Find the local maximum and local minimum values of the function $f$ given by $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

(N/A) Given the function $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.
First,find the derivative $f'(x) = 12x^3 + 12x^2 - 24x$.
Factorizing the derivative,we get $f'(x) = 12x(x^2 + x - 2) = 12x(x - 1)(x + 2)$.
Setting $f'(x) = 0$,we find the critical points $x = 0, x = 1, x = -2$.
Now,find the second derivative $f''(x) = 36x^2 + 24x - 24$.
Evaluating the second derivative at the critical points:
$f''(0) = -24 < 0$,so $x = 0$ is a point of local maxima. The local maximum value is $f(0) = 12$.
$f''(1) = 36(1)^2 + 24(1) - 24 = 36 > 0$,so $x = 1$ is a point of local minima. The local minimum value is $f(1) = 3(1)^4 + 4(1)^3 - 12(1)^2 + 12 = 3 + 4 - 12 + 12 = 7$.
$f''(-2) = 36(-2)^2 + 24(-2) - 24 = 144 - 48 - 24 = 72 > 0$,so $x = -2$ is a point of local minima. The local minimum value is $f(-2) = 3(-2)^4 + 4(-2)^3 - 12(-2)^2 + 12 = 48 - 32 - 48 + 12 = -20$.