Maximum and minimum values

The temperature in a laboratory is monitored for $5$ hours. After $t$ hours, it is $T(t)=t^3-6t^2+9t+20$ (°C), $0\le t\le 5$ . At the start the room is at $20$ °C, after $1$ hour it is $24$ °C, and after $3$ hours it is back at $20$ °C. Over those $5$ hours, what are the highest and the lowest temperatures?

You might think straight away of the local maximum $24$ °C at $t=1$ . But the answer is not $24$ .

Explore: is a local maximum the highest value?

We compute the temperature at whole hours:

$t$	$0$	$1$	$2$	$3$	$4$	$5$
$T(t)$	$20$	$24$	$22$	$20$	$24$	$40$

The derivative $T'(t)=3t^2-12t+9=3(t-1)(t-3)$ tells us that $T$ has a local maximum of $24$ at $t=1$ and a local minimum of $20$ at $t=3$ . Yet by the end of the session, $T(5)=40$ , far above the local maximum. A local maximum is compared only with the moments near it; "the highest" must be compared with every moment.

Laboratory temperature

T(t)=t^3-6t^2+9t+20

The condition "there is $x_0$ " matters. For example, $f(x)=x^2\le 4$ for every $x\in(0, 2)$ , but no $x$ in that interval gives $f(x)=4$ ; so $4$ is not the maximum value of $f$ on $(0, 2)$ , and in fact $f$ has no maximum value on this interval.

On a closed interval

The maximum can only be reached in two places: at an endpoint, or at an interior point where the graph "turns round", that is, where $f'=0$ or $f'$ is undefined. At any other interior point, $f'$ is not zero, so the function is still increasing or decreasing and there is a larger value nearby. So we only need to compare a few values:

Find the points $x_1, x_2, \dots$ in the open interval $(a, b)$ where $f'(x)=0$ or $f'(x)$ is undefined.
Compute $f(a)$ , $f(x_1)$ , $f(x_2)$ , …, $f(b)$ .
The largest of these numbers is $\max\limits_{[a, b]} f(x)$ , and the smallest is $\min\limits_{[a, b]} f(x)$ .

Example A cubic on a closed interval

Find the maximum and minimum values of $f(x)=x^3-3x+1$ on $[0, 2]$ .

Solution. A polynomial is continuous on $[0, 2]$ . We find the roots of $f'$ and keep only those in $(0, 2)$ :

$f'(x)=3x^2-3=0\iff x=1\ \text{or}\ x=-1\ (\text{rejected since } -1\notin(0, 2)).$

We compare the values at the critical point and at the two endpoints:

$f(0)=1,\qquad f(1)=-1,\qquad f(2)=8-6+1=3.$

So $\max\limits_{[0, 2]} f(x)=3$ (at $x=2$ ) and $\min\limits_{[0, 2]} f(x)=-1$ (at $x=1$ ).

Rejecting a root outside the interval is an easy step to forget. The next example has one rejected root and one irrational root.

Example A quartic

Find the maximum and minimum values of $f(x)=x^4-4x^2+5$ on $[-1, 3]$ .

Solution. We solve $f'(x)=0$ and keep the roots in $(-1, 3)$ :

$f'(x)=4x^3-8x=4x(x^2-2)=0\iff x\in\{0, \sqrt2, -\sqrt2\},\quad -\sqrt2\notin(-1, 3).$

We compute the values to compare. For $x=\sqrt2$ we have $x^2=2$ and $x^4=4$ :

$f(-1)=1-4+5=2,\quad f(0)=5,\quad f(\sqrt2)=4-8+5=1,\quad f(3)=81-36+5=50.$

So $\max\limits_{[-1, 3]} f(x)=50$ at $x=3$ and $\min\limits_{[-1, 3]} f(x)=1$ at $x=\sqrt2$ .

With a square root, we solve $f'(x)=0$ carefully, with a condition before squaring.

Example A function with a square root

Find the maximum and minimum values of $f(x)=x+\sqrt{4-x^2}$ .

Solution. The domain is $[-2, 2]$ and $f$ is continuous on it. On $(-2, 2)$ :

$f'(x)=1-\frac{x}{\sqrt{4-x^2}}=0\iff \sqrt{4-x^2}=x.$

This equation needs $x\ge 0$ ; we square both sides:

$4-x^2=x^2\iff x^2=2\iff x=\sqrt2\quad(\text{since } x\ge 0).$

Comparing: $f(-2)=-2$ , $f(\sqrt2)=\sqrt2+\sqrt2=2\sqrt2$ , $f(2)=2$ .

