Increasing, Decreasing and the First Derivative Test

Q: What is the first derivative test?

At a critical point, if f′ changes from positive to negative, f has a local maximum; from negative to positive, a local minimum; if the sign does not change, neither.

Q: Does f′(c) = 0 mean there is a maximum or minimum at c?

No. f(x) = x³ has f′(0) = 0 but keeps increasing through 0. The sign of f′ must change.

Q: Why do points where f′ is undefined matter?

A corner or cusp can be a maximum or minimum, as |x| has a minimum at 0 where f′ does not exist.

Quick answer

A function is increasing where f′ > 0 and decreasing where f′ < 0. The sign can change only at critical points, where f′ = 0 or f′ does not exist, so a sign chart built on the critical points shows the whole pattern. The first derivative test classifies each critical point: a change from + to − is a local maximum, from − to + a local minimum, and no change is neither. A zero derivative alone proves nothing: x³ has f′(0) = 0 and no extremum.

What you'll learn

Find critical points, including points where f′ does not exist
Build a sign chart for f′ and read off where f increases and decreases
Classify critical points with the first derivative test

The sign of f′ gives the direction

On an interval, the Mean Value Theorem gives:

if $f'(x) > 0$ throughout, $f$ is increasing there;
if $f'(x) < 0$ throughout, $f$ is decreasing there.

So the question “where does $f$ rise?” becomes “where is $f'$ positive?”

Critical points

A critical point of $f$ is a number $c$ in the domain of $f$ where

f'(c) = 0 \quad\text{or}\quad f'(c) \text{ does not exist}

A derivative that comes from a formula can change sign only where it is $0$ or undefined. So the critical points cut the number line into intervals on which $f'$ keeps one sign. Test one value in each interval, and you have the sign of $f'$ everywhere.

f(x) = x³ − 3x² − 9x + 5

Rising, falling and rising again

The first derivative test

At a critical point $c$ :

Sign of $f'$ around $c$	Conclusion
$+$ then $-$	local maximum at $c$
$-$ then $+$	local minimum at $c$
no change	no local extremum at $c$

Why the sign has to change

At a local maximum the function climbs up to the peak and descends after it, so $f'$ must be positive right before it and negative right after. A zero derivative is only a momentary pause.

$f(x) = x^3$ shows the difference: $f'(x) = 3x^2$ is $0$ at $x = 0$ , yet $f' \ge 0$ on both sides, so the function keeps rising straight through. A critical point is only a candidate; the change of sign is what makes it a maximum or minimum.

Example 1 — A full sign chart

Find the intervals of increase and decrease and the local extrema of $f(x) = x^3 - 3x^2 - 9x + 5$ .

$f'(x) = 3x^2 - 6x - 9 = 3(x + 1)(x - 3)$ , so the critical points are $-1$ and $3$ .

Interval	$x < -1$	$-1 < x < 3$	$x > 3$
test value	$-2$	$0$	$4$
sign of $f'$	$+$	$-$	$+$
$f$	increasing	decreasing	increasing

The sign goes $+$ to $-$ at $-1$ : a local maximum, $f(-1) = 10$ . It goes $-$ to $+$ at $3$ : a local minimum, $f(3) = -22$ .

Worked examples

Common mistakes

Practice problems

Find the critical points of $f(x) = x^3 - 12x$ .

Answer

$x = -2$ and $x = 2$

Full solution

$f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$ , which is $0$ at $\pm 2$ .
For $f(x) = x^3 - 12x$ , find where $f$ increases and decreases, and classify the critical points.

Answer

Increasing on $(-\infty, -2)$ and $(2, \infty)$ ; decreasing on $(-2, 2)$ . Local max at $-2$ , local min at $2$ .

Full solution

Test values $-3$ , $0$ , $3$ give $f'$ signs $+$ , $-$ , $+$ . Local max $f(-2) = 16$ ; local min $f(2) = -16$ .
Classify the critical point of $f(x) = (x - 1)^4$ .

Answer

A local minimum at $x = 1$

Full solution

$f'(x) = 4(x - 1)^3$ is negative for $x < 1$ and positive for $x > 1$ .
Does $f(x) = x^5$ have a local extremum at $0$ ?

Answer

No

Full solution

$f'(x) = 5x^4 \ge 0$ on both sides of $0$ , so there is no sign change.
Find the local extrema of $f(x) = x^4 - 2x^2$ .

Answer

Local max at $x = 0$ ( $f = 0$ ); local minima at $x = \pm 1$ ( $f = -1$ )

Full solution

$f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$ . Signs on $(-\infty, -1)$ , $(-1, 0)$ , $(0, 1)$ , $(1, \infty)$ : $-$ , $+$ , $-$ , $+$ .
Find the critical points of $f(x) = |x - 3|$ and classify them.

Answer

$x = 3$ , a local minimum

Full solution

$f'$ is $-1$ for $x < 3$ and $1$ for $x > 3$ , and does not exist at $3$ . The sign changes from $-$ to $+$ .
Where is $f(x) = \ln(x^2 + 1)$ increasing?

Answer

On $(0, \infty)$

Full solution

$f'(x) = \tfrac{2x}{x^2 + 1}$ , which has the sign of $x$ . So $f$ decreases on $(-\infty, 0)$ and increases on $(0, \infty)$ .
Find the local extrema of $f(x) = x + \tfrac{4}{x}$ .

Answer

Local max at $x = -2$ ( $f = -4$ ); local min at $x = 2$ ( $f = 4$ )

Full solution

$f'(x) = 1 - \tfrac{4}{x^2} = \tfrac{x^2 - 4}{x^2}$ , zero at $\pm 2$ . ( $x = 0$ is not in the domain, so it is not a critical point.) Signs: $+$ before $-2$ , $-$ between $-2$ and $0$ and between $0$ and $2$ , $+$ after $2$ .
The graph of $f'$ is positive on $(-\infty, 1)$ , negative on $(1, 4)$ and positive on $(4, \infty)$ . Where does $f$ have local extrema?

Answer

A local max at $x = 1$ and a local min at $x = 4$

Full solution

$f'$ changes from $+$ to $-$ at $1$ and from $-$ to $+$ at $4$ .
A student says $f(x) = x^3 + 3x$ has no critical points, so it has no local extrema, and also says $g(x) = x^3$ has an extremum at $0$ because $g'(0) = 0$ . Which claim is wrong, and why?

Hint

Check the sign of $g'$ on both sides of $0$ .

Answer

The second: $g'$ does not change sign at $0$ , so there is no extremum.

Full solution

The first claim is right: $f'(x) = 3x^2 + 3 > 0$ everywhere, so $f$ always increases and has no local extrema.

The second is wrong: $g'(x) = 3x^2 \ge 0$ on both sides of $0$ , so $g$ keeps increasing. A zero derivative without a sign change is not an extremum.

Frequently asked questions

What is a critical point?

A number c in the domain of f where f′(c) = 0 or f′(c) does not exist. Local extrema can occur only at critical points.

How do I find where a function is increasing?

Find the critical points, then test the sign of f′ in each interval between them. Where f′ > 0 the function increases; where f′ < 0 it decreases.

What is the first derivative test?