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Second Derivative
Test

Learn how to use concavity and the second derivative to classify critical points as local maxima or minima: faster and more elegant than the First Derivative Test.

AP Calculus AB

What Is a Critical Point?

A critical point of a function \( f \) occurs at \( x = c \) where the first derivative is zero or undefined:

At these points, the tangent line is horizontal (or absent). The function could have a local max, a local min, or neither: the critical point alone does not tell us which.

Concavity and \( f'' \)

\( f''(x) > 0 \)
Graph is concave up: shaped like a cup. Slopes are increasing.
\( f''(x) < 0 \)
Graph is concave down: shaped like a cap. Slopes are decreasing.
Concave Up \( f''(x) > 0 \) min Concave Down \( f''(x) < 0 \) max

Why Use the Second Derivative?

The First Derivative Test requires checking the sign of \( f' \) on both sides of every critical point. The Second Derivative Test gives the answer in one computation: just evaluate \( f''(c) \).

The Second Derivative Test

Let \( f \) be a function where \( f'(c) = 0 \) and \( f'' \) is continuous near \( c \).

1
If \( f''(c) > 0 \), then \( f \) has a local minimum at \( x = c \).
2
If \( f''(c) < 0 \), then \( f \) has a local maximum at \( x = c \).
3
If \( f''(c) = 0 \), the test is inconclusive: use the First Derivative Test instead.

Theorem Visualized

Local Min \( f''(c) > 0 \) Local Max \( f''(c) < 0 \)

First vs Second Derivative Test

First Derivative Test Second Derivative Test
What you check Sign of \( f' \) on each side of \( c \) Value of \( f''(c) \)
Number of evaluations At least 2 (left and right) Just 1
Always conclusive? Yes No: fails when \( f''(c)=0 \)
Best for All cases, including cusps Polynomial/rational functions

How to Apply the Test

1
Find \( f'(x) \): differentiate the function.
2
Solve \( f'(x) = 0 \): locate all critical points.
3
Find \( f''(x) \): differentiate again.
4
Evaluate \( f''(c) \) at each critical point and classify.
5
If \( f''(c) = 0 \), fall back to the First Derivative Test.

Example 1: \( f(x) = x^3 - 3x \)

1
\( f'(x) = 3x^2 - 3 = 3(x^2 - 1) \)
2
Set \( f'(x) = 0 \): \( x^2 - 1 = 0 \Rightarrow x = -1, \; x = 1 \)
3
\( f''(x) = 6x \)
4
\( f''(-1) = -6 < 0 \) → local max  |  \( f''(1) = 6 > 0 \) → local min

Graph of \( f(x) = x^3 - 3x \)

Example 2: \( g(x) = -2x^2 + 8x - 5 \)

1
\( g'(x) = -4x + 8 \)
2
Set \( g'(x) = 0 \): \( -4x + 8 = 0 \Rightarrow x = 2 \)
3
\( g''(x) = -4 \) (constant: always negative)
4
\( g''(2) = -4 < 0 \) → local maximum at \( x = 2 \)

Example 3: Inconclusive Case

Consider \( h(x) = x^4 \). Find and classify its critical point.

1
\( h'(x) = 4x^3 \), so \( h'(0) = 0 \). Critical point at \( x = 0 \).
2
\( h''(x) = 12x^2 \), so \( h''(0) = 0 \). Test is inconclusive!
3
First Derivative Test: \( h'(x) < 0 \) for \( x < 0 \) and \( h'(x) > 0 \) for \( x > 0 \). Sign changes from − to +: local minimum.

Example 4: \( f(x) = x + \frac{4}{x} \)

1
\( f'(x) = 1 - \frac{4}{x^2} \). Set equal to 0: \( x^2 = 4 \Rightarrow x = \pm 2 \).
2
\( f''(x) = \frac{8}{x^3} \)
3
\( f''(2) = \frac{8}{8} = 1 > 0 \) → local min at \( x = 2 \)
4
\( f''(-2) = \frac{8}{-8} = -1 < 0 \) → local max at \( x = -2 \)

Common Mistakes

✕ Forgetting the Inconclusive Case
When \( f''(c) = 0 \), students wrongly conclude there is no extremum. The test simply gives no answer: you must use another method.
✕ Confusing Signs
Positive second derivative means concave up (minimum), not maximum. Remember: cup = up = min.
✕ Applying at Non-Critical Points
The test only works where \( f'(c) = 0 \). If \( f'(c) \neq 0 \), the point is not a candidate for an extremum.

