Second Derivative Test: Rules, Examples & When It Fails

What Is The Second Derivative Test?

The Second Derivative Test is a rule for deciding whether a critical point of a function is a local maximum, a local minimum, or neither, by looking at the sign of the second derivative there. It applies to a function $f$ that is twice differentiable near the point.

The setup is always the same. First find the critical points by solving $f'(x) = 0$. At each such point $c$, compute $f''(c)$ and read off the result:

$$f''(c) < 0 ;\Rightarrow; \text{local maximum at } c$$

$$f''(c) > 0 ;\Rightarrow; \text{local minimum at } c$$

$$f''(c) = 0 ;\Rightarrow; \text{test inconclusive}$$

The sign of $f''(c)$ is a measure of concavity. A negative value means the curve bends downward like the top of a hill, so a flat spot there must be a peak. A positive value means the curve bends upward like the floor of a valley, so a flat spot there must be a trough. When $f''(c) = 0$ the curve is momentarily flat in its bending too, and that single number can no longer tell peaks from valleys.

For the broader family of shape tools this belongs to, see applications of derivatives.

How Do You Apply The Second Derivative Test? Step By Step

The test is a short, fixed procedure. Every worked example below follows these four steps.

• Step 1. Differentiate to get $f'(x)$, set $f'(x) = 0$, and solve for the critical points. These are the only candidates for a local maximum or minimum.

• Step 2. Differentiate again to get $f''(x)$.

• Step 3. Substitute each critical point into $f''(x)$ and check the sign.

• Step 4. Classify: negative gives a maximum, positive gives a minimum, zero sends you to the first derivative test.

A word on Step 1: only points where the tangent is horizontal count here. The Second Derivative Test says nothing about points where $f'(c) \neq 0$, and it says nothing at points where $f'$ does not exist, such as a sharp corner. Those stationary points with $f'(c) = 0$ are the whole domain of the test.

Why Does The Sign Of The Second Derivative Decide It?

The second derivative measures how the slope is changing, and that is exactly concavity. Pairing the algebra with the picture makes the rule impossible to forget.

• Positive second derivative means concave up. If $f''(c) > 0$, the slope $f'$ is increasing as it passes through $c$. At a critical point the slope is zero, so it moves from negative (falling) to positive (rising). A curve that falls then rises has a valley, which is a local minimum.

• Negative second derivative means concave down. If $f''(c) < 0$, the slope $f'$ is decreasing through $c$, moving from positive (rising) to negative (falling). A curve that rises then falls has a peak, which is a local maximum.

• Zero second derivative hides the shape. If $f''(c) = 0$, the slope might still be changing sign, or it might not be. The second derivative alone cannot see which, so the test goes quiet.

This is why the test is really the second derivative reading concavity at a horizontal tangent. The geometry and the algebra are the same statement: the way a curve bends at a flat spot decides whether that spot is a high point or a low point.

How Do You Use The Second Derivative Test? Worked Examples

Each example follows the four steps, and each maximum or minimum value is computed in full.

Example 1: A cubic with a clear max and min.

Classify the critical points of $f(x) = x^3 - 3x$.

Step 1, find the critical points:

$$f'(x) = 3x^2 - 3 = 0 ;\Rightarrow; x^2 = 1 ;\Rightarrow; x = \pm 1$$

Step 2, the second derivative:

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

Step 3 and 4, test each point:

$$f''(-1) = 6(-1) = -6 < 0 ;\Rightarrow; \text{local maximum at } x = -1$$

$$f''(1) = 6(1) = 6 > 0 ;\Rightarrow; \text{local minimum at } x = 1$$

The values are $f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$ and $f(1) = 1 - 3 = -2$.

Final answer: local maximum $(-1, 2)$ and local minimum $(1, -2)$.

Example 2: A richer cubic.

Classify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 4$.

$$f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) = 0 ;\Rightarrow; x = 2 \text{ or } x = -1$$

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

$$f''(-1) = -18 < 0 ;\Rightarrow; \text{local maximum at } x = -1, \quad f(-1) = 11$$

$$f''(2) = 18 > 0 ;\Rightarrow; \text{local minimum at } x = 2, \quad f(2) = -16$$

Final answer: local maximum $(-1, 11)$ and local minimum $(2, -16)$.

