Applications of Derivatives · Curve Analysis

Second Derivative Test

Q: How do you use the Second Derivative Test?

Worked example. Classify the critical points of f(x) = x⁴ - 4x². f'(x) = 4x³ - 8x = 4x(x²-2). Critical points: x = 0, x = ±√2. f''(x) = 12x² - 8. f''(0) = -8 0 → local min at x = √2. f''(-√2) = 16 > 0 → local min at x = -√2. Answer: Local max at x = 0, local min at x = ±√2.

f'(c) = 0: \quad f''(c) > 0 \Rightarrow \text{local min}; \quad f''(c) < 0 \Rightarrow \text{local max}

At a critical point where f'(c) = 0: if f''(c) > 0, c is a local minimum (concave up). If f''(c) < 0, c is a local maximum (concave down). If f''(c) = 0, the test is inconclusive.

2 worked examples
15 in Applications of Derivatives

Worked examples

Classify the critical points of f(x) = x⁴ - 4x².

f'(x) = 4x³ - 8x = 4x(x²-2). Critical points: x = 0, x = ±√2
f''(x) = 12x² - 8
f''(0) = -8 < 0 → local max at x = 0
f''(√2) = 24-8 = 16 > 0 → local min at x = √2
f''(-√2) = 16 > 0 → local min at x = -√2

Answer: Local max at x = 0, local min at x = ±√2.

1 more worked examplePremium

Classify the critical points of f(x) = x³ - 3x² + 1.

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First Derivative Test Concavity

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Common questions

What is the Second Derivative Test?

At a critical point where f'(c) = 0: if f''(c) > 0, c is a local minimum (concave up). If f''(c) < 0, c is a local maximum (concave down). If f''(c) = 0, the test is inconclusive. It is one of the applications of derivatives formulas in the CalcRef reference.

How do you use the Second Derivative Test?

Worked example. Classify the critical points of f(x) = x⁴ - 4x². f'(x) = 4x³ - 8x = 4x(x²-2). Critical points: x = 0, x = ±√2. f''(x) = 12x² - 8. f''(0) = -8 < 0 → local max at x = 0. f''(√2) = 24-8 = 16 > 0 → local min at x = √2. f''(-√2) = 16 > 0 → local min at x = -√2. Answer: Local max at x = 0, local min at x = ±√2.

Practice

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