Topic 5.7 Notes – Using the Second Derivative Test to Determine Extrema

1. The Second Derivative Test

A critical point occurs at $x = c$ where:

• $f'(c) = 0$, or
• $f'(c)$ does not exist (but $f$ does).

The second derivative tells you about concavity:

• $f''(x) > 0$ → concave up
• $f''(x) < 0$ → concave down

Here’s the connection that makes this test work:

If $f'(c) = 0$ and $f''(c)$ exists, then:

• If $f''(c) > 0$ → local minimum
• If $f''(c) < 0$ → local maximum
• If $f''(c) = 0$ → inconclusive

Why? Because concavity tells you the shape at that point. In the figure below, the left graph is concave up with a minimum at its vertex, and the right graph is concave down with a maximum at its vertex.

• Concave up looks like a bowl → bottom → minimum.
• Concave down looks like a hill → top → maximum.

This is a faster alternative to the First Derivative Test. Instead of checking sign changes of $f'$, you check the sign of $f''$ at the critical point.

2. The Full Procedure

When a problem says “Use the Second Derivative Test,” here’s the flow:

1. Find $f'(x)$.
2. Find critical points by solving $f'(x) = 0$ (and checking where it doesn’t exist, if relevant).
3. Find $f''(x)$.
4. Plug each critical point into $f''(x)$.
5. Classify using the sign of $f''(c)$.

Quick example:

Let $f(x) = x^3 - 6x^2 + 5x$.

• $f'(x) = 3x^2 - 12x + 5$
• Solve $3x^2 - 12x + 5 = 0$ → gives critical points (you’d solve with quadratic formula).
• Then compute $f''(x) = 6x - 12$.
• Plug each critical value into $f''$ and classify by sign.

On a no-calculator section, algebra accuracy matters. A small mistake in $f''(x)$ flips max to min.

Also remember:
You must have $f'(c) = 0$. If that’s not true, you cannot use this test.

3. When the Test Is Inconclusive

If:

• $f''(c) = 0$, or
• $f''(c)$ does not exist

the test gives you no classification.

This does not mean there isn’t a max or min.

Example idea:
For $f(x) = x^4$,

• $f'(x) = 4x^3$
• Critical point at $x = 0$
• $f''(x) = 12x^2$
• $f''(0) = 0$

The test fails. But $x=0$ is clearly a minimum. You can see the flat bottom at the origin in the graph below.

Graph of $f(x) = x^4$

When this happens, switch to the First Derivative Test and check if $f'(x)$ changes sign around the point.

The second derivative only checks concavity at the point. The first derivative checks behavior around it.

4. Local vs Absolute Extrema on an Interval

Here’s a statement the AP exam loves in justification questions:

If a function is continuous on an interval and has exactly one critical point on that interval, and that point is a local extremum, then it is also the absolute extremum on that interval.

Why this works:

• A continuous function can only change direction at critical points.
• If there’s only one, there’s no second peak or valley competing.

On an FRQ, you must actually state:

• The function is continuous on the interval.
• There is exactly one critical point.
• Therefore, the local extremum is absolute.

If you skip continuity, you can lose justification points.

5. Common Mistakes That Cost Points

• Plugging a value into $f''(x)$ without verifying $f'(c) = 0$.
• Mixing up the signs. Positive second derivative means minimum.
• Calling something absolute without mentioning continuity and “only one critical point.”
• Forgetting that endpoints matter when a problem asks for absolute extrema on a closed interval. The Second Derivative Test does not check endpoints.