Example 3: An inconclusive case that turns out to be a minimum.

Classify the critical point of $f(x) = x^4$.

$$f'(x) = 4x^3 = 0 ;\Rightarrow; x = 0, \qquad f''(x) = 12x^2, \qquad f''(0) = 0$$

The Second Derivative Test is inconclusive here. Fall back on the first derivative test: $f'(x) = 4x^3$ is negative for $x < 0$ and positive for $x > 0$, so the slope changes from falling to rising. That sign change is a local minimum.

Final answer: the test is inconclusive, but $x = 0$ is a local minimum $(0, 0)$.

Example 4: An inconclusive case that is not an extremum at all.

Classify the critical point of $f(x) = x^3$.

$$f'(x) = 3x^2 = 0 ;\Rightarrow; x = 0, \qquad f''(x) = 6x, \qquad f''(0) = 0$$

Again the test is inconclusive, and this time the answer is different. The first derivative $f'(x) = 3x^2$ is positive on both sides of $0$ and never changes sign, so the slope does not switch direction. There is no peak and no valley; the curve simply flattens for an instant and keeps rising through an inflection point.

Final answer: the test is inconclusive, and $x = 0$ is neither a maximum nor a minimum.

Examples 3 and 4 are the heart of the topic. Both have $f''(0) = 0$, yet one is a local minimum and the other is an inflection point that is no extremum at all. That is precisely why $f''(c) = 0$ can never be a verdict on its own.

What Do The Three Outcomes Look Like Together?

Table: The Second Derivative Test at a critical point $c$ where $f'(c) = 0$.

The first two rows are the reason the test is popular: one substitution settles the question. The third row is the reason it is not the whole story, and it is where most marks are lost.

Second Derivative Test Or First Derivative Test — Which Should You Use?

Both tests classify critical points, and they never disagree when both give an answer. They differ in what they need and when they work.

Table: Comparing the two tests for classifying a critical point.

Reach for the first derivative test when the second derivative is awkward to compute, when it comes out zero, or when the critical point is a sharp corner. Reach for the Second Derivative Test when $f''$ is quick to find and nonzero, which is the common case for polynomials and many standard functions. The two together cover every situation, which is why a full treatment of maxima and minima teaches both.

Why Does The Second Derivative Test Fail When It Fails?

The failure is not a flaw in the rule; it is the rule reaching the edge of what a single number can tell you.

• A zero says "flat," not "which shape." The value $f''(c)$ measures curvature. When it is zero, the curve is not bending at that instant, so there is no downward or upward cup to read a maximum or minimum from.

• Higher-order behaviour takes over. For $f(x) = x^4$ the fourth derivative is what finally makes the point a minimum; for $f(x) = x^3$ the odd power makes it an inflection. The second derivative is simply blind to that finer detail.

• The first derivative test still sees everything. Checking the sign of $f'$ on each side of $c$ reads the actual rise-and-fall of the graph, so it succeeds exactly where the Second Derivative Test goes quiet.

Seen this way, "inconclusive" is an honest report, not a dead end. It means the fast tool has run out of information and the reliable tool, the sign chart of $f'$, needs one more line of work.

What Is The History Of The Second Derivative Test?

The idea of using slopes to hunt for high and low points is older than calculus notation itself, and the curvature refinement came later.

Two later figures supplied the curvature half:

• Colin Maclaurin (1698–1746, Scotland) set out, in his Treatise of Fluxions (1742), how the sign of higher derivatives distinguishes maxima, minima, and points where the simple test fails.

• Joseph-Louis Lagrange (1736–1813, Italy and France) placed the higher-derivative criteria on a firmer footing, generalising the check to the first nonzero derivative when the second derivative vanishes.

Where Is The Second Derivative Test Used In The Real World?

Anywhere a "best" or "worst" value hides at a flat spot, the test is the quick way to confirm which it is.

• Physics and motion: a particle's velocity is the derivative of position and its acceleration is the second derivative, so the sign of acceleration at a moment of zero velocity says whether the particle is at a farthest or nearest point.

• Economics: profit is maximised where marginal profit is zero, and a negative second derivative confirms the flat spot is a peak rather than a costly trough.

