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Reading Time: 7 min
Last Updated: February 27, 2026
Main Ideas: 5
Reading Time: 7 min
Last Updated: February 27, 2026
Main Ideas: 5

Topic 5.4 Notes – Using the First Derivative Test to Determine Relative (Local) Extrema

Verified for 2027 AP® Calculus AB Exam
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You’ll use the sign of f′(x) f'(x) around critical points to justify how the original function behaves. This is one of the core ways derivatives explain function behavior.

1. What the First Derivative Test Says About Local Extrema

Everything starts with a critical point.

A critical point happens where:

  • f′(x)=0 f'(x) = 0 , or
  • f′(x) f'(x) is undefined (and f f exists there).

Now remember what the derivative tells you:

  • f′(x)>0 f'(x) > 0 → function is increasing
  • f′(x)<0 f'(x) < 0 → function is decreasing

The First Derivative Test uses this idea:

  • If f′ f' changes positive → negative, the function goes increasing → decreasing → local maximum
  • If f′ f' changes negative → positive, the function goes decreasing → increasing → local minimum
  • If f′ f' does not change sign, there is no local extremum

Think in motion:

  • Climbing then descending = hill (max)
  • Descending then climbing = valley (min)

Here’s what that sign change looks like on a sign chart:

First Derivative sign chart

At point a, the sign changes from positive to negative, so there is a local maximum. At b, the sign stays negative on both sides, so there is no extremum. At c, the sign changes from negative to positive, so there is a local minimum.

Notice that what matters is the sign change, not just whether f′(x)=0 f'(x) = 0 .

2. The Full Classification at a Critical Point

At any critical point, only four things can happen:

Sign of f′ f' Function BehaviorClassification
+→− + \to - Increasing → DecreasingRelative maximum
−→+ - \to + Decreasing → IncreasingRelative minimum
+→+ + \to + Increasing both sidesNot an extremum
−→− - \to - Decreasing both sidesNot an extremum

On tests, you are not guessing from the graph. You are justifying using derivative behavior.

A correct AP-style sentence sounds like:

Since f′(x) f'(x) changes from negative to positive at x=2 x = 2 , f f has a relative minimum at x=2 x = 2 .

That explanation earns the point.

3. How to Apply the First Derivative Test

Let’s walk through the process cleanly.

Step 1: Differentiate

Find f′(x) f'(x) .

Step 2: Find Critical Points

Solve:

  • f′(x)=0 f'(x) = 0
  • Where f′(x) f'(x) is undefined (but f f is defined)

These are the only possible locations of local extrema.

Step 3: Create a Sign Chart

Place critical points on a number line.

Then:

  • Either plug in test values
  • Or analyze signs from factored form

For example, if

f′(x)=(x−1)(x+3) f'(x) = (x-1)(x+3)

Critical points: x=1 x=1 , x=−3 x=-3

Test intervals:

  • Left of −3 → both factors negative → positive
  • Between −3 and 1 → one positive, one negative → negative
  • Right of 1 → both positive → positive

Sign pattern: +→−→+ + \to - \to +

So:

  • x=−3 x=-3 : +→− + \to - → local max
  • x=1 x=1 : −→+ - \to + → local min

That full reasoning is what earns FRQ credit.

4. Patterns That Show Up Often on Tests

Factored derivatives

When f′(x) f'(x) is already factored, use sign logic instead of plugging numbers. It’s faster and cleaner.

Repeated factors (multiplicity)

If

f′(x)=(x−2)2(x+1) f'(x) = (x-2)^2(x+1)

The squared factor has even multiplicity, so its sign does not change.

Rule:

  • Even power → no sign change
  • Odd power → sign changes

This lets you classify points quickly without test values.

When a graph of f′ f' is given

Very common on AP exams. Instead of an equation, you get the graph of f′(x) f'(x) and must decide what f f is doing.

Graph of f′(x) f'(x) showing sign changes

Notice in the graph above:

  • At x=−2 x=-2 , f′ f' crosses from positive to negative → local maximum of f f
  • At x=1 x=1 , f′ f' touches 0 but does not change sign → not an extremum
  • At x=3 x=3 , f′ f' crosses from negative to positive → local minimum of f f

You are analyzing sign change, not height of the graph.

5. Common Mistakes That Cost Points

  • Saying “f′(a)=0 f'(a)=0 , so there’s a max/min.” That is incomplete. You must show a sign change.
  • Checking only one side of the critical point.
  • Mixing up max and min. If you forget, picture walking along the graph.
  • Forgetting this only gives local extrema. Absolute extrema require checking endpoints on closed intervals.

The entire idea of this topic is simple but powerful:

The behavior of the derivative determines the behavior of the function.

Key Takeaways

A critical point occurs where f′(x)=0 f'(x)=0 or f′(x) f'(x) is undefined and f f exists there.
A local maximum happens where f′ f' changes from positive to negative.
A local minimum happens where f′ f' changes from negative to positive.
If f′ f' does not change sign at a critical point, there is no local extremum.
On FRQs, you must explicitly reference the sign change of f′(x) f'(x) to justify your answer.

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Notes

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