## Calculus: Early Transcendentals (2nd Edition)

$-a$ is the critical point. There is a local minimum at x = −a.
$f'(x) = 4(x + a)^3$, which is $0$ for $x = −a$. Note that $f''(x) = 12(x + a)^2$, which is $0$ at $x = −a$, so the test is inconclusive. The first derivative test shows that there is a local minimum at x = −a.