Determining Intervals on Which a Function Is Increasing or Decreasing

AP Calculus · Analytical Applications of Differentiation

Intervals of Increase and Decrease

The sign of $f'$ decides everything. Its size decides nothing.

key idea$f'\gt 0\Rightarrow f\text{ increasing};\qquad f'\lt 0\Rightarrow f\text{ decreasing}$

Reading the graph

Increase and decrease are decided entirely by the sign of $f'$, never by its size. A large positive derivative and a tiny positive one both mean the same thing: the function is going up.

$f'(x)=3x^{2}-12$

The derivative $f'(x)=3x^{2}-12$. Where this curve sits above the axis $f$ is increasing; where it sits below, $f$ is decreasing. The two zeros are the only places the behaviour can switch.

sign chart — $f'(x)=3(x-2)(x+2)$

	$x\lt -2$	$-2\lt x\lt 2$	$x\gt 2$
$(x+2)$	−	+	+
$(x-2)$	−	−	+
$f'$	+	−	+
verdict	$f$ increasing	$f$ decreasing	$f$ increasing

the four steps

Find every critical point — where $f'=0$ or $f'$ is undefined.
Those points divide the number line into test intervals.
Test one point in each interval; the sign holds across the whole interval.
Report intervals in terms of $x$, using the notation the question uses.

Worked examples

EX 1

Build the sign chart

$f(x)=x^{3}-12x$

Find where $f(x)=x^{3}-12x$ is increasing and decreasing.

Differentiate

$f'(x)=3x^{2}-12=3(x-2)(x+2)$.

Find the critical points

$f'=0$ at $x=\pm2$; $f'$ is defined everywhere else.

Test each interval

$f'(-3)=15\gt 0$; $f'(0)=-12\lt 0$; $f'(3)=15\gt 0$.

Translate to $f$

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

On the notation

Open intervals are the safe convention. Some texts include the endpoints; open intervals are never marked wrong.

EX 2

A factor that never changes sign

$f'(x)=x^{2}(x-3)$

A function has $f'(x)=x^{2}(x-3)$. Where is $f$ increasing?

Find the critical points

$f'=0$ at $x=0$ and $x=3$.

Examine each factor

$x^{2}\ge0$ always — it never changes sign. Only $(x-3)$ can flip.

Deduce the signs

For $x\lt 3$ (and $x\neq0$): positive times negative $=$ negative. For $x\gt 3$: positive.

Answer

$f$ is increasing on $(3,\infty)$ and decreasing on $(-\infty,3)$, with a momentary flat spot at $x=0$.

The shortcut

An even-power factor contributes a zero but no sign change. Spotting that saves testing an interval that cannot differ.

EX 3

Where $f'$ is undefined

$f(x)=x^{2/3}(x-5)$

Find the intervals of increase for $f(x)=x^{2/3}(x-5)$.

Differentiate with the product rule

$f'(x)=\tfrac23 x^{-1/3}(x-5)+x^{2/3}$.

Combine over a common denominator

$f'(x)=\dfrac{2(x-5)+3x}{3x^{1/3}}=\dfrac{5x-10}{3x^{1/3}}$.

Find all critical points

Numerator zero at $x=2$; denominator zero at $x=0$, where $f'$ is undefined but $f(0)=0$ exists. Both are critical.

Test the three intervals

$f'(-1)=\dfrac{-15}{-3}\gt 0$; $f'(1)=\dfrac{-5}{3}\lt 0$; $f'(8)=\dfrac{30}{6}\gt 0$.

Answer

Increasing on $(-\infty,0)$ and $(2,\infty)$; decreasing on $(0,2)$. Missing $x=0$ would merge two intervals and give the wrong answer.

EX 4

From the graph of $f'$ alone

graph → conclusion

Using the figure above, state where $f$ increases — without any formula for $f$.

Read where the curve is above the axis

$f'\gt 0$ for $x\lt -2$ and for $x\gt 2$.

Read where it is below

$f'\lt 0$ on $(-2,2)$.

Translate

$f$ increases on $(-\infty,-2)$ and $(2,\infty)$, and decreases on $(-2,2)$.

Resist reading heights as values

The graph shown is $f'$, so its $y$-values are slopes of $f$, not heights of $f$. The minimum of $f'$ at $x=0$ is where $f$ is steepest downward, not where $f$ is smallest.

Watch out

Sign, not size

$f'=0.001$ and $f'=1000$ both mean $f$ is increasing. Magnitude describes how fast, never whether.

Include points where $f'$ is undefined

In Example 3 the interval breaks at $x=0$ even though $f'$ never equals zero there. Testing only $f'=0$ merges intervals incorrectly.

Even-power factors never flip the sign

$x^{2}$, $(x-3)^{4}$ contribute zeros without sign changes. Recognise them and skip the redundant test.

A graph of $f'$ is not a graph of $f$

Its heights are slopes. The lowest point on $f'$ marks the steepest descent of $f$, not a minimum of $f$.

Course map. Topic 5.3 · LO FUN-4.A — justify conclusions about the behaviour of a function based on the behaviour of its derivatives (EK FUN-4.A.1, FUN-4.A.2).