5.3 Determining Intervals on Which a Function Is Increasing or Decreasing

5.3 Determining Intervals on Which a Function Is Increasing or Decreasing

A derivative can reveal whether a graph is climbing or falling without requiring us to see the graph itself. The essential question is simple: what does the sign of $f'(x)$ say about the behavior of $f(x)$?

The sign of the derivative controls direction

If $f'(x)>0$ throughout an open interval, then $f$ is increasing on that interval. If $f'(x)<0$ throughout an open interval, then $f$ is decreasing there. The derivative acts like a slope meter: positive slopes push the graph upward as $x$ increases, while negative slopes pull it downward.

Derivative sign rule

$$f'(x)>0 \Longrightarrow f \text{ is increasing}$$

$$f'(x)<0 \Longrightarrow f \text{ is decreasing}$$

The word throughout matters. To justify that $f$ is increasing on $(a,b)$, you must establish that $f'(x)>0$ for every $x$ in $(a,b)$, not merely calculate one positive derivative value.

From a derivative formula to intervals

Use the critical numbers identified earlier—where $f'(c)=0$ or does not exist—to divide the domain into intervals for a sign chart. Then determine the sign of $f'$ on each interval. One convenient test value is enough when the sign cannot change inside that interval, but the conclusion must refer to the entire interval.

Consider

$$f'(x)=5+3x.$$

Set the derivative equal to zero:

$$5+3x=0 \quad\Longrightarrow\quad x=-\frac{5}{3}.$$

This divides the real line into

$$\left(-\infty,-\frac{5}{3}\right) \quad\text{and}\quad \left(-\frac{5}{3},\infty\right).$$

Choose one test value from each interval:

Interval	Test value	Sign of $f'(x)=5+3x$	Behavior of $f$
$\left(-\infty,-\frac{5}{3}\right)$	$x=-3$	$5+3(-3)=-4<0$	Decreasing
$\left(-\frac{5}{3},\infty\right)$	$x=0$	$5+3(0)=5>0$	Increasing

Therefore, $f$ is decreasing on $\left(-\infty,-\frac{5}{3}\right)$ because $f'(x)<0$ there, and increasing on $\left(-\frac{5}{3},\infty\right)$ because $f'(x)>0$ there. Notice how evaluating at $x=-3$ is not the final conclusion; it supplies evidence for the sign on the whole interval containing $-3$.

Worked contextual example

Suppose $P(t)$ represents the population of a microorganism colony, and its derivative satisfies

$$P'(t)=(t-2)(t-5), \qquad 0\le t\le 7.$$

The critical numbers in the domain are $t=2$ and $t=5$. Test one value in each resulting interval:

Time interval	Test value	Sign of $P'(t)$	Population behavior
$(0,2)$	$t=1$	$(1-2)(1-5)>0$	Increasing
$(2,5)$	$t=3$	$(3-2)(3-5)<0$	Decreasing
$(5,7)$	$t=6$	$(6-2)(6-5)>0$	Increasing

Therefore, $P$ increases on $(0,2)$, decreases on $(2,5)$, and increases on $(5,7)$. At $t=2$, the sign changes from positive to negative, so the colony has a relative maximum. At $t=5$, the sign changes from negative to positive, so the colony has a relative minimum.

The interval conclusions are this topic's work. The sign changes at $t=2$ and $t=5$ are what Topic 5.4's First Derivative Test reads to classify those points as relative extrema.

These conclusions demonstrate FUN-4.A: Justify conclusions about the behavior of a function based on the behavior of its derivatives: the first derivative shows the intervals on which a function is increasing or decreasing. A complete justification gives the sign of $f'$ on the whole interval as the reason, not merely the answer.

Retrieval check: Suppose $f'(x)=(x-1)(x+4)$. Which critical numbers divide the real line, and on which intervals is $f$ increasing? Explain using the sign of $f'$. (Here $f'>0$ on $(-\infty,-4)$ and on $(1,\infty)$, so $f$ is increasing there; $f'<0$ on $(-4,1)$, so $f$ is decreasing there.)