## Calculus: Early Transcendentals (2nd Edition)

$0$ and $1$ are critical points. Critical point at $x=1$ yields a local minimum and critical point at $x=0$ yields a local maximum.
$f'(x) = 6x^2 − 6x = 6x(x − 1)$, so $x = 0$ and $x = 1$ are critical points. $f''(x) = 12x − 6$, so $f''(1) = 6 > 0$, so the critical point at $x = 1$ yields a local minimum. Also, $f''(0) = −6 < 0$, so the critical point at $0$ yields a local maximum.