Quick Reference Summary

Condition at \( x = c \) Concavity Conclusion
\( f'(c) = 0 \), \( f''(c) > 0 \) Concave Up Local Minimum
\( f'(c) = 0 \), \( f''(c) < 0 \) Concave Down Local Maximum
\( f'(c) = 0 \), \( f''(c) = 0 \) Unknown Inconclusive
\( f'(c) \neq 0 \) N/A Not a critical point

Worked practice: Section 9, Practice Problem 4 — Applied

Work through this original lesson problem, then reveal the solution and illustrations.

A manufacturer designs an open-top box by cutting equal squares of side length x inches from each corner of a 12-inch by 18-inch sheet of metal and folding up the sides. (a) Show that the volume of the box is V(x) = x(12 − 2x)(18 − 2x). (b) Find all critical points of V on the interval (0, 6). (c) Use the Second Derivative Test to classify each critical point. (d) State the dimensions that maximize the volume and give the maximum volume.

Show answer and explanation

Rubric (5 points): Point 1 (Part a): After cutting squares of side x from each corner, the base has dimensions (12 − 2x) by (18 − 2x), and the height is x. Therefore V(x) = x(12 − 2x)(18 − 2x). Expanding: V(x) = x(216 − 24x − 36x + 4x²) = 4x³ − 60x² + 216x. Point 2 (Part b): V′(x) = 12x² − 120x + 216 = 12(x² − 10x + 18). Using the quadratic formula: x = (10 ± √(100 − 72))/2 = (10 ± √28)/2 = 5 ± √7. So x = 5 − √7 ≈ 2.354 and x = 5 + √7 ≈ 7.646. Only x = 5 − √7 lies in (0, 6). Point 3 (Part c — computing f″): V″(x) = 24x − 120. At x = 5 − √7: V″(5 − √7) = 24(5 − √7) − 120 = 120 − 24√7 − 120 = −24√7 ≈ −63.5. Point 4 (Part c — conclusion): Since V′(5 − √7) = 0 and V″(5 − √7) = −24√7 < 0, by the Second Derivative Test, V has a local maximum at x = 5 − √7. Because this is the only critical point on (0, 6) and V(0) = V(6) = 0, this local maximum is also the absolute maximum on [0, 6]. Point 5 (Part d): With x = 5 − √7, the dimensions are: height = 5 − √7 ≈ 2.354 in, width = 12 − 2(5 − √7) = 2 + 2√7 ≈ 7.292 in, length = 18 − 2(5 − √7) = 8 + 2√7 ≈ 13.292 in. Maximum volume: V(5 − √7) = (5 − √7)(2 + 2√7)(8 + 2√7). Computing numerically: V ≈ 2.354 × 7.292 × 13.292 ≈ 228.2 in³.

An 18 in by 12 in sheet with four equal square corner cuts and a schematic folded open box. Height is x and base sides are 18−2x and 12−2x.
An 18 in by 12 in sheet produces an open box of volume V=x(18−2x)(12−2x), maximized at x=5−√7 in.
Volume in cubic inches versus cut length in inches for an 18 in by 12 in sheet. The maximum occurs at x=5−√7, about 2.354 in.
An 18 in by 12 in sheet produces an open box of volume V=x(18−2x)(12−2x), maximized at x=5−√7 in.

Question 1 of 5

Let \( f(x) = x^3 - 12x + 1 \). Using the Second Derivative Test, classify the critical point at \( x = 2 \).

Question 2 of 5

If \( f''(c) = 0 \) and \( f'(c) = 0 \), what does the Second Derivative Test tell you?

Question 3 of 5

Let \( g(x) = -x^4 + 4x^2 \). Find \( g''(x) \) and determine the classification at \( x = 0 \).

Question 4 of 5

For \( h(x) = \sin(x) \), the Second Derivative Test at \( x = \frac{\pi}{2} \) yields:

Question 5 of 5

Which of the following is an advantage of the First Derivative Test over the Second Derivative Test?