Monotonicity and extrema

A drone flies for $7$ minutes. Its height after $t$ minutes is $h(t)=t^3-12t^2+36t+10$ (metres). It takes off from a rooftop at a height of $10$ m, after $2$ minutes it is at $42$ m, after $6$ minutes it is back down at $10$ m, and it reaches $17$ m when time is up. So during which time intervals is the drone climbing, during which is it descending, and at which moments does it "turn round"?

To answer exactly, we do not need the height at lots of moments. We only need the sign of the derivative $h'(t)$ .

Monotonicity

Explore: when does the graph go up?

We compute the height at whole minutes, and the derivative $h'(t)=3t^2-24t+36=3(t-2)(t-6)$ :

$t$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$h(t)$	$10$	$35$	$42$	$37$	$26$	$15$	$10$	$17$
$h'(t)$	$36$	$15$	$0$	$-9$	$-12$	$-9$	$0$	$15$

From $t=0$ to $t=2$ the height increases and $h'>0$ . From $t=2$ to $t=6$ the height decreases and $h'<0$ . After $t=6$ the height increases again and $h'>0$ . This makes sense: $h'(t)$ is the gradient of the tangent, and a positive gradient means the graph is going up.

Height of the drone

h(t)=t^3-12t^2+36t+10

Why is it true? The derivative $f'(x)$ is the limit of the ratio $\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$ . If the graph has an upward-sloping tangent at every point of $K$ , it cannot go down anywhere in $K$ . A few points with a horizontal tangent (like $y=x^3$ at $x=0$ ) do not make the graph turn round.

To study monotonicity, we work in this order:

Find the domain.
Compute $f'(x)$ and find the points where $f'(x)=0$ (the stationary points) or $f'(x)$ is undefined.
Draw the table of variations: find the sign of $f'$ on each interval.
State the intervals where $f$ increases and decreases.

Example A cubic function

Find the intervals on which $y=x^3-6x^2+9x+1$ is increasing or decreasing.

Solution. The function is defined on $\mathbb{R}$ . We differentiate and factorise to see the sign:

$y'=3x^2-12x+9=3(x-1)(x-3),\qquad y'=0\iff x=1\ \text{or}\ x=3.$

The quadratic $y'$ has $a=3>0$ , so it is positive outside the roots and negative between them. We have $y(1)=5$ , $y(3)=1$ and the table of variations:

$x$	$-\infty$		$1$		$3$		$+\infty$
$y'$		$+$	$0$	$-$	$0$	$+$
$y$	$-\infty$	$\nearrow$	$5$	$\searrow$	$1$	$\nearrow$	$+\infty$

So the function is increasing on the intervals $(-\infty, 1)$ and $(3, +\infty)$ , and decreasing on the interval $(1, 3)$ .

The next example shows that $f'(x_0)=0$ does not always make the function turn round.

Extrema of a function

In the drone problem, the moment $t=2$ gives a height greater than at all nearby moments, although it is not necessarily the highest of the whole flight. We say that $h$ has a local maximum at $t=2$ .

Definition Local maximum and local minimum

Let $f$ be defined on a set $D$ and $x_0\in D$ .

$f$ has a local maximum at $x_0$ (and $x_0$ is a local maximum point of $f$ ) if there is an interval $(a, b)\subset D$ containing $x_0$ such that $f(x)<f(x_0)$ for every $x\in(a, b)$ , $x\neq x_0$ . Then $f(x_0)$ is the local maximum value, written $y_{\text{max}}$ .
$f$ has a local minimum at $x_0$ ( $x_0$ is a local minimum point) if " $<$ " is replaced by " $>$ ". Then $f(x_0)$ is the local minimum value, written $y_{\text{min}}$ .
The point $M(x_0, f(x_0))$ on the graph is a turning point (a local maximum or local minimum of the graph). Local maxima and minima together are called extrema.

The reason: a change from $+$ to $-$ means the graph goes up just before $x_0$ and down just after it, so $f(x_0)$ is greater than the nearby values. In Example 1, $y'$ changes sign from $+$ to $-$ at $x=1$ , so $f$ has a local maximum at $x=1$ with $y_{\text{max}}=5$ ; likewise $f$ has a local minimum at $x=3$ with $y_{\text{min}}=1$ .

