Answer
$4$ and $-3$ are critical points. There is a local minimum at $x = −3$, but the test is inconclusive for $x = 4$. The first derivative test shows that there is neither a maximum nor a minimum at $x = 4$.
Work Step by Step
$f'(x) = x^3−5x^2−8x+48 = (x−4)^2(x+3)$. (This can be obtained by using trial-and-error to determine that $x = 4$ is a root, and then using long division of polynomials to see that $f'(x) = (x − 4)(x^2 − x − 12)$.) Note that $f''(x) = 3x^2−10x−8$, so $f''(−3) = 49 > 0$ and $f''(4) = 0$. So there is a local minimum at $x = −3$, but the test is inconclusive for $x = 4$. The first derivative test shows that there is neither a maximum nor a minimum at $x = 4$.