• Engineering design: minimising material, stress, or energy reduces to finding a critical point and checking that the second derivative is positive, guaranteeing a genuine minimum.

• Optimisation problems: whenever a real quantity is squeezed to its smallest or stretched to its largest, the Second Derivative Test verifies the answer, which is why it sits at the centre of optimization problems.

One curvature check separates a true optimum from a false one across physics, money, and design, which is why the test survives long after the exam that taught it.

What Are The Most Common Mistakes With The Second Derivative Test?

These three errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and course common-error handouts.

Using the test where $f'(c) \neq 0$.

Where it slips in:

A student picks any $x$ value, plugs it into $f''$, and reports a maximum or minimum from the sign, without first checking that the tangent is horizontal there.

Don't do this:

Do not classify a point that is not a critical point. The test only speaks at points where $f'(c) = 0$.

The correct way:

Always solve $f'(x) = 0$ first. Only feed those critical points into $f''$. A positive or negative $f''$ at a non-critical point tells you about concavity, not about a peak or valley.

Concluding "no extremum" when $f''(c) = 0$.

Where it slips in:

A student computes $f''(c) = 0$ and decides the point cannot be a maximum or minimum, or automatically calls it a point of inflection.

Don't do this:

Do not read $f''(c) = 0$ as a verdict. It is a non-answer, not a "no." Recall that $x^4$ has $f''(0) = 0$ and is still a minimum.

The correct way:

Treat $f''(c) = 0$ as inconclusive and switch to the first derivative test. Check the sign of $f'$ just left and just right of $c$: a sign change is an extremum, no sign change is not.

Reversing the sign rule.

Where it slips in:

A student remembers that one sign gives a maximum and the other a minimum, but pairs them the wrong way, calling a positive second derivative a maximum.

Don't do this:

Do not guess the pairing. A positive $f''(c)$ is a minimum, not a maximum.

The correct way:

Anchor it to the shape: positive means concave up means a valley means a minimum; negative means concave down means a peak means a maximum. The cup that holds water is the positive one.

Practice Problems On The Second Derivative Test

Work each one with the four steps, then check against the answer. Answers are verified.

1. Classify the critical point of $f(x) = x^2 - 4x + 1$.
   (Answer: $f'(x) = 2x - 4 = 0$ at $x = 2$; $f''(2) = 2 > 0$, local minimum $(2, -3)$.)

2. Classify the critical points of $f(x) = x^3 - 12x$.
   (Answer: $x = \pm 2$; $f''(x) = 6x$, so maximum $(-2, 16)$ and minimum $(2, -16)$.)

3. Classify the critical points of $f(x) = x^4 - 2x^2$.
   (Answer: $x = 0, \pm 1$; $f''(x) = 12x^2 - 4$, so maximum $(0, 0)$ and minima $(-1, -1)$ and $(1, -1)$.)

4. Classify the critical point of $f(x) = -x^4$.
   (Answer: $x = 0$, $f''(0) = 0$ inconclusive; first derivative test gives a local maximum $(0, 0)$.)

5. Classify the critical point of $f(x) = x^5$.
   (Answer: $x = 0$, $f''(0) = 0$ inconclusive; $f'(x) = 5x^4$ never changes sign, so it is neither a maximum nor a minimum.)

6. Classify the critical points of $f(x) = x^3 - 3x^2 + 2$.
   (Answer: $x = 0, 2$; $f''(x) = 6x - 6$, so maximum $(0, 2)$ and minimum $(2, -2)$.)

Where Should You Go Next After The Second Derivative Test?

Classifying critical points is one move inside the larger job of understanding a graph, and several natural doors open from here.

1. The first derivative test. Master the sign-chart method that rescues every inconclusive case and works at corners the second derivative cannot see.

2. Maxima and minima. Put both tests to work finding the highest and lowest points of real functions, including absolute extrema on an interval.

3. Curve sketching. Combine turning points with concavity and inflection points to draw a full graph from its derivatives.

If your child is meeting the Second Derivative Test for the first time, a live Bhanzu trainer teaches it from the concavity picture up, so the sign rule and the inconclusive case feel like one idea, in the Bhanzu math classes.