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

Max: 12; Min: -20, 7

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

We have

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

$\text{or } \quad f'(x) = 12x^3 + 12x^2 - 24x = 12x(x - 1)(x + 2)$

$\text{or } \quad f'(x) = 0 \text{ at } x = 0, x = 1 \text{ and } x = -2.$

$\text{Now } \quad f''(x) = 36x^2 + 24x - 24 = 12(3x^2 + 2x - 2)$

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

Therefore, by second derivative test, $x = 0$ is a point of local maxima and local maximum value of $f$ at $x = 0$ is $f(0) = 12$ 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.