Increasing and Decreasing Functions, calculus lesson

On a hiking map, the trail goes up where the ground slopes upward and down where it slopes downward. For a function the slope is the derivative, so its sign should tell you where the graph rises and where it falls. The mean value theorem makes that reasoning airtight, as long as you stay on one interval at a time.

Key terms

Strict and nonstrict monotonicity

Definition Let $f$ be defined on an interval $I$ . Then $f$ is

strictly increasing on $I$ if $x_1 < x_2$ in $I$ implies $f(x_1) < f(x_2)$ ,
nondecreasing on $I$ if $x_1 < x_2$ in $I$ implies $f(x_1) \le f(x_2)$ ,
strictly decreasing on $I$ if $x_1 < x_2$ in $I$ implies $f(x_1) > f(x_2)$ ,
nonincreasing on $I$ if $x_1 < x_2$ in $I$ implies $f(x_1) \ge f(x_2)$ .

A function with any of these properties is called monotonic on $I$ . Many books say “increasing” for strictly increasing, and some use it for nondecreasing, so check the convention.

In plain words Strictly increasing means always going up. Nondecreasing allows flat stretches but never a step down.

Example The function $x^3$ is strictly increasing on $\mathbb{R}$ . The floor function $\lfloor x \rfloor$ , the greatest integer $\le x$ , is nondecreasing but not strictly increasing, because $\lfloor 0.2 \rfloor = \lfloor 0.7 \rfloor = 0$ .

Derivative sign tests

Definition Let $f$ be continuous on an interval $I$ and differentiable at every interior point of $I$ .

If $f'(x) > 0$ at every interior point, then $f$ is strictly increasing on $I$ .
If $f'(x) < 0$ at every interior point, then $f$ is strictly decreasing on $I$ .
If $f'(x) \ge 0$ at every interior point, then $f$ is nondecreasing on $I$ , and if $f'(x) \le 0$ , nonincreasing.

Each follows from the mean value theorem. For $x_1 < x_2$ in $I$ there is a $c$ in $(x_1, x_2)$ , an interior point of $I$ , with

f(x_2) - f(x_1) = f'(c)(x_2 - x_1),

and the sign of $f'(c)$ gives the sign of the difference. Conversely, if $f$ is differentiable and nondecreasing on an open interval, then $f' \ge 0$ there, but not necessarily $f' > 0$ .

In plain words Where the slope is positive, the function goes up. Where it is negative, the function goes down. To find the intervals, locate where $f'$ is zero or undefined and test the sign in between.

Example For $f(x) = x^3 - 3x$ the derivative is $f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)$ .

Interval	$(-\infty, -1)$	$(-1, 1)$	$(1, \infty)$
Sign of $f'$	$+$	$-$	$+$
Behavior of $f$	increasing	decreasing	increasing

Since $f$ is continuous, it is strictly increasing on $(-\infty, -1]$ and $[1, \infty)$ and strictly decreasing on $[-1, 1]$ .

Isolated zero derivatives

Definition Let $f$ be continuous on an interval $I$ and differentiable at its interior points with $f'(x) \ge 0$ there. If the set of points where $f'(x) = 0$ contains no open interval, then $f$ is strictly increasing on $I$ . In particular this holds when $f' > 0$ except at isolated points. The same statement with $f' \le 0$ gives strictly decreasing.

Here is why. By the sign test $f$ is nondecreasing on $I$ . Suppose $f(x_1) = f(x_2)$ for some $x_1 < x_2$ in $I$ . Every $x$ in $[x_1, x_2]$ then satisfies

f(x_1) \le f(x) \le f(x_2) = f(x_1),

so $f$ is constant on $[x_1, x_2]$ and $f' = 0$ on the whole open interval $(x_1, x_2)$ . That contradicts the assumption on the zero set, so $f(x_1) < f(x_2)$ whenever $x_1 < x_2$ .

In plain words A few flat instants do not stop a function from strictly increasing. Only a whole flat stretch does.

Example For $f(x) = x^3$ the derivative $f'(x) = 3x^2$ is positive except at $x = 0$ , so $f$ is strictly increasing on $\mathbb{R}$ . For $g(x) = x - \sin x$ the derivative is $g'(x) = 1 - \cos x \ge 0$ , with zeros only at the isolated points $x = 2k\pi$ , so $g$ is strictly increasing on $\mathbb{R}$ too.

