The Second Derivative Test Explained (With Examples)

You already know that setting the first derivative to zero helps you find critical points—the places where a function temporarily flattens out. But once you find a critical point, how do you know if it’s the peak of a mountain (a local maximum) or the bottom of a valley (a local minimum)?

The First Derivative Test works by checking the slope on either side of the point. But checking multiple test numbers can be tedious. Enter The Second Derivative Test. By simply taking the derivative of the derivative, you can analyze the concavity (how the graph bends) to instantly classify your critical points.

Here is exactly how the Second Derivative Test works, the rules to remember it by, and five examples showing when to use it (and what to do when it fails).

What is the Second Derivative?

If the first derivative $f'(x)$ tells you the slope of a function, the second derivative $f''(x)$ tells you the rate of change of the slope. In plain English, it tells you how the curve is bending.

If $f''(x) > 0$ , the slopes are increasing. The graph bends upward like a smile. This is called concave up.
If $f''(x) < 0$ , the slopes are decreasing. The graph bends downward like a frown. This is called concave down.

Before we calculate it, you can experiment with how the shape of a curve determines its second derivative using this interactive graph:

The Golden Rule of the Test: If you are standing at a critical point where the slope is zero, and the graph is smiling (concave up), you must be at the bottom of a valley (local minimum). If the graph is frowning (concave down), you must be at the peak of a mountain (local maximum).

The Second Derivative Test Rules

To classify a critical point $x = c$ , follow these steps:

Find the critical points by setting $f'(x) = 0$ and solving for $x$ .
Find the second derivative, $f''(x)$ .
Plug your critical point(s) into $f''(x)$ and check the sign:

If $f''(c) > 0 \implies$ Concave Up $\implies$ Local Minimum
If $f''(c) < 0 \implies$ Concave Down $\implies$ Local Maximum
If $f''(c) = 0 \implies$ Inconclusive (You must go back and use the First Derivative Test).

5 Examples of the Second Derivative Test

Let’s see how this rule cuts down on algebra and speeds up optimization problems.

Example 1: The Simple Quadratic

Let’s find and classify the optimum of $f(x) = x^2 - 6x + 5$ .

Step 1: Find the first derivative and set to zero.

$f'(x) = 2x - 6$
$2x - 6 = 0 \implies x = 3$

Our critical point is $x = 3$ .

Step 2: Find the second derivative.

$f''(x) = 2$
Step 3: Plug in $x = 3$ .
Because $f''(x)$ is a constant 2, $f''(3) = 2$ .
Since $2 > 0$ , the graph is concave up (smiling).
Therefore, at $x = 3$ , there is a local minimum.

Example 2: The Cubic Curve

Higher-powered functions have varying concavity.

$f(x) = x^3 - 12x$
First derivative:

$f'(x) = 3x^2 - 12$
$3x^2 - 12 = 0 \implies x^2 = 4 \implies x = 2, x = -2$
Second derivative:

$f''(x) = 6x$
Now, test both critical points in the second derivative:
Test $x = -2$ : $f''(-2) = 6(-2) = -12$ . Since $-12 < 0$ , it’s concave down (local maximum).
Test $x = 2$ : $f''(2) = 6(2) = 12$ . Since $12 > 0$ , it’s concave up (local minimum).

Example 3: When the Test Fails (Inconclusive)

The Second Derivative Test is fantastic, but it has a blind spot. What happens when $f''(c) = 0$ ?

$f(x) = x^4$
First derivative:

$f'(x) = 4x^3$
$4x^3 = 0 \implies x = 0$
Second derivative:

$f''(x) = 12x^2$
Plug in our critical point: $f''(0) = 0$ .

The test is inconclusive. It tells us absolutely nothing. To find out if $x=0$ is a maximum, minimum, or neither, you must use the First Derivative Test. (By checking slopes around 0, you would find it falls then rises, meaning it is actually a local minimum).

Example 4: Trigonometry

Because trig functions cycle endlessly, the Second Derivative Test is often much faster than testing intervals on a number line. Let’s look at $f(x) = \cos(x)$ on the interval $[0, 2\pi]$ .

First derivative:

$f'(x) = -\sin(x)$
$-\sin(x) = 0 \implies x = \pi$

(ignoring the endpoints for this test).

Second derivative:

$f''(x) = -\cos(x)$
Test the critical point:

$f''(\pi) = -\cos(\pi) = -(-1) = 1$

Since $1 > 0$ , the curve is concave up. There is a local minimum at $x = \pi$ .

Example 5: Rational Powers

Let’s look at a function with fractional coefficients:

$f(x) = \frac{4}{3}x^3 - 16x$
First derivative:

$f'(x) = 4x^2 - 16$
$4x^2 - 16 = 0 \implies x^2 = 4 \implies x = 2 \text{ and } x = -2$
Second derivative:

$f''(x) = 8x$
Test $x = 2$ : $f''(2) = 16$ . Positive, so concave up $\implies$ Local minimum.
Test $x = -2$ : $f''(-2) = -16$ . Negative, so concave down $\implies$ Local maximum.

Why This Rule Matters

The Second Derivative Test is a massive time-saver. By skipping the number line and jumping straight to the second derivative, you can often classify critical points in your head in seconds.

Beyond saving time on calculus tests, the second derivative represents acceleration in physics. In economics, it represents the rate of diminishing returns. Knowing whether a curve is bending upward or downward is the key to understanding whether a moving object is speeding up or whether a business strategy is losing its momentum.

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The Second Derivative Test Explained (With Examples)

What is the Second Derivative?

The Second Derivative Test Rules