What is the difference between the First Derivative Test and the Second Derivative Test?

Both methods identify whether a critical point is a local extremum. The First Derivative Test checks whether f'(x) changes sign around the critical point: a sign change from positive to negative indicates a local maximum, and negative to positive indicates a local minimum. The Second Derivative Test evaluates f''(c) at the critical point: if f''(c) > 0, you have a local minimum (concave up), and if f''(c) < 0, a local maximum (concave down). The Second Derivative Test is often quicker, but it fails when f''(c) = 0, whereas the First Derivative Test always works.

What should I do if f''(c) = 0?

The Second Derivative Test is inconclusive when the second derivative is zero. The critical point could be a local extremum, an inflection point, or neither. Use the First Derivative Test instead: evaluate the sign of f'(x) just to the left and right of c to determine whether f'(x) changes sign, which would confirm the presence of a local extremum.

Can the Second Derivative Test be used at the endpoints of the domain?

No, the Second Derivative Test requires a critical point in the interior of the domain where f'(c) = 0. Endpoints are not critical points. To determine whether an endpoint is a local extremum, compare the function value at the endpoint to nearby points within the domain.

Why does f''(c) > 0 mean a local minimum instead of a maximum?

Because f''(c) > 0 means the graph is concave up—it bends upward like a smile. At a critical point where f'(c) = 0 (the tangent line is horizontal), a concave-up shape means the function descends from both sides down to the critical point and then rises away, creating a valley or local minimum. Conversely, f''(c) < 0 creates a concave-down shape (like a frown), which forms a peak or local maximum at a horizontal tangent.

Second Derivative Test

Q: What should I do if f''(c) = 0?

The Second Derivative Test is inconclusive when the second derivative is zero. The critical point could be a local extremum, an inflection point, or neither. Use the First Derivative Test instead: evaluate the sign of f'(x) just to the left and right of c to determine whether f'(x) changes sign, which would confirm the presence of a local extremum.

Q: Can the Second Derivative Test be used at the endpoints of the domain?

No, the Second Derivative Test requires a critical point in the interior of the domain where f'(c) = 0. Endpoints are not critical points. To determine whether an endpoint is a local extremum, compare the function value at the endpoint to nearby points within the domain.

The Second Derivative Test determines whether a critical point is a local minimum or maximum using the concavity at that point.

f'(c) = 0, ; f''(c) > 0 \implies \text{local min}

Solve a problem with Second Derivative Test

Type the problem. The solver will use Second Derivative Test where Second Derivative Test is the right tool, and tell you when it is not.

What each symbol means

What Second Derivative Test takes

Symbol	Meaning
$f$	The function being analyzed; the condition $f'(c) = 0$ identifies $c$ as a critical point, and $f''(c)$ reveals the concavity at that point.
$c$	A critical point in the domain of $f$ where $f'(c) = 0$ ; confusing it with the function value $f(c)$ is a common error.

When to use it

When you have found a critical point and want to know if it is a local extremum.

Level

Usually taught in: Calculus I

Worked examples

1. Classify critical points of a cubic polynomial

Problem

Find and classify all local extrema of

f(x) = x^3 - 3x + 2

using the Second Derivative Test.

$f'(x) = 3x^2 - 3$
We differentiate using the Power Rule: $\frac{d}{dx}(x^3) = 3x^2$ and $\frac{d}{dx}(-3x) = -3$ .
$3x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1$
We set $f'(x) = 0$ , add 3 to both sides, and divide by 3 to find critical points at $x = -1$ and $x = 1$ .
$f''(x) = 6x$
We differentiate the first derivative to obtain the second derivative.
$f''(-1) = -6 < 0 \implies \text{local maximum}$
Since the second derivative is negative at $x = -1$ , the curve is concave down there, confirming a local maximum.
$f''(1) = 6 > 0 \implies \text{local minimum}$
Since the second derivative is positive at $x = 1$ , the curve is concave up there, confirming a local minimum.

Answer: $x = -1 \text{ (local maximum)}, \quad x = 1 \text{ (local minimum)}$

The Second Derivative Test is efficient here because the second derivative is simple to compute and evaluate. By testing the concavity at each critical point, we immediately classify them without needing to track sign changes of $f'(x)$ across intervals.

2. Classify critical points with fractions and factoring

Problem

Find and classify all local extrema of

f(x) = -2x^3 + 3x^2 + 12x - 5

using the Second Derivative Test.

$f'(x) = -6x^2 + 6x + 12$
We apply the Power Rule to each term: $\frac{d}{dx}(-2x^3) = -6x^2$ , $\frac{d}{dx}(3x^2) = 6x$ , and $\frac{d}{dx}(12x) = 12$ .
$x^2 - x - 2 = 0$
We set $f'(x) = 0$ and then divide the equation by $-6$ to simplify.
$(x - 2)(x + 1) = 0 \implies x = 2 \text{ or } x = -1$
We factor the quadratic and solve for the critical points.
$f''(x) = -12x + 6$
We differentiate the first derivative to find the second derivative.
$f''(-1) = -12(-1) + 6 = 12 + 6 = 18 > 0 \implies \text{local minimum}$
At $x = -1$ , the second derivative equals 18, which is positive, so the curve is concave up and we have a local minimum.
$f''(2) = -12(2) + 6 = -24 + 6 = -18 < 0 \implies \text{local maximum}$
At $x = 2$ , the second derivative equals $-18$ , which is negative, so the curve is concave down and we have a local maximum.

