Topic 5.3 Notes – Determining Intervals on Which a Function Is Increasing or Decreasing

How the First Derivative Controls Increasing and Decreasing

Remember what the derivative represents.

$f'(x)$ is the instantaneous rate of change, which is the slope of the tangent line to $f$.

That slope tells you everything about direction:

• If $f'(x) > 0$ → slope positive → function increasing
• If $f'(x) < 0$ → slope negative → function decreasing
• If $f'(x) = 0$ → horizontal tangent (possible turning point)

Think of it like driving:

• Positive slope = driving uphill → graph going up.
• Negative slope = driving downhill → graph going down.

Here’s how that looks on a sign diagram and the corresponding graph.

On the number line, the plus and minus signs show where $f'(x)$ is positive or negative. On the graph, those same intervals match where the function rises or falls.

Notice: we don’t look at whether $f(x)$ itself is positive or negative. Only the sign of $f'(x)$ matters.

Critical Numbers and Where Behavior Can Change

A function can only switch from increasing to decreasing (or vice versa) at specific x-values.

Those are critical numbers, defined as values in the domain of $f$ where:

1. $f'(x) = 0$
2. $f'(x)$ does not exist

Important detail: the value must be in the domain of $f$ to count as a critical number.

Also, if the function itself is undefined at some point, that still splits the number line into intervals. Behavior can’t “flow through” a discontinuity.

These special x-values divide the number line into intervals. On each interval, the derivative cannot change sign unless you cross one of these points.

That’s why testing one point per interval works.

The Sign Chart Method

This is the standard justification method you’ll use on quizzes and FRQs.

Step 1: Find $f'(x)$

Differentiate carefully. Algebra mistakes here ruin everything.

Step 2: Find critical numbers

Solve:

$$f'(x) = 0$$

and determine where $f'(x)$ is undefined (while $f(x)$ exists).

Step 3: Split the number line

Use those x-values (and any discontinuities of $f$) to form intervals.

Step 4: Test one value in each interval

Plug test values into $f'(x)$, not $f(x)$.

Example:

Let $f'(x) = (x-4)(x+1)$

Critical numbers: $x=4$, $x=-1$

Intervals:

• $(-\infty,-1)$
• $(-1,4)$
• $(4,\infty)$

Test signs:

• Pick $x=-2$: $(-)(-) = (+)$ → increasing
• Pick $x=0$: $(-)(+) = (-)$ → decreasing
• Pick $x=5$: $(+)(+) = (+)$ → increasing

So:

• Increasing on $(-\infty,-1)$ $\cup$ $(4,\infty)$
• Decreasing on $(-1,4)$

On a free-response question, you must explicitly connect sign to conclusion: “Since $f'(x) > 0$ on (a,b), $f$ is increasing on (a,b).”

Reading Behavior from a Graph of $f'$

Sometimes you’re given the graph of the derivative instead of a formula.

Here’s how to read it:

• Above x-axis → $f'(x) > 0$ → $f$ increasing
• Below x-axis → $f'(x) < 0$ → $f$ decreasing
• Where $f'$ crosses x-axis → possible change in direction

In the graph below, $f'(x)$ is positive for $x < -2$, negative on $(-2, 3)$, and positive again for $x > 3$. That means $f$ is increasing, then decreasing, then increasing.

If $f'$ touches zero but does not change sign, the function does not change direction there. That shows up a lot in multiple-choice questions.

Common Mistakes I See Every Year

• Plugging test points into $f(x)$ instead of $f'(x)$
• Forgetting to consider where the function is undefined
• Calling every critical number a max or min (you need a sign change)
• Giving single x-values instead of interval notation

The AP graders care about justification. The phrase “because $f'(x)$ is positive” is what earns the point.