# How do you graph f(x)=1/3x^3-6x using the information given by the first derivative?

Aug 17, 2016

From $f ' \left(x\right) = 0 , f \left(x\right)$ has turning points at $x = \pm \sqrt{6}$

#### Explanation:

$f \left(x\right) = \frac{1}{3} {x}^{3} - 6 x$

$f ' \left(x\right) = \cancel{3} {x}^{2} / \cancel{3} - 6$

$f \left(x\right)$ has tuning points where $f ' \left(x\right) = 0$
I.e. when ${x}^{2} = 6 \to x = \pm \sqrt{6}$

From the graph below, $f \left(- \sqrt{6}\right)$ has curvature concave down $\to$ a local maximum and $f \left(+ \sqrt{6}\right)$ has curvature concave up $\to$ a local minimum.

Also, $f \left(x\right) = 0$ when $x \left({x}^{2} / 3 - 6\right) = 0$
I.e. when $x = 0$ or $x = \pm \sqrt{18} = \pm 3 \sqrt{2}$

graph{1/3x^3-6x [-25.67, 25.65, -12.83, 12.84]}