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

What the Second Derivative Test Says

First, quick grounding.

A critical point occurs where:

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

The second derivative tells you about concavity:

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

Here’s the key idea:

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

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

Why this makes sense visually:

On the left, the graph is concave up. Notice how the slope is negative before the vertex and positive after it, so the flat point is a minimum.

On the right, the graph is concave down. The slope is positive before the vertex and negative after it, so the flat point is a maximum.

This is often faster than building a full sign chart for $f'$.

How to Use the Second Derivative Test

Let’s walk through the mechanics with a concrete example.

Suppose
$$f(x) = x^3 - 6x^2 + 9x$$

1. Find $f'(x)$

$$f'(x) = 3x^2 - 12x + 9$$

2. Find critical points

Set $f'(x)=0$:

$$3x^2 - 12x + 9 = 0$$

Divide by 3:

$$x^2 - 4x + 3 = 0$$

Factor:

$$(x-1)(x-3)=0$$

Critical points: $x=1$, $x=3$

3. Find $f''(x)$

$$f''(x) = 6x - 12$$

4. Evaluate $f''(c)$

• $f''(1) = 6(1) - 12 = -6$ → negative → local max
• $f''(3) = 6(3) - 12 = 6$ → positive → local min

On a free-response question, don’t just say “max” or “min.” Write something like:

“Since $f''(1) < 0$, the function is concave down at $x=1$, so $f$ has a local maximum there.”

That justification language earns the point.

When the Test Is Inconclusive

The test only works when:

• $f'(c)=0$, and
• $f''(c) \neq 0$

If $f''(c)=0$, you cannot conclude anything.

Example:

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

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

The test fails. But $x=0$ is actually a local minimum. You’d need the First Derivative Test to confirm.

Important:
$f''(c)=0$ does not automatically mean inflection point. Concavity must change sign for that.

On multiple choice, they love giving a point where $f''(c)=0$ and seeing if you jump to the wrong conclusion.

Connecting Local and Absolute Extrema

Here’s a powerful theorem you’re expected to use in explanations.

If a function is:

• Continuous on an interval, and
• Has exactly one critical point in that interval, and
• That point is a local extremum,

then that point is also the absolute extremum on that interval.

Why this works:

If there’s only one place where the slope is zero, there’s no other peak or valley competing with it.

This shows up a lot in optimization FRQs. You’ll often see wording like:

“Explain why this value gives the absolute maximum.”

A strong response mentions:

• continuity,
• one critical point,
• and that it is a local maximum,
• therefore it must be absolute.

If endpoints are involved, remember that this theorem applies to the interval given. Always pay attention to whether the interval is open or closed.