Example A quartic with three extrema

Find the extrema of $y=x^4-2x^2+3$ .

Solution. We find the roots of $y'$ and then the sign of each factor:

$y'=4x^3-4x=4x(x-1)(x+1),\qquad y'=0\iff x\in\{-1, 0, 1\}.$

Substituting: $y(-1)=2$ , $y(0)=3$ , $y(1)=2$ . The table of variations:

$x$	$-\infty$		$-1$		$0$		$1$		$+\infty$
$y'$		$-$	$0$	$+$	$0$	$-$	$0$	$+$
$y$	$+\infty$	$\searrow$	$2$	$\nearrow$	$3$	$\searrow$	$2$	$\nearrow$	$+\infty$

So the function has a local minimum at $x=\pm1$ with $y_{\text{min}}=2$ , and a local maximum at $x=0$ with $y_{\text{max}}=3$ .

The graph of

y=x^4-2x^2+3

For a rational function, a value that makes the denominator zero is not in the domain, but it must still be written in the table of variations to split the intervals.

Example A rational function

Find the intervals on which $y=\dfrac{x^2+3}{x-1}$ is increasing or decreasing, and its extrema.

Solution. The domain is $D=\mathbb{R}\setminus\{1\}$ . We use the quotient rule:

$y'=\frac{2x(x-1)-(x^2+3)}{(x-1)^2}=\frac{x^2-2x-3}{(x-1)^2}=\frac{(x+1)(x-3)}{(x-1)^2}.$

The denominator is positive on $D$ , so $y'$ has the sign of $(x+1)(x-3)$ . We have $y(-1)=-2$ , $y(3)=6$ :

$x$	$-\infty$		$-1$		$1$		$3$		$+\infty$
$y'$		$+$	$0$	$-$	$\Vert$	$-$	$0$	$+$
$y$	$-\infty$	$\nearrow$	$-2$	$\searrow$	$\Vert$	$\searrow$	$6$	$\nearrow$	$+\infty$

So the function is increasing on $(-\infty, -1)$ and $(3, +\infty)$ , decreasing on $(-1, 1)$ and $(1, 3)$ ; the local maximum is $y_{\text{max}}=y(-1)=-2$ and the local minimum is $y_{\text{min}}=y(3)=6$ .

Look back. Here $y_{\text{max}}<y_{\text{min}}$ . Nothing is wrong: a local maximum is only greater than the values near it.

The graph of

y=\dfrac{x^2+3}{x-1}

Using the second derivative

When the sign of $f'$ is hard to find (often with trigonometric functions), we can use the second derivative (grade 11).

Example A trigonometric function

Find the extrema of $f(x)=x-2\sin x$ on the interval $(0, 2\pi)$ .

Solution. We solve $f'(x)=0$ on $(0, 2\pi)$ , then find the sign of $f''$ at each root:

$f'(x)=1-2\cos x=0\iff \cos x=\frac12\iff x=\frac{\pi}{3}\ \text{or}\ x=\frac{5\pi}{3}.$

The second derivative is $f''(x)=2\sin x$ :

$f''\left(\frac{\pi}{3}\right)=\sqrt3>0,\qquad f''\left(\frac{5\pi}{3}\right)=-\sqrt3<0.$

So $f$ has a local minimum at $x=\dfrac{\pi}{3}$ with $f_{\text{min}}=\dfrac{\pi}{3}-\sqrt3$ , and a local maximum at $x=\dfrac{5\pi}{3}$ with $f_{\text{max}}=\dfrac{5\pi}{3}+\sqrt3$ .

Finally, we answer the question about the drone from the start of the lesson.

Example The drone goes up and down

The height of the drone is $h(t)=t^3-12t^2+36t+10$ (metres), $0\le t\le 7$ (minutes). When is the drone climbing, and when is it descending? Find the extrema.

Solution. We find the sign of $h'(t)=3(t-2)(t-6)$ on $(0, 7)$ : the quadratic has $a=3>0$ , so it is negative between the roots $2$ and $6$ and positive outside them.

