# Question #bf7f4

Dec 3, 2017

That is correct. Furthermore, the function increases for $x \ge \frac{1}{2}$

#### Explanation:

A function is increasing if $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$.

$\frac{\mathrm{dy}}{\mathrm{dx}} > 0$
$8 x - \frac{1}{x} ^ 2 > 0$

Since ${x}^{2} > 0$, multiplying it to both sides of the inequality retains the sign.

$8 {x}^{3} - 1 > 0$
$8 {x}^{3} > 1$
${x}^{3} > \frac{1}{8}$

Since the cube root is an increasing function, taking the cube root on both sides does not change the sign too.

$x > \frac{1}{2}$

The point $\left(\frac{1}{2} , 3\right)$ is included as well.