Interval assumptions

Definition The derivative sign tests are statements about a single interval. On a domain made of several separate intervals, a sign of $f'$ gives monotonicity on each interval separately, not on their union. An endpoint of $I$ can be included whenever $f$ is continuous there, even if $f$ is not differentiable at that endpoint.

In plain words A gap in the domain lets the function jump back down. Always state the interval when you say a function is increasing.

Example For $f(x) = -\frac{1}{x}$ the derivative is

f'(x) = \frac{1}{x^2} > 0 \qquad \text{for all } x \neq 0.

So $f$ is strictly increasing on $(-\infty, 0)$ and on $(0, \infty)$ . It is not increasing on its whole domain, because $-1 < 1$ while $f(-1) = 1 > -1 = f(1)$ . In contrast, $\sqrt{x}$ is strictly increasing on $[0, \infty)$ , including the endpoint $0$ where it has no derivative.

First derivative test for extrema

Definition Let $c$ be a critical point of $f$ , let $f$ be continuous at $c$ , and let $f$ be differentiable on an open interval containing $c$ , except possibly at $c$ itself.

If $f'$ changes from positive to negative at $c$ , then $f$ has a local maximum at $c$ .
If $f'$ changes from negative to positive at $c$ , then $f$ has a local minimum at $c$ .
If $f'$ is positive on both sides of $c$ , or negative on both sides, then $f$ has no local extremum at $c$ .

Here “changes from positive to negative” means that for some $\delta > 0$ we have $f' > 0$ on $(c - \delta, c)$ and $f' < 0$ on $(c, c + \delta)$ . The other cases are read the same way.

In plain words Up then down gives a peak, and down then up gives a valley. If the function keeps going the same way, the critical point is just a pause.

Example Let $f(x) = x^{5/3} - 5x^{2/3}$ . For $x \neq 0$ ,

f'(x) = \frac{5}{3} x^{2/3} - \frac{10}{3} x^{-1/3} = \frac{5(x - 2)}{3 x^{1/3}},

and $f'(0)$ does not exist. The critical points are $0$ and $2$ .

Interval	$(-\infty, 0)$	$(0, 2)$	$(2, \infty)$
Sign of $f'$	$+$	$-$	$+$

So $f$ has a local maximum $f(0) = 0$ at the singular point $0$ and a local minimum $f(2) = -3 \sqrt[3]{4} \approx -4.762$ at the stationary point $2$ .

Check yourself

Find the intervals where $f(x) = x^4 - 2x^2$ is increasing or decreasing, and its local extrema.

Show the answer

The derivative $f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$ is negative on $(-\infty, -1)$ , positive on $(-1, 0)$ , negative on $(0, 1)$ and positive on $(1, \infty)$ . So $f$ decreases on $(-\infty, -1]$ , increases on $[-1, 0]$ , decreases on $[0, 1]$ and increases on $[1, \infty)$ . It has local minima $f(\pm 1) = -1$ and a local maximum $f(0) = 0$ .

Is $f(x) = x + \cos x$ strictly increasing on $\mathbb{R}$ ?

Show the answer

Yes. The derivative $f'(x) = 1 - \sin x$ is never negative, and it is zero only at the isolated points $x = \frac{\pi}{2} + 2k\pi$ , so $f$ is strictly increasing.

Since $\frac{d}{dx} \tan x = \sec^2 x > 0$ , is $\tan x$ increasing on its domain?

Show the answer

No. It is strictly increasing on each interval $\left(-\frac{\pi}{2} + k\pi, \frac{\pi}{2} + k\pi\right)$ , but the domain is not an interval. For example $\tan \frac{\pi}{4} = 1 > -1 = \tan \frac{3\pi}{4}$ .

Increasing and Decreasing Functions

Theory, part 71

Practice, part 72

Key terms

Strict and nonstrict monotonicity

Derivative sign tests

Isolated zero derivatives

Interval assumptions

First derivative test for extrema

Check yourself

How this lesson connects

Builds on

Used later in