Increasing and Decreasing Functions

The sign of the derivative on an interval tells us whether the function is increasing or decreasing on that interval. Positive derivative means the graph rises; negative derivative means it falls.

Condition	Conclusion
f'(x)>0 for all $x \in (a, b)$	$f$ is increasing on $[a, b]$
f'(x)<0 for all $x \in (a, b)$	$f$ is decreasing on $[a, b]$
f'(x)=0 for all $x \in (a, b)$	$f$ is constant on $[a, b]$

Review of Definitions

A function is increasing on an interval if its value increases as the independent variable increases. Mathematically, $f$ is increasing on $I$ if:

f (x_{1}) < f (x_{2}) whenever x_{1} < x_{2} in I .

Graph of an increasing function: as x increases, f(x) increases

A function is decreasing on $I$ if:

f (x_{1}) > f (x_{2}) whenever x_{1} < x_{2} in I .

Graph of a decreasing function: as x increases, f(x) decreases

A function is constant on $I$ if:

f (x_{1}) = f (x_{2}) for every x_{1} and x_{2} in I .

Graph of a constant function: f(x) has the same value throughout the interval

A function that is increasing or decreasing is called monotonic.

A function may be increasing on some intervals and decreasing on others. For example, the graph of $f (x) = 2 x^{3} - 12 x^{2} + 18 x - 2$ is shown below.

Graph of f(x) = 2x^3 - 12x^2 + 18x - 2 showing increasing and decreasing intervals — Graph of $f (x) = 2 x^{3} - 12 x^{2} + 18 x - 2$ . The function increases until $x = 1$ , decreases between $x = 1$ and $x = 3$ , and increases again after $x = 3$ .

The Increasing/Decreasing Test

If the tangent line makes a positive acute angle with the positive $x$ -axis, then \text{slope}=f'(x)>0 and the graph rises. If the tangent makes an obtuse angle, f'(x)<0 and the graph falls.

Graph showing positive slope (rising) and negative slope (falling) tangent lines

Increasing/Decreasing Test. Assume $f$ is continuous on the closed interval $[a, b]$ and is differentiable on the open interval $(a, b)$ .

(a) If f'(x)>0 for every $x$ in $(a, b)$ , then $f$ is increasing ( $↗$ ) on $[a, b]$ .

(b) If f'(x)<0 for every $x$ in $(a, b)$ , then $f$ is decreasing ( $↘$ ) on $[a, b]$ .

Proof

Suppose

x_{1}

and

x_{2}

are any two points in

[a, b]

with

x_{1} < x_{2}

. To prove (a), we must show

f (x_{1}) < f (x_{2})

. Applying the Mean Value Theorem on

[x_{1}, x_{2}]

: f(x_2)-f(x_1)=f'(c)(x_2-x_1) for some

c

between

x_{1}

and

x_{2}

. Since f'(c)>0 by hypothesis and

x_{2} - x_{1} > 0

f (x_{2}) - f (x_{1}) > 0 \Rightarrow f (x_{2}) > f (x_{1}) .

The proofs of (b) and (c) are similar.

Although stated for closed intervals, the theorem applies to any interval $I$ : if $f$ is continuous on $I$ and has a derivative at every interior point, then f'(x)>0 on the interior implies $f$ is increasing on $I$ , and f'(x)<0 implies $f$ is decreasing on $I$ .

Determining the Sign of the Derivative

To apply the Increasing/Decreasing Test, we need the sign of f'(x) on each subinterval. Two methods are available.

Method 1, Factor analysis. If f'(x) can be factored, determine the sign of each factor. A factor $(x - a)$ is negative when $x < a$ and positive when $x > a$ . For example:

f'(x)=(x+2)(x-1)(x-3).

The zeros of f' divide the real line into four intervals. In each, determine the sign of the product:

Interval	Sign of $(x + 2)$	Sign of $(x - 1)$	Sign of $(x - 3)$	Sign of f'
$x < - 2$	$-$	$-$	$-$	$-$
$- 2 < x < 1$	$+$	$-$	$-$	$+$
$1<x<3$	$+$	$+$	$-$	$-$
$x > 3$	$+$	$+$	$+$	$+$

Sign diagram for f'(x) = (x+2)(x-1)(x-3)

Method 2, Test numbers. Critical numbers divide the domain into subintervals. Pick one number from each subinterval and substitute it into the derivative. Since the sign of the derivative does not change within each subinterval, the sign of your result determines the sign of the derivative throughout that subinterval.

For example, to determine the sign of f'(x)=(x+2)(x-1)(x-3) on $(- 2, 1)$ , choose $x = 0$ : f'(0)=2(-1)(-3)=6>0, so f'(x)>0 on $(- 2, 1)$ .

Examples

Determine the intervals on which the following functions are increasing and decreasing.

(a) $f (x) = 2 x^{2} - 5 x - 7$

(b) $f (x) = x^{3}$

Solution

**(a)** f'(x)=4x-5=4\!\left(x-\frac{5}{4}\right).

Sign diagram for f'(x) = 4x - 5, negative left of 5/4 and positive right of 5/4

f'(x)<0 when

x < 5 / 4

, so

f

is decreasing on

(- \infty, 5 / 4]

. f'(x)>0 when

x > 5 / 4

, so

f

is increasing on

[5 / 4, \infty)

Graph of y = 2x^2 - 5x - 7 showing the parabola decreasing then increasing — Graph of $y = 2 x^{2} - 5 x - 7$ .