So the maximum value is $2\sqrt2$ (at $x=\sqrt2$ ) and the minimum value is $-2$ (at $x=-2$ ).

With a trigonometric function, a substitution often turns the problem into a polynomial on a closed interval.

Example A trigonometric function and a substitution

Find the maximum and minimum values of $f(x)=2\sin x+\cos 2x$ on $[0, \pi]$ .

Solution. We use the double-angle formula $\cos 2x=1-2\sin^2x$ and put $t=\sin x$ :

$f(x)=2\sin x+1-2\sin^2 x=g(t)=-2t^2+2t+1.$

As $x$ runs over $[0, \pi]$ , $t=\sin x$ runs over $[0, 1]$ . We find the maximum and minimum of $g$ on $[0, 1]$ :

$g'(t)=-4t+2=0\iff t=\frac12;\qquad g(0)=1,\quad g\left(\frac12\right)=\frac32,\quad g(1)=1.$

So $\max\limits_{[0, \pi]} f(x)=\dfrac32$ (when $\sin x=\dfrac12$ , that is $x=\dfrac{\pi}{6}$ or $x=\dfrac{5\pi}{6}$ ) and $\min\limits_{[0, \pi]} f(x)=1$ (when $x\in\left\{0, \dfrac{\pi}{2}, \pi\right\}$ ).

Look back. The range of the new variable must be right: $t\in[0, 1]$ , not $[-1, 1]$ , because $\sin x\ge 0$ on $[0, \pi]$ .

On an open interval

On an open interval (with no endpoints), a function may have no maximum or no minimum value. Then we draw the table of variations and read the result from it.

Example A minimum on an open interval

Find the minimum value of $f(x)=x+\dfrac{9}{x}$ on $(0, +\infty)$ .

Solution. We differentiate and write $f'$ as one fraction to find its sign:

$f'(x)=1-\frac{9}{x^2}=\frac{(x-3)(x+3)}{x^2},\qquad f'(x)=0\iff x=3\ \ (x>0).$

For $x>0$ , $f'$ has the sign of $x-3$ . As $x\to0^+$ and as $x\to+\infty$ , $f(x)\to+\infty$ :

$x$	$0$		$3$		$+\infty$
$f'(x)$		$-$	$0$	$+$
$f(x)$	$+\infty$	$\searrow$	$6$	$\nearrow$	$+\infty$

So $\min\limits_{(0, +\infty)} f(x)=f(3)=6$ , and $f$ has no maximum value on $(0, +\infty)$ .

The graph of

y=x+\dfrac{9}{x}

for

x>0

Real-life problems

Now we answer the temperature question from the start. In real-life problems the endpoints are often the start and the end of a period, so they must not be ignored.

Watch out Common mistakes

Taking a local maximum as the maximum. Wrong: " $T$ has a local maximum of $24$ , so the highest temperature is $24$ °C". Why it is wrong: the endpoints were not compared, and $T(5)=40$ . Right: always compute $f(a)$ and $f(b)$ as well.
Keeping a root outside the interval. Wrong: finding the minimum of $x^3-3x+1$ on $[0, 2]$ and including $f(-1)=3$ . Why it is wrong: $-1$ is not in the interval. Right: only use roots in $(a, b)$ .
The wrong range for a substitution. Wrong: putting $t=\sin x$ for $x\in[0, \pi]$ and then using $t\in[-1, 1]$ , giving the minimum $g(-1)=-3$ . Why it is wrong: on $[0, \pi]$ , $\sin x\ge 0$ . Right: $t\in[0, 1]$ and the minimum is $1$ .
Claiming a maximum that is never reached. Wrong: " $x+\dfrac9x$ has maximum value $+\infty$ ". Why it is wrong: $+\infty$ is not a number and no $x$ gives it. Right: the function has no maximum value on $(0, +\infty)$ .

Key terms

English	Tiếng Việt
maximum (greatest) value	giá trị lớn nhất
minimum (least) value	giá trị nhỏ nhất
closed interval	đoạn
open interval	khoảng
endpoint	đầu mút
critical point	điểm tới hạn
continuous function	hàm số liên tục
substitution (new variable)	ẩn phụ

Summary

$M=\max\limits_D f$ when $f(x)\le M$ for every $x\in D$ and $M$ is reached at some point of $D$ . Likewise for $\min$ .
A function continuous on $[a, b]$ always has a maximum and a minimum value there.
On a closed interval: roots of $f'$ (and points where $f'$ is undefined) inside $(a, b)$ → compute $f$ there and at $a$ , $b$ → compare.
On an open interval: draw the table of variations; there may be no maximum or no minimum.
After a substitution, find the correct range of the new variable.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Find the maximum and minimum values of $y=x^3-3x^2$ on $[-1, 3]$ .