$h'(t)>0 \text{ for } t\in(0, 2)\cup(6, 7),\qquad h'(t)<0 \text{ for } t\in(2, 6).$

$h'$ changes sign from $+$ to $-$ at $t=2$ and from $-$ to $+$ at $t=6$ .

So the drone climbs during the first $2$ minutes and during the last minute (from minute $6$ to minute $7$ ), and descends from minute $2$ to minute $6$ . It has a local maximum of $42$ m at $t=2$ and a local minimum of $10$ m at $t=6$ .

Watch out Common mistakes

Joining two intervals with a union. Wrong: " $y=\dfrac{x+2}{x-1}$ is decreasing on $(-\infty, 1)\cup(1, +\infty)$ ". Why it is wrong: $y(0)=-2<y(2)=4$ although $0<2$ . Right: the function is decreasing on each interval $(-\infty, 1)$ and $(1, +\infty)$ .
Thinking $f'(x_0)=0$ means an extremum. Wrong: " $y=x^3-3x^2+3x+5$ has $y'(1)=0$ , so it has an extremum at $x=1$ ". Why it is wrong: $y'=3(x-1)^2$ does not change sign at $x=1$ . Right: the function has no extremum.
Mixing up the point and the value. Wrong: "the local maximum point of $y=x^4-2x^2+3$ is $3$ ". Why it is wrong: the local maximum point is a value of $x$ . Right: local maximum point $x=0$ , local maximum value $y_{\text{max}}=3$ , turning point of the graph $(0, 3)$ .
Forgetting the domain. Wrong: drawing the table of variations of $y=\dfrac{x^2+3}{x-1}$ without $x=1$ . Why it is wrong: the sign of $y'$ may change at $x=1$ , and the function is undefined there. Right: always write the excluded points in the table.

Key terms

English	Tiếng Việt
increasing	đồng biến
decreasing	nghịch biến
monotonicity	tính đơn điệu
table of variations	bảng biến thiên
local maximum point	điểm cực đại
local minimum point	điểm cực tiểu
local maximum value	giá trị cực đại
local minimum value	giá trị cực tiểu
extremum (turning point)	cực trị
second derivative	đạo hàm cấp hai

Summary

$f'>0$ on $K$ means $f$ is increasing on $K$ ; $f'<0$ means decreasing. If $f'\ge0$ and it is zero at only finitely many points, $f$ is still increasing.
Steps: domain → $f'$ and its roots → table of variations → conclusion for each interval.
$f'$ changes sign $+\to-$ at $x_0$ : local maximum; $-\to+$ : local minimum. No sign change: no extremum.
Second-derivative test: $f'(x_0)=0$ and $f''(x_0)<0$ gives a local maximum, $f''(x_0)>0$ a local minimum.
Distinguish the point $x_0$ where the extremum occurs, the extreme value $f(x_0)$ , and the turning point of the graph $(x_0, f(x_0))$ .

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 intervals on which $y=-x^3+3x^2+1$ is increasing or decreasing.

What you need and how to start

What you need: if $y'>0$ on an interval, the function is increasing there; if $y'<0$ , it is decreasing.

Steps:

Compute $y'$ and factorise it.
Find the sign of $y'$ (mind the negative coefficient).

Show solution

$y'=-3x^2+6x=-3x(x-2)$ , with roots $0$ and $2$ . The quadratic has $a=-3<0$ , so it is positive between the roots and negative outside them.

Answer. Increasing on $(0, 2)$ ; decreasing on $(-\infty, 0)$ and $(2, +\infty)$ .

Exercise 2

Find the extrema of $y=x^4-8x^2+1$ .

What you need and how to start

What you need: $y'$ changes sign from $-$ to $+$ at a local minimum and from $+$ to $-$ at a local maximum.

Steps:

Compute $y'=4x^3-16x$ and find its three roots.
Draw the table of variations.
Compute $y$ at these points.

