Answer
a. $-2\lt x\lt0$ or $4\lt x\lt \infty $
b. See proof below.
Work Step by Step
a. See the graph below.
$\frac{x}{2}\gt1+\frac{4}{x}$ when $-2\lt x\lt0$ or $4\lt x\lt \infty$
b. We have $\frac{x}{2}\gt1+\frac{4}{x}$
for $ x\gt0$
$x^{2}\gt 2x+8$
$x^{2}-2x-8\gt 0$
$(x-4)(x+2)\gt0$
and this happens only when $4\lt x\lt \infty $
For $x\lt0$
$x^{2}\lt 2x+8$
$x^{2}-2x-8\lt 0$
$(x-4)(x+2)\lt0$
and this happens only when $-2\lt x\lt0$