What you need and how to start

What you need: on a closed interval, compare the values at the roots of $y'$ in $(-1, 3)$ and at the two endpoints.

Steps:

Solve $y'=3x^2-6x=0$ .
Compute $y$ at the roots and at $-1$ , $3$ .

Show solution

$y'=3x(x-2)=0\iff x=0$ or $x=2$ , both in $(-1, 3)$ . $y(-1)=-4$ , $y(0)=0$ , $y(2)=8-12=-4$ , $y(3)=27-27=0$ .

Answer. Maximum value $0$ (at $x=0$ and $x=3$ ); minimum value $-4$ (at $x=-1$ and $x=2$ ).

Exercise 2

Find the maximum and minimum values of $y=\dfrac{x-1}{x+2}$ on $[0, 3]$ .

What you need and how to start

What you need: if a function is monotonic on a closed interval, its maximum and minimum are at the endpoints.

Steps:

Compute $y'$ and look at its sign.
Compute $y(0)$ and $y(3)$ .

Show solution

$y'=\dfrac{1\cdot2-(-1)\cdot1}{(x+2)^2}=\dfrac{3}{(x+2)^2}>0$ , so the function is increasing on $[0, 3]$ . $y(0)=-\dfrac12$ , $y(3)=\dfrac25$ .

Answer. Maximum value $\dfrac25$ ; minimum value $-\dfrac12$ .

Exercise 3

Find the maximum and minimum values of $y=x^4-2x^2+3$ on $[0, 2]$ .

What you need and how to start

What you need: keep only the roots of $y'$ in the open interval $(0, 2)$ .

Steps:

Solve $y'=4x^3-4x=0$ .
Reject the roots not in $(0, 2)$ , then compare the values.

Show solution

$y'=4x(x-1)(x+1)$ ; the only root in $(0, 2)$ is $x=1$ . $y(0)=3$ , $y(1)=2$ , $y(2)=16-8+3=11$ .

Answer. Maximum value $11$ (at $x=2$ ); minimum value $2$ (at $x=1$ ).

Exercise 4

Find the minimum value of $y=x^2+\dfrac{16}{x}$ on $(0, +\infty)$ .

What you need and how to start

What you need: on an open interval, draw the table of variations to read off the minimum.

Steps:

Compute $y'=2x-\dfrac{16}{x^2}=\dfrac{2(x^3-8)}{x^2}$ .
Find the positive root of $y'$ and the sign on each side.

Show solution

$y'=0\iff x^3=8\iff x=2$ . For $0<x<2$ , $y'<0$ ; for $x>2$ , $y'>0$ . So the minimum value is $y(2)=4+8=12$ .

Answer. $\min\limits_{(0, +\infty)} y=12$ at $x=2$ .

Exercise 5

Consider $f(x)=-x^3+3x^2+2$ on $[-1, 3]$ . Is each claim true or false?

a) The equation $f'(x)=0$ has two roots in $[-1, 3]$ .

b) The maximum value of $f$ on $[-1, 3]$ is $6$ .

c) $f$ reaches its maximum value only at $x=2$ .

d) The minimum value of $f$ on $[-1, 3]$ is $2$ .

What you need and how to start

What you need: compare the values at the roots of $f'$ and at the two endpoints; a maximum can be reached at several points.

Steps:

Solve $f'(x)=-3x^2+6x=0$ .
Compute $f(-1)$ , $f(0)$ , $f(2)$ , $f(3)$ .

Show solution

$f'(x)=-3x(x-2)=0\iff x=0$ or $x=2$ . $f(-1)=1+3+2=6$ , $f(0)=2$ , $f(2)=-8+12+2=6$ , $f(3)=-27+27+2=2$ .

a) True: $0$ and $2$ are both in $[-1, 3]$ .

b) True: the largest of the values is $6$ .

c) False: $f(-1)=6$ as well, so the maximum is reached at both $x=-1$ and $x=2$ .

d) True: the smallest value is $2$ , reached at $x=0$ and $x=3$ .

Answer. a) True; b) True; c) False; d) True.

Exercise 6

Find the maximum value of $y=x\sqrt{9-x^2}$ on $[0, 3]$ (give the result as a decimal).

What you need and how to start

What you need: $\left(\sqrt{u}\right)'=\dfrac{u'}{2\sqrt u}$ and comparing values on a closed interval.