Show solution

$y'=4x(x-2)(x+2)$ , with roots $-2$ , $0$ , $2$ . The signs of $y'$ on the four intervals are $-, +, -, +$ . So there is a local minimum at $x=\pm2$ with $y=16-32+1=-15$ , and a local maximum at $x=0$ with $y=1$ .

Answer. $y_{\text{min}}=-15$ at $x=\pm2$ ; $y_{\text{max}}=1$ at $x=0$ .

Exercise 3

Determine where $y=\dfrac{x+2}{x-1}$ is increasing or decreasing.

What you need and how to start

What you need: $\left(\dfrac{ax+b}{cx+d}\right)'=\dfrac{ad-bc}{(cx+d)^2}$ .

Steps:

Find the domain.
Compute $y'$ and look at the sign of the numerator.

Show solution

$D=\mathbb{R}\setminus\{1\}$ and $y'=\dfrac{1\cdot(-1)-2\cdot1}{(x-1)^2}=\dfrac{-3}{(x-1)^2}<0$ for every $x\in D$ .

Answer. The function is decreasing on each interval $(-\infty, 1)$ and $(1, +\infty)$ , and has no extremum.

Exercise 4

Find the extrema of $y=x+\dfrac{4}{x}$ .

What you need and how to start

What you need: write $y'$ as one fraction to find its sign; remember $x=0$ is not in the domain.

Steps:

Compute $y'=1-\dfrac{4}{x^2}=\dfrac{x^2-4}{x^2}$ .
Draw the table of variations with the values $-2$ , $0$ , $2$ marked.

Show solution

$y'=\dfrac{(x-2)(x+2)}{x^2}$ has the sign of $(x-2)(x+2)$ : positive when $|x|>2$ , negative when $0<|x|<2$ . So $y'$ changes sign $+\to-$ at $x=-2$ and $-\to+$ at $x=2$ .

Answer. $y_{\text{max}}=y(-2)=-4$ ; $y_{\text{min}}=y(2)=4$ .

Exercise 5

A function $y=f(x)$ has derivative $f'(x)=x(x-1)^2(x+2)$ for every $x\in\mathbb{R}$ . Is each claim true or false?

a) $f$ is increasing on the interval $(0, +\infty)$ .

b) $f$ has three turning points.

c) $f$ has a local maximum at $x=-2$ .

d) $f$ is decreasing on the interval $(-2, 1)$ .

What you need and how to start

What you need: the factor $(x-1)^2$ is non-negative, so it does not make $f'$ change sign; extrema occur only where $f'$ changes sign.

Steps:

Make a sign table of $f'$ with the values $-2$ , $0$ , $1$ marked.
Read off where $f$ increases or decreases and where $f'$ changes sign.

Show solution

$f'$ has the sign of $x(x+2)$ (except at $x=1$ , where $f'=0$ ): positive on $(-\infty, -2)$ , negative on $(-2, 0)$ , positive on $(0, 1)$ and $(1, +\infty)$ .

a) True: $f'\ge 0$ on $(0, +\infty)$ and it is zero only at $x=1$ .

b) False: $f'$ changes sign only at $x=-2$ and $x=0$ , so there are two turning points.

c) True: $f'$ changes sign from $+$ to $-$ at $x=-2$ .

d) False: on $(0, 1)$ we have $f'>0$ .

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

Exercise 6

The function $y=x^3-3x^2-9x+4$ has a local maximum value $y_{\text{max}}$ and a local minimum value $y_{\text{min}}$ . Compute $y_{\text{max}}-y_{\text{min}}$ .

What you need and how to start

What you need: for a cubic with two extrema and $a>0$ , the local maximum occurs at the smaller root of $y'$ .

Steps:

Solve $y'=3x^2-6x-9=0$ .
Compute $y$ at the two roots.

Show solution

$y'=3(x+1)(x-3)$ . Since $a=3>0$ , $y'$ changes sign $+\to-$ at $x=-1$ and $-\to+$ at $x=3$ . $y(-1)=-1-3+9+4=9$ , $y(3)=27-27-27+4=-23$ . The difference is $9-(-23)=32$ .

Answer. $32$ .

