## Calculus: Early Transcendentals 8th Edition

Using the First Derivative Test or the Second Derivative Test, we can determine that $f$ has a local minimum at $x = 0$ and $f$ has a local maximum at $x = 1$
$f(x) = 1+3x^2-2x^3$ First Derivative Test: $f'(x) = 6x-6x^2 = 0$ $6x~(1-x) = 0$ $x = 0, 1$ As $x \to 0^-,$ then $f'(x) \lt 0$ As $x \to 0^+,$ then $f'(x) \gt 0$ $f$ has a local minimum at $x = 0$ As $x \to 1^-,$ then $f'(x) \gt 0$ As $x \to 1^+,$ then $f'(x) \lt 0$ $f$ has a local maximum at $x = 1$ Second Derivative Test: $f''(x) = 6-12x$ At $x = 0,~~f''(x) \gt 0$ $f$ has a local minimum at $x = 0$ At $x = 1,~~f''(x) \lt 0$ $f$ has a local maximum at $x = 0$