Steps:

Compute $y'$ and write it as one fraction: the numerator is $9-2x^2$ .
Find the root in $(0, 3)$ and compare with $y(0)$ , $y(3)$ .

Show solution

$y'=\sqrt{9-x^2}-\dfrac{x^2}{\sqrt{9-x^2}}=\dfrac{9-2x^2}{\sqrt{9-x^2}}=0\iff x^2=\dfrac92$ , that is $x=\dfrac{3}{\sqrt2}$ . Then $y=\dfrac{3}{\sqrt2}\cdot\sqrt{\dfrac92}=\dfrac92$ . The endpoints give $y(0)=y(3)=0$ .

Answer. $4.5$ .

Exercise 7

A workshop makes $x$ chairs a week ( $10\le x\le 100$ ) at a total cost of $C(x)=x^2+100x+2500$ (thousand đồng). The average cost per chair is $A(x)=\dfrac{C(x)}{x}$ . How many chairs a week should the workshop make to have the lowest average cost? What is that cost?

What you need and how to start

What you need: find the minimum of $A(x)=x+100+\dfrac{2500}{x}$ on $[10, 100]$ .

Steps:

Compute $A'(x)$ and solve $A'(x)=0$ .
Compare the values at the root and at $10$ , $100$ .

Show solution

$A'(x)=1-\dfrac{2500}{x^2}=0\iff x=50$ . $A(10)=10+100+250=360$ , $A(50)=50+100+50=200$ , $A(100)=100+100+25=225$ .

Answer. $50$ chairs a week; the lowest average cost is $200$ thousand đồng per chair.

Exercise 8 Review

Find the maximum and minimum values of $y=\cos 2x+2\cos x$ on $[0, \pi]$ .

What you need and how to start

What you need: the double-angle formula $\cos 2x=2\cos^2x-1$ (grade 11); for $x\in[0, \pi]$ , $\cos x\in[-1, 1]$ .

Steps:

Put $t=\cos x$ and write $y=g(t)$ .
Find the maximum and minimum of $g$ on $[-1, 1]$ .

Show solution

$y=2t^2+2t-1=g(t)$ with $t\in[-1, 1]$ . $g'(t)=4t+2=0\iff t=-\dfrac12$ . $g(-1)=-1$ , $g\left(-\dfrac12\right)=\dfrac12-1-1=-\dfrac32$ , $g(1)=3$ .

Answer. Maximum value $3$ (at $x=0$ ); minimum value $-\dfrac32$ (at $x=\dfrac{2\pi}{3}$ ).

Exercise 9 Review

Find the maximum and minimum values of $y=-x^2+4x+1$ on $[0, 5]$ in two ways: with the vertex of the parabola (grade 10) and with the derivative.

What you need and how to start

What you need: the parabola $y=ax^2+bx+c$ with $a<0$ has its highest point at the vertex $x=-\dfrac{b}{2a}$ (see the quadratic function).

Steps:

Method 1: find the vertex, check that it is in $[0, 5]$ , then compare with the two endpoints.
Method 2: solve $y'=0$ and compare as usual.

Show solution

Method 1: the vertex is at $x=-\dfrac{4}{2\cdot(-1)}=2\in[0, 5]$ , $y(2)=5$ . Method 2: $y'=-2x+4=0\iff x=2$ . Both methods compare $y(0)=1$ , $y(2)=5$ , $y(5)=-25+20+1=-4$ .

Answer. Maximum value $5$ (at $x=2$ ); minimum value $-4$ (at $x=5$ ).

Exercise 10 Challenge

Find the maximum and minimum values of $y=x+\sqrt2\cos x$ on $\left[0, \dfrac{\pi}{2}\right]$ .

What you need and how to start

What you need: comparing values on a closed interval; to compare irrational numbers, use approximations.

Steps:

Solve $y'=1-\sqrt2\sin x=0$ on $\left(0, \dfrac{\pi}{2}\right)$ .
Compute $y$ at the root and at both endpoints, then compare using approximations.

Show solution

$\sin x=\dfrac{1}{\sqrt2}$ gives $x=\dfrac{\pi}{4}$ . $y(0)=\sqrt2\approx1.414$ ; $y\left(\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}+1\approx1.785$ ; $y\left(\dfrac{\pi}{2}\right)=\dfrac{\pi}{2}\approx1.571$ .

Answer. Maximum value $\dfrac{\pi}{4}+1$ (at $x=\dfrac{\pi}{4}$ ); minimum value $\sqrt2$ (at $x=0$ ).

Maximum and minimum values