Determine intervals of increase and decrease

The learner interprets the sign of a function’s derivative: where $f'(x)>0$ , $f$ increases, and where $f'(x)<0$ , $f$ decreases. By identifying critical numbers and points where the derivative is undefined, partitioning the domain, and constructing a derivative sign chart, the learner determines intervals of monotonic behavior and distinguishes changes in sign that indicate local extrema from points where $f'(x)=0$ without a change. The treatment is limited to single-variable functions and standard algebraic or graphical representations.

Detailed Explanation: 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.

Learn by doing: Determine intervals of increase and decrease

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Critical Points - Derivative Chart to Intervals of Increase or Decrease

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