Question

The function $f(x) = 2x^3 - 3x^2 - 12x + 4$, has

• A. two points of local maximum
• B. two points of local minimum
• C. one maxima and one minima
• D. no maxima or minima

Answer 1

Answer: (C)

$f(x) = 2x^3 - 3x^2 - 12x + 4$
$\Rightarrow f'(x) = 6x^2-6 x -1 2$
$= 6(x^2 - x - 2 )$
For maxima and minima, $f'(x) = 0$
$ \therefore 6(x - 2)(x + 1) = 0$
$\Rightarrow x = 2$, $-1$
Now, $f''(x) = 12x - 6$
At $x = 2$
$f'' (x) = 24 - 6 = 18 > 0$
$ \therefore x = 2$, local min. point
At $x = -1$
$f''(x) = 12(- 1) - 6 = - 18 < 0$
$ \therefore x = - 1$ local max. point