JAMB 1992 · UME · Q39

Obtain the maximum value of the function $f(x) = x^3 - 12x + 11$ .

Worked solution (try it first)

At a turning point $f'(x) = 0$ : $3x^2 - 12 = 0$ , so $x^2 = 4$ and $x = 2$ or $x = -2$ .
$f''(x) = 6x$ .
At $x = -2$ , $f'' = -12 < 0$ , so this is the maximum.
At $x = 2$ it is the minimum.
$f(-2) = -8 + 24 + 11 = 27$ , so the maximum value is 27, option D.