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

Local Max: $12$; Local Min: $-20, 7$

The correct answer is Option (1) → Local Max: $12$; Local Min: $-20, 7$ ##

We have

Now $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$

$f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2)$

$f'(x) = 0$, critical point at $x = 0, x = 1$ and $x = -2$. Again differentiate $f'(x)$ w.r.t $x$

$ f''(x) = 36x^2 + 24x - 24 = 12(3x^2 + 2x - 2) $

Using second-order derivatives test

$\begin{cases} f''(0) = -24 < 0 \\ f''(1) = 36 > 0 \\ f''(-2) = 72 > 0 \end{cases}$

Therefore, by second derivative test, $x = 0$ is a point of local maxima.

Local maximum value $f(0) = 12$

While $x = 1$ and $x = -2$ are the points of local minima and local minimum values of $f$ at $x = 1$ and $-2$ are $f(1) = 7$ and $f(-2) = -20$, respectively.