#### Answer

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 $x = 1$

#### Work Step by Step

$f(x) = 1+3x^2-2x^3$
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 $x = 0$