Lesson 2: Second Derivative and Concavity

Estimated time: 35-45 minutes

Learning Objectives

Define concave up and concave down
Find inflection points where concavity changes
Apply the Second Derivative Test to classify critical points

Concavity

Concavity

f is concave up on an interval if f''(x) > 0 (the graph bends upward, like a cup).

f is concave down on an interval if f''(x) < 0 (the graph bends downward, like a frown).

Inflection Point

A point where the concavity changes (from up to down or vice versa). At an inflection point, f''(c) = 0 or f''(c) DNE, and f'' changes sign.

Example 1: Finding Concavity and Inflection Points

Analyze f(x) = x³ − 3x² + 2.

Step 1: f''(x) = 6x − 6 = 6(x − 1).

Step 2: f'' = 0 at x = 1. f'' < 0 for x < 1 (concave down), f'' > 0 for x > 1 (concave up).

Step 3: Concavity changes at x = 1, so (1, f(1)) = (1, 0) is an inflection point.

Second Derivative Test

Second Derivative Test

If f'(c) = 0 and f''(c) exists:

f''(c) > 0 ⇒ f has a local minimum at c (concave up at the critical point).
f''(c) < 0 ⇒ f has a local maximum at c (concave down at the critical point).
f''(c) = 0 ⇒ inconclusive; use the First Derivative Test instead.

Example 2: Applying the Second Derivative Test

Classify the critical points of f(x) = x⁴ − 4x³ + 6.

f'(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0, x = 3.

f''(x) = 12x² − 24x.

f''(0) = 0: inconclusive (use FDT: f' does not change sign at 0, so no extremum).

f''(3) = 108 − 72 = 36 > 0: local minimum at x = 3, f(3) = −21.

Example 3: Full Analysis

f(x) = x e^−x. f'(x) = (1 − x)e^−x. Critical point: x = 1.

f''(x) = (x − 2)e^−x. f''(1) = −e⁻¹ < 0 ⇒ local max at x = 1.

Inflection: f'' = 0 at x = 2. f'' changes sign ⇒ inflection point at (2, 2e⁻²).

Key Takeaways

f'' > 0: concave up (cup). f'' < 0: concave down (frown).
Inflection points occur where f'' changes sign.
Second Derivative Test: f''(c) > 0 at a critical point means local min; f''(c) < 0 means local max.
When the Second Derivative Test is inconclusive (f''(c) = 0), fall back to the First Derivative Test.

Check Your Understanding

1. Find the inflection points of f(x) = x⁴ − 6x².

Answer: f''(x) = 12x² − 12 = 12(x − 1)(x + 1). f'' = 0 at x = ±1. Sign changes at both, so inflection points at x = −1 and x = 1.

2. Use the Second Derivative Test on f(x) = x³ − 12x at its critical points.

Answer: f'(x) = 3x² − 12 = 0 at x = ±2. f''(x) = 6x. f''(2) = 12 > 0: local min. f''(−2) = −12 < 0: local max.

3. Where is f(x) = sin x concave down on [0, 2π]?

Answer: f''(x) = −sin x < 0 when sin x > 0, i.e., (0, π).

4. Find and classify inflection points of g(x) = x⁵ − 5x⁴.

Answer: g''(x) = 20x³ − 60x² = 20x²(x − 3). g'' = 0 at x = 0 and x = 3. Sign change at x = 3 only (g'' is negative on both sides of x = 0). Inflection point at x = 3.

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