Stationary points

Stationary points of $f(x)$ occur at values of $x$ in the interior of the domain where the derivative is zero. To locate stationary points, set $f'(x) = 0$ and solve for $x.$

Example
Let $f(x) = x^{3} - 12x.$
Then to locate the stationary points, set $f'(x) = 0$ and solve for $x.$ This gives

$3x^{2} - 12 = 0,$

so $f$ has stationary points at $x = \pm 2.$ The corresponding points on the graph are

$(-2, f(-2)) = (-2, 16)$ and $(2, f(2)) = (2, -16).$