Use derivatives to sketch the graph of a function

Let

$f(x) = x^3 - 4x.$

Find all points such that $f'(x) = 0$ ;
Determine the intervals on which $f$ is monotonic by examining the sign of $f'$ ;
Determine the intervals on which $f'$ is monotonic by examining the sign of $f''$ ;
Sketch the graph of $f$ .

We take the derivative,
$f(x) = x^3 - 4x \quad \implies \quad f'(x) = 3x^2 - 4.$

Thus,

$f'(x) = 0 \quad \implies \quad 3x^2 - 4 = 0 \quad \implies \quad x = \pm \frac{2}{\sqrt{3}}.$
Since $f'(x) > 0$ when $|x| > \frac{2}{\sqrt{3}}$ and $f'(x) < 0$ when $|x| < \frac{2}{\sqrt{3}}$ we have $f$ is increasing if $|x| > \frac{2}{\sqrt{3}}$ and $f$ is decreasing if $|x| < \frac{2}{\sqrt{3}}$ .
Taking the second derivative,
$f'(x) = 3x^2 - 4 \quad \implies \quad f''(x) = 6x.$

Thus, $f'$ is increasing for $x > 0$ ; and decreasing for $x < 0$ .
We sketch the curve,