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Second Derivative Test

Second Derivative Test

A method for determining whether a critical point is a relative minimum or maximum.

 

Three rules for the Second Derivative Test: positive f''=relative min, negative f''=relative max, zero/undefined=inconclusive.
Example using y=x³-3x+4: critical points (1,2) and (-1,6); y″=6x confirms (1,2) minimum, (-1,6) maximum.

 

 

See also

Second derivative, first derivative test, absolute minimum, absolute maximum

Key Formula

If f′(c)=0, then:{f′′(c)>0  ⟹  f(c) is a relative minimumf′′(c)<0  ⟹  f(c) is a relative maximumf′′(c)=0  ⟹  test is inconclusive\text{If } f'(c) = 0 \text{, then:}\begin{cases} f''(c) > 0 \implies f(c) \text{ is a relative minimum} \\ f''(c) < 0 \implies f(c) \text{ is a relative maximum} \\ f''(c) = 0 \implies \text{test is inconclusive} \end{cases}
Where:
  • f(x)f(x) = A function that is twice differentiable near the critical point
  • cc = A critical point where f'(c) = 0
  • f′(c)f'(c) = The first derivative of f evaluated at c (must equal zero)
  • f′′(c)f''(c) = The second derivative of f evaluated at c, which determines concavity

Worked Example

Problem: Use the Second Derivative Test to find and classify all critical points of f(x) = x³ − 6x² + 9x + 2.
Step 1: Find the first derivative and set it equal to zero to locate critical points.
f′(x)=3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3)=0f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3) = 0
Step 2: Solve for the critical points.
x=1andx=3x = 1 \quad \text{and} \quad x = 3
Step 3: Find the second derivative.
f′′(x)=6x−12f''(x) = 6x - 12
Step 4: Evaluate the second derivative at x = 1.
f′′(1)=6(1)−12=−6<0  ⟹  relative maximum at x=1f''(1) = 6(1) - 12 = -6 < 0 \implies \text{relative maximum at } x = 1
Step 5: Evaluate the second derivative at x = 3.
f′′(3)=6(3)−12=6>0  ⟹  relative minimum at x=3f''(3) = 6(3) - 12 = 6 > 0 \implies \text{relative minimum at } x = 3
Answer: There is a relative maximum at (1, 6) and a relative minimum at (3, 2).

Another Example

This example demonstrates the inconclusive case where f''(c) = 0, showing when the test fails and a fallback method is needed.

Problem: Use the Second Derivative Test to classify the critical points of f(x) = x⁴.
Step 1: Find the first derivative and set it equal to zero.
f′(x)=4x3=0  ⟹  x=0f'(x) = 4x^3 = 0 \implies x = 0
Step 2: Find the second derivative.
f′′(x)=12x2f''(x) = 12x^2
Step 3: Evaluate the second derivative at x = 0.
f′′(0)=12(0)2=0f''(0) = 12(0)^2 = 0
Step 4: Since f''(0) = 0, the Second Derivative Test is inconclusive. You must use another method, such as the First Derivative Test. Checking sign changes: f'(x) changes from negative to positive at x = 0, so x = 0 is a relative minimum.
f′(−1)=−4<0andf′(1)=4>0f'(-1) = -4 < 0 \quad \text{and} \quad f'(1) = 4 > 0
Answer: The Second Derivative Test is inconclusive at x = 0. The First Derivative Test confirms x = 0 is a relative minimum with f(0) = 0.

Frequently Asked Questions

What happens when the second derivative equals zero at a critical point?
When f''(c) = 0, the Second Derivative Test is inconclusive — it does not tell you anything about whether the point is a minimum, maximum, or neither. You need to use another method, typically the First Derivative Test, which examines sign changes in f'(x) around the critical point. For example, f(x) = x³ has f''(0) = 0 and the point is neither a min nor max (it is an inflection point), while f(x) = x⁴ also has f''(0) = 0 but the point is a minimum.
What is the difference between the First Derivative Test and the Second Derivative Test?
The First Derivative Test checks how the sign of f'(x) changes around a critical point: positive-to-negative means a relative max, negative-to-positive means a relative min. The Second Derivative Test instead evaluates f''(c) at the critical point directly. The Second Derivative Test is often quicker when the second derivative is easy to compute, but it can be inconclusive when f''(c) = 0. The First Derivative Test always gives a definitive answer.
Does the Second Derivative Test find absolute maxima or minima?
No. The Second Derivative Test only identifies relative (local) extrema, not absolute (global) extrema. To find absolute extrema on a closed interval, you also need to evaluate f(x) at the endpoints and compare those values with the values at any relative extrema. On an open interval or the entire real line, additional analysis of the function's behavior at the boundaries or at infinity is required.

Second Derivative Test vs. First Derivative Test

Second Derivative TestFirst Derivative Test
What it usesThe value of f''(c) at the critical pointThe sign of f'(x) on intervals around the critical point
Relative max conditionf''(c) < 0f'(x) changes from positive to negative
Relative min conditionf''(c) > 0f'(x) changes from negative to positive
Inconclusive caseWhen f''(c) = 0; must use another methodNever inconclusive (always gives a result if f'(c) = 0 or is undefined)
SpeedOften faster — just plug c into f''(x)Requires testing multiple points or analyzing sign intervals
Requirementf must be twice differentiable near cf must be continuous at c; f' must exist near c (not necessarily at c)

Why It Matters

The Second Derivative Test is one of the most frequently used tools in AP Calculus and college-level calculus courses for optimization problems. When you need to minimize cost, maximize area, or find the optimal value of any quantity, you first find critical points and then use this test to classify them. It also builds intuition about concavity — understanding that a curve bending upward at a critical point creates a valley (minimum) and bending downward creates a peak (maximum).

Common Mistakes

Mistake: Applying the Second Derivative Test at a point where f'(c) ≠ 0.
Correction: The test only applies at critical points where f'(c) = 0. If f'(c) ≠ 0, the point is not a critical point and cannot be a relative extremum. Always verify that f'(c) = 0 before using the test.
Mistake: Concluding that f''(c) = 0 means the point is an inflection point or that there is no extremum.
Correction: When f''(c) = 0, the test is simply inconclusive — the point could be a relative min (like x⁴ at x = 0), a relative max, or neither (like x³ at x = 0). You must use the First Derivative Test or higher-order derivative tests to classify the point.

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