Skill: Determine intervals of increase and decrease

To determine where a function increases or decreases, use the sign of its derivative:

• If $f'(x)>0$, then $f$ is increasing.
• If $f'(x)<0$, then $f$ is decreasing.

Example

Determine the intervals where

$$f(x)=x^4-\frac{8}{3}x^3$$

is increasing or decreasing.

1. Find the derivative

Differentiate:

$$f'(x)=4x^3-8x^2$$

Factor the derivative:

$$f'(x)=4x^2(x-2)$$

2. Find the critical numbers

Critical numbers occur where $f'(x)=0$ or where $f'(x)$ is undefined.

Set the derivative equal to zero:

$$4x^2(x-2)=0$$

So,

$$x=0 \quad \text{or} \quad x=2$$

The derivative is a polynomial, so it is defined for every real number. Therefore, the critical numbers are $0$ and $2$.

These numbers divide the number line into three intervals:

$$(-\infty,0), \qquad (0,2), \qquad (2,\infty)$$

3. Make a derivative sign chart

Test one number from each interval.

Interval	Test value	Sign of $4x^2(x-2)$	Behavior of $f$
$(-\infty,0)$	$-1$	Negative	Decreasing
$(0,2)$	$1$	Negative	Decreasing
$(2,\infty)$	$3$	Positive	Increasing

For example, at $x=1$,

$$f'(1)=4(1)^2(1-2)=-4<0$$

so $f$ is decreasing on $(0,2)$.

4. State the answer

The function is:

• Decreasing on $(-\infty,0)$ and $(0,2)$
• Increasing on $(2,\infty)$

Notice that $f'(x)=0$ at $x=0$, but the derivative is negative on both sides of $0$. Since the sign does not change there, the function does not change from increasing to decreasing or vice versa at $x=0$.

At $x=2$, the derivative changes from negative to positive, so the function changes from decreasing to increasing. Therefore, $x=2$ is a local minimum.