Answer: $x = -1 \text{ (local minimum)}, \quad x = 2 \text{ (local maximum)}$

Even though finding the critical points required factoring, the Second Derivative Test quickly classifies them by evaluating concavity. This is faster than the First Derivative Test when computing the second derivative is straightforward but checking sign changes of $f'(x)$ would be tedious.

3. Optimize garden dimensions using the Second Derivative Test

Problem

A rectangular garden plot is to be fenced on three sides with 120 feet of fencing, with one side against a barn. If the width is

w

feet, the area is

A(w) = 120w - 2w^2

. Use the Second Derivative Test to classify the critical point and find the dimensions that optimize the area.

$A'(w) = 120 - 4w$
We differentiate the area function with respect to width using the Power Rule.
$120 - 4w = 0 \implies w = 30$
We set $A'(w) = 0$ to find the critical point where the area stops increasing or decreasing.
$A''(w) = -4$
We differentiate the first derivative; the second derivative is the constant $-4$ .
$A''(30) = -4 < 0 \implies \text{local maximum}$
Since the second derivative is negative everywhere, including at $w = 30$ , the area function is concave down, confirming that $w = 30$ gives a local maximum.
$l = 120 - 2(30) = 120 - 60 = 60$
We find the length using the relationship $l = 120 - 2w$ from the constraint that three sides use 120 feet of fencing.

Answer: $w = 30 \text{ feet (width)}, \quad l = 60 \text{ feet (length)}$

Real-world optimization problems often involve finding the maximum or minimum of a function modeling cost, area, or profit. The Second Derivative Test confirms that our critical point represents a maximum, meaning this width and length maximize the garden area subject to the fencing constraint.

Common mistakes

Where Second Derivative Test usually goes wrong

The mistake	Why it is wrong	The fix
Applying the Second Derivative Test without first verifying that $f'(c) = 0$ .	The test only identifies extrema at critical points; if $f'(c) \neq 0$ , then $c$ is not a critical point and the test is meaningless.	Always solve $f'(c) = 0$ and verify that you have found critical points before applying the Second Derivative Test.
When $f''(c) = 0$ , concluding that the Second Derivative Test proves $x = c$ is an inflection point.	A zero second derivative makes the test inconclusive; the critical point could be a local extremum, an inflection point, or neither.	When $f''(c) = 0$ , use the First Derivative Test by checking whether $f'(x)$ changes sign on either side of $c$ .
Interpreting $f''(c) > 0$ as meaning the function is decreasing at $c$ , so $x = c$ is a local maximum.	The second derivative measures concavity (how the curve bends), not the direction of the function; $f''(c) > 0$ means the graph is concave up, which produces a local minimum at a critical point with a horizontal tangent.	Remember: $f''(c) > 0$ with $f'(c) = 0$ gives a local minimum (concave up); $f''(c) < 0$ with $f'(c) = 0$ gives a local maximum (concave down).

Tips and when to use something else

Always verify that $f'(c) = 0$ before applying the Second Derivative Test; if the point is not critical, the test does not apply.
The Second Derivative Test is inconclusive when $f''(c) = 0$ ; in that case, use the First Derivative Test to determine whether the critical point is a local extremum.
If computing the second derivative is complicated, the First Derivative Test may be faster even though it requires checking sign changes around the critical point.
Connect the test to concavity: a concave-up graph ( $f''(c) > 0$ ) with a horizontal tangent ( $f'(c) = 0$ ) looks like a smile and gives a local minimum; a concave-down graph ( $f''(c) < 0$ ) looks like a frown and gives a local maximum.

Related formulas

Critical PointsCritical points mark where a function's derivative is zero or undefined, revealing local extrema and where the graph changes direction.First Derivative TestClassify critical points as local maxima or minima by checking whether the derivative changes sign before and after the point.Concavity and Inflection PointsPinpoint where a function's curve changes from bending upward to downward, revealing the transition point between concave up and concave down.

Frequently asked questions

What is the difference between the First Derivative Test and the Second Derivative Test?: Both methods identify whether a critical point is a local extremum. The First Derivative Test checks whether $f'(x)$ changes sign around the critical point: a sign change from positive to negative indicates a local maximum, and negative to positive indicates a local minimum. The Second Derivative Test evaluates $f''(c)$ at the critical point: if $f''(c) > 0$ , you have a local minimum (concave up), and if $f''(c) < 0$ , a local maximum (concave down). The Second Derivative Test is often quicker, but it fails when $f''(c) = 0$ , whereas the First Derivative Test always works.
What should I do if $f''(c) = 0$ ?: The Second Derivative Test is inconclusive when the second derivative is zero. The critical point could be a local extremum, an inflection point, or neither. Use the First Derivative Test instead: evaluate the sign of $f'(x)$ just to the left and right of $c$ to determine whether $f'(x)$ changes sign, which would confirm the presence of a local extremum.
Can the Second Derivative Test be used at the endpoints of the domain?: No, the Second Derivative Test requires a critical point in the interior of the domain where $f'(c) = 0$ . Endpoints are not critical points. To determine whether an endpoint is a local extremum, compare the function value at the endpoint to nearby points within the domain.
Why does $f''(c) > 0$ mean a local minimum instead of a maximum?: Because $f''(c) > 0$ means the graph is concave up—it bends upward like a smile. At a critical point where $f'(c) = 0$ (the tangent line is horizontal), a concave-up shape means the function descends from both sides down to the critical point and then rises away, creating a valley or local minimum. Conversely, $f''(c) < 0$ creates a concave-down shape (like a frown), which forms a peak or local maximum at a horizontal tangent.

Need a different method?

The full solver is not scoped to one formula — type any problem and it will pick the method.

Open the math solver

Reviewed 2026-09-18