Exercise 7

The concentration of a medicine in the blood $t$ hours after an injection is $C(t)=\dfrac{30t}{t^2+9}$ (mg/l), $t>0$ . During which time interval does the concentration increase, and when does it decrease? When does it reach its local maximum, and what is that value?

What you need and how to start

What you need: the quotient rule $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}$ .

Steps:

Compute $C'(t)$ and simplify the numerator.
Find the sign of $C'(t)$ for $t>0$ .

Show solution

$C'(t)=\dfrac{30(t^2+9)-30t\cdot 2t}{(t^2+9)^2}=\dfrac{30(9-t^2)}{(t^2+9)^2}$ . For $t>0$ : $C'>0$ when $0<t<3$ and $C'<0$ when $t>3$ . So $C$ has a local maximum at $t=3$ , $C(3)=\dfrac{90}{18}=5$ .

Answer. It increases during the first $3$ hours and decreases afterwards; the local maximum is $5$ mg/l at $t=3$ hours.

Exercise 8 Review

Find the extrema of $y=x+2\cos x$ on the interval $(0, \pi)$ .

What you need and how to start

What you need: the equation $\sin x=\dfrac12$ (basic trigonometric equations, grade 11) and the second-derivative test.

Steps:

Solve $y'=1-2\sin x=0$ on $(0, \pi)$ .
Compute $y''=-2\cos x$ at each root.

Show solution

$\sin x=\dfrac12$ gives $x=\dfrac{\pi}{6}$ or $x=\dfrac{5\pi}{6}$ . $y''\left(\dfrac{\pi}{6}\right)=-\sqrt3<0$ : local maximum; $y''\left(\dfrac{5\pi}{6}\right)=\sqrt3>0$ : local minimum.

Answer. $y_{\text{max}}=\dfrac{\pi}{6}+\sqrt3$ at $x=\dfrac{\pi}{6}$ ; $y_{\text{min}}=\dfrac{5\pi}{6}-\sqrt3$ at $x=\dfrac{5\pi}{6}$ .

Exercise 9 Review

Prove that $y=x^3+x^2+x+1$ is increasing on $\mathbb{R}$ .

What you need and how to start

What you need: a quadratic expression with $\Delta<0$ always has the sign of $a$ (see the sign of a quadratic expression).

Steps:

Compute $y'$ .
Compute the discriminant of $y'$ and look at the sign of $a$ .

Show solution

$y'=3x^2+2x+1$ has $\Delta=4-12=-8<0$ and $a=3>0$ , so $y'>0$ for every $x$ .

Answer. $y'>0$ for every $x\in\mathbb{R}$ , so the function is increasing on $\mathbb{R}$ .

Exercise 10 Challenge

Find the intervals on which $y=x\sqrt{4-x^2}$ is increasing or decreasing, and its extrema.

What you need and how to start

What you need: $\left(\sqrt{u}\right)'=\dfrac{u'}{2\sqrt{u}}$ ; a square root is defined only when the expression under it is non-negative.

Steps:

Find the domain.
Compute $y'$ and write it as a single fraction.
Find the sign of the numerator on $(-2, 2)$ .

Show solution

$D=[-2, 2]$ . On $(-2, 2)$ : $y'=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}},$ which is zero when $x=\pm\sqrt2$ . The numerator is positive when $|x|<\sqrt2$ and negative when $\sqrt2<|x|<2$ . $y(\sqrt2)=\sqrt2\cdot\sqrt2=2$ , $y(-\sqrt2)=-2$ .

Answer. Increasing on $(-\sqrt2, \sqrt2)$ , decreasing on $(-2, -\sqrt2)$ and $(\sqrt2, 2)$ ; $y_{\text{min}}=-2$ at $x=-\sqrt2$ , $y_{\text{max}}=2$ at $x=\sqrt2$ .

Monotonicity

Explore: when does the graph go up?

Extrema of a function

Using the second derivative

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $f'(x)=6x^2-6x-12$ .	True	False
b) $f$ is increasing on the interval $(-1, 2)$ .	True	False
c) The local maximum value of $f$ is $8$ .	True	False
d) The line through the two turning points of the graph has gradient $9$ .	True	False