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Answer 1 · 2017-12-03T06:22:45

That is correct. Furthermore, the function increases for $x \geq \frac{1}{2}$ $x>=1/2$

A function is increasing if $\frac{d y}{d x} > 0$ $frac{dy}{dx} > 0$ .

$\frac{d y}{d x} > 0$ $frac{dy}{dx} > 0$
$8 x - \frac{1}{x^{2}} > 0$ $8x - 1/x^2 >0$

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

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

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

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

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