The Second Derivative Test

Using the Second Derivative to Classify Critical Points

The Second Derivative Test

Suppose \(c\) is a critical point of \(f\) where \(f'(c) = 0\) and \(f''(x)\) is continuous near \(x=c\).

1. If \(f''(c) > 0\), then \(f(x)\) has a relative minimum at \(x=c\).

2. If \(f''(c) < 0\), then \(f(x)\) has a relative maximum at \(x=c\).

3. If \(f''(c) = 0\), then the test is inconclusive.

Example 1

Classifying critical points using the second derivative test

In Example 1 from Critical Points, we found that the critical points of

\[f(x)=2x^3-15x^2+36x+20\]

were \(x=2\) and \(x=3\). Classify each critical point using the Second Derivative Test.

Step 1: Compute \(f''(x)\).

\[\begin{align*} f''(x) &= \frac{d}{dx}(6x^2 - 30x+36) & \hbox{using $f'(x)$ from previous example}\\ &= 12x - 30 \end{align*}\]

Step 2: Classify each critical point.

Since \(f'(2) = 0\) and \(f''(2) = -6 < 0\), \(f(x)\) has a relative maximum at \(x=2\).

Since \(f'(3) = 0\) and \(f''(3) = 6 > 0\), \(f(x)\) has a relative minimum at \(x=3\).

Example 2

Classifying critical points using the second derivative test

In Example 2 from Critical Points, we found that the only critical point of

\[f(x)=\frac{1}{x^2-1}\]

was \(x=0\). Classify the critical point using the Second Derivative Test.

Step 1: Compute \(f''(x)\).

\[\begin{align*} f''(x) &= \frac{d}{dx}\left(\frac{-2x}{(x^2-1)^2}\right) & \hbox{using $f'(x)$ from previous example}\\ &= \frac{-2(x^2-1)^2 - (-2x)\cdot 2(x^2-1)2x}{(x^2-1)^4}\\ &= \frac{2(x^2-1)[-(x^2-1) + 4x^2]}{(x^2-1)^4}\\ &= \frac{2(1 + 3x^2)}{(x^2-1)^3} \end{align*}\]

Step 2: Classify each critical point.

Since \(f'(0) = 0\) and \(f''(0) = -2 < 0\), \(f(x)\) has a relative maximum at \(x=0\).