Find the local maximum value of the function $g(x) = x^3 - 3x$

The correct answer is 2

Given $g(x) = x^3 - 3x$

$g'(x) = 3x^2 - 3$

$g''(x) = 6x$


To obtain critical point, put $g'(x) = 0$

∴ $g'(x) = 3x^2 - 3 = 0$

∴ $x = \pm 1$


At $x = -1$, $g''(x) = 6x = 6(-1) = -6 < 0$

Hence there is a local maxima at $x = -1$


Value of the function at local maxima = $g(x) = x^3 - 3x$ - $(-1)^3 - 3(-1)$ = $2$