**(b)** f'(x)=3x^2.

Sign diagram for f'(x) = 3x^2, zero at x=0 and positive elsewhere

f'(x)\geq0 for all

x

, with equality only at

x = 0

. So

f

is increasing on

(- \infty, 0]

and on

[0, \infty)

, and therefore increasing on

(- \infty, \infty)

Graph of y = x^3, always increasing with a horizontal tangent at the origin — Graph of $y = x^{3}$ . The function is always increasing.

Find where $f (x) = x^{3} - 3 x^{2} + 1$ is increasing and where it is decreasing.

Solution

f'(x)=3x^2-6x=3x(x-2).

Sign diagram for f'(x) = 3x(x-2) with zeros at 0 and 2

f'(x)>0 on

(- \infty, 0)

and

(2, \infty)

, so

f

is increasing on

(- \infty, 0] \cup [2, + \infty)

. f'(x)<0 on $(0,2)$, so

f

is decreasing on $[0,2]$.

Graph of y = x^3 - 3x^2 + 1 with local max at x=0 and local min at x=2 — Graph of $y = x^{3} - 3 x^{2} + 1$ .

Determine where $f (x) = x^{2} e^{- x}$ is increasing and decreasing.

Solution

f'(x)=2xe^{-x}-x^2e^{-x}=e^{-x}(2x-x^2)=e^{-x}x(2-x). Since

e^{- x} > 0

for all

x

, the sign of f'(x) is determined by

x (2 - x)

Sign diagram showing f'(x) negative on (-inf,0) and (2,inf), positive on (0,2)

f

is decreasing on

(- \infty, 0] \cup [2, + \infty)

and increasing on $[0,2]$.

Graph of y = x^2 e^{-x} showing the function decreasing, then increasing, then decreasing again — Graph of $y = x^{2} e^{- x}$ .

Determine where $f (x) = \sqrt[3]{x} (x - 2)$ is increasing and decreasing.

Solution

Rewrite as

f (x) = x^{4 / 3} - 2 x^{1 / 3}

, then: \begin{aligned} f'(x) &= \frac{4}{3}x^{1/3}-\frac{2}{3}x^{-2/3} \\ &= \frac{2}{3}x^{-2/3}(2x-1) \\ &= \frac{2(2x-1)}{3x^{2/3}}. \end{aligned} So f'(1/2)=0 and f'(0) does not exist. Since

x^{2 / 3} \geq 0

, the sign of f'(x) is determined by

2 x - 1

Sign diagram for f'(x) with zeros at x=0 (undefined) and x=1/2

f

is decreasing on

(- \infty, 0]

and on

[0, 1 / 2]

, and therefore decreasing on

(- \infty, 1 / 2]

(since

f

is continuous at

x = 0

f

is increasing on

[1 / 2, + \infty)

Graph of y = x^(1/3)(x-2) showing the minimum at x = 1/2 — Graph of $y = x^{1 / 3} (x - 2)$ . Decreasing on $(- \infty, 1 / 2]$ , increasing on $[1 / 2, + \infty)$ .

When the Test May Fail

The Increasing/Decreasing Test may fail if $f$ is not continuous on $I$ .

For example, $f (x) = 1 / x$ has f'(x)=-1/x^2<0 for all $x \neq 0$ , but $f$ is not decreasing on $(- \infty, \infty)$ (since $f (- 1) = - 1$ and $f (1) = 1$ , with $- 1 < 1$ but $f (- 1) < f (1)$ ). The failure occurs because $f$ is discontinuous at $x = 0$ . However, $f$ is decreasing on each of $(- \infty, 0)$ and $(0, \infty)$ .

Graph of f(x) = 1/x showing it is not monotonic on all of (-inf, inf) due to the discontinuity at x=0 — $f (x) = 1 / x$ is not monotonic on $(- \infty, + \infty)$ because $x_{0} < x_{1}$ but $f (x_{0}) > f (x_{1})$ .

Frequently Asked Questions

Can a function be increasing even if the derivative equals zero at some points?

Yes.

f (x) = x^{3}

has f'(0)=0, but

f

is increasing on all of

(- \infty, \infty)

. A single point where the derivative is zero does not break the increasing behavior, as long as the derivative is non-negative throughout the interval and not identically zero on any subinterval.

Why does the Increasing/Decreasing Test require continuity on the closed interval?

Without continuity, a function might jump in a way that violates the increasing or decreasing behavior. The function

1 / x

is a classic example: f'(x)<0 everywhere it is defined, yet

f

is not globally decreasing due to the discontinuity at

x = 0

How do I find all intervals on which a function is increasing or decreasing?

Find all critical points (where f'=0 or f' is undefined). These divide the domain into subintervals. Use a test number in each subinterval to determine the sign of f', then apply the Increasing/Decreasing Test.

Review of Definitions

The Increasing/Decreasing Test

Determining the Sign of the Derivative

Examples

When the Test May Fail

More Advanced Topic

Frequently Asked Questions

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