Classify critical points using the Second Derivative Test

Consider the function $f(x) = x^3 - 6x^2 + 9x + 1$ .

Find the critical numbers of $f(x) = x^3 - 6x^2 + 9x + 1$ , and use the Second Derivative Test to classify each one as a local maximum, local minimum, or neither.

Hint

First solve $f'(x) = 0$ to locate the critical numbers. Then compute $f''(x)$ and evaluate it at each critical number: a positive value indicates a local minimum, a negative value indicates a local maximum, and a zero value means the test is inconclusive.

Step-by-step solution

Compute the first derivative: $f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$ . Setting $f'(x) = 0$ gives critical numbers $x = 1$ and $x = 3$ .

Compute the second derivative: $f''(x) = 6x - 12$ . Evaluating at $x=1$ : $f''(1) = 6(1) - 12 = -6 < 0$ , so $x = 1$ is a local maximum. Evaluating at $x=3$ : $f''(3) = 6(3) - 12 = 6 > 0$ , so $x = 3$ is a local minimum.

Classify critical points using the Second Derivative Test

Related problems