19$ $0+3=3$

Absolute maximum value $19$ (at $x=-3$ ⁠); absolute minimum value $3$ (at $x=1$ ⁠).

Direct check: for $-3\le x\le1$ ⁠, $-4\le x-1\le0$ ⁠, so $0\le(x-1)^2\le16$ and $3\le f(x)\le19$ ⁠, with $3$ reached at $x=1$ and $19$ at $x=-3$ ⁠.

Working Rule, steps 2–4.
A note on the answer key

The book's answer for (iv) reads "Absolute minimum value = 19, absolute maximum value = 3". The two numbers are right but the words are swapped: a minimum value cannot be larger than the maximum value, and the table shows $f(-3)=19$ is the largest value and $f(1)=3$ the smallest.

In parts (i)–(iii) the key lists the minimum first and the maximum second, as here; in (iv) the two numbers were simply put in the wrong order.

$x$	$-3$ (end)	$1$ (end, critical)
$(x-1)^2+3$	$16+3=19$	$0+3=3$

(x-1)^2+3

[-3,1]

is the left half of a parabola: it falls from

(-3,19)

to its bottom

(1,3)

⁠, which is also the right end point.

Result

(i) Absolute maximum value $8$ ⁠, absolute minimum value $-8$ ⁠.

(ii) Absolute maximum value $\sqrt2$ ⁠, absolute minimum value $-1$ ⁠.

(iii) Absolute maximum value $8$ ⁠, absolute minimum value $-10$ ⁠.

(iv) Absolute maximum value $19$ ⁠, absolute minimum value $3$ ⁠.

(Textbook answer: (i) −8 and 8, (ii) −1 and √2, (iii) −10 and 8, as above. (iv) The key prints "Absolute minimum value = 19, absolute maximum value = 3", with maximum and minimum interchanged; see Part (iv), Step 3.)

Check it with numbers

(i) $f(1.9)=6.859\lt8$ ⁠, $f(-1.9)=-6.859\gt-8$ ⁠. ✓

(ii) $f(0.7)\approx1.409060$ ⁠, below $\sqrt2\approx1.414214$ ⁠; $f(3)\approx0.141120-0.989992$ $=-0.848872\gt-1$ ⁠. ✓

(iii) $f(3.9)=15.6-7.605=7.995\lt8$ ⁠, $f(-1.9)=-7.6-1.805=-9.405\gt-10$ ⁠. ✓

(iv) $f(-2.9)=18.21\lt19$ ⁠, $f(0.9)=3.01\gt3$ ⁠. ✓

Common mistakes

Forgetting the end points. In (i), (ii) and (iv) at least one extreme is at an end point.
Using a critical point outside the interval. In (ii), $\frac{5\pi}{4}$ also solves $\tan x=1$ but is not in $[0,\pi]$ ⁠.
Using the second derivative test for absolute extremes. It only gives local information; the rule is to compare values.
Copying the key's wording in (iv). $19$ is the maximum and $3$ the minimum.

What to take away

On a closed interval: list the end points and the interior critical points, evaluate, compare.
A table of values makes the comparison clear and easy to check.
A critical point need not be an extreme (part (i)); it just has to be checked.

Every fact used, and where it comes from

Fact	Used in	Source
A continuous function on $[a,b]$ has absolute maximum and minimum values	Every part	Theorem 5, Section 6.4.1 (Lesson: absolute extrema)
At an interior absolute extreme, $f'(c)=0$	Every part	Theorem 6, Section 6.4.1
Working Rule (critical points, end points, values, compare)	Every part	Working Rule, Section 6.4.1; compare Examples 27 and 28
$f'\lt0$ on $(a,b)$ ⇒ $f$ decreasing on $[a,b]$	Part (iv), Step 1	Theorem 1, Section 6.3
Values of $\sin$ ⁠, $\cos$ ⁠, $\tan$ at $0$ ⁠, $\frac{\pi}{4}$ ⁠, $\frac{\pi}{2}$ ⁠, $\pi$	Part (ii)	Class XI, Chapter 3 (Basics)
Answer for (iv) with maximum and minimum interchanged	Part (iv), Step 3	Answers, Exercise 6.3, Q5(iv)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 6.3 · Q5: Absolute maximum and minimum of x3x^3x3⁠, sin⁡x+cos⁡x\sin x+\cos xsinx+cosx⁠, 4x−12x24x-\tfrac12x^24x−21​x2⁠, (x−1)2+3(x-1)^2+3(x−1)2+3 on closed intervals

Basics you need

Part (i): f(x)=x3f(x)=x^3f(x)=x3 on [−2,2][-2,2][−2,2]

Critical points in the interval

Tabulate and compare

Part (ii): f(x)=sin⁡x+cos⁡xf(x)=\sin x+\cos xf(x)=sinx+cosx on [0,π][0,\pi][0,π]

Critical points in [0,π][0,\pi][0,π]

Tabulate and compare

Part (iii): f(x)=4x−12x2f(x)=4x-\frac12x^2f(x)=4x−21​x2 on [−2,92]\left[-2,\frac92\right][−2,29​]

Critical points in the interval

Tabulate and compare

Part (iv): f(x)=(x−1)2+3f(x)=(x-1)^2+3f(x)=(x−1)2+3 on [−3,1][-3,1][−3,1]

Critical points

Tabulate and compare

A note on the answer key

Every fact used, and where it comes from

Exercise 6.3 · Q5: Absolute maximum and minimum of $x^3$ ⁠, $\sin x+\cos x$ ⁠, $4x-\tfrac12x^2$ ⁠, $(x-1)^2+3$ on closed intervals

Part (i): $f(x)=x^3$ on $[-2,2]$

Part (ii): $f(x)=\sin x+\cos x$ on $[0,\pi]$

Critical points in $[0,\pi]$

Part (iii): $f(x)=4x-\frac12x^2$ on $\left[-2,\frac92\right]$

Part (iv): $f(x)=(x-1)^2+3$ on $[-3,1]$