$f\left(x\right)=-100{(x+10)}^{-3}+40$

$x=-10$ and $y=40$.

${\text{D}}_f=⟨\leftarrow ,-10⟩\cup ⟨-10,\to ⟩$ and ${\text{B}}_f=⟨\leftarrow ,40⟩\cup ⟨40,\to ⟩$

$\frac{100}{{(x+10)}^3}=40$ gives you ${(x+10)}^3=2\frac{1}{2}$ and therefore $x=-10+\sqrt[3]{2\frac{1}{2}}$.
The x-intersect is $(-10+\sqrt[3]{2\frac{1}{2}},0)$.

Determine the intersects with the graphing calculator and reading off the solution to the inequality: $\mathrm{-8,73}\le x<\mathrm{40,00}$.

$16=\frac{1}{2}{x}^5$ gives you $x^5=32$ and $x=2$.
Solution for the inequality: $x<0\vee 0<x\le 2$.

$\frac{2x}{x-10}=80$ gives you $2x=80x-800$ and therefore $x=\frac{400}{39}$.
Solution for the inequality: $x<10\vee x>\frac{400}{39}$.

$g\left(x\right)=20x^{2\frac{1}{2}}-100$

${\text{D}}_f=[0,\to ⟩$ and ${\text{B}}_f=[-100,\to ⟩$.

$x^{2\frac{1}{2}}=5$ gives you $x=5^{\frac{2}{5}}$, so $(5^{\frac{2}{5}},0)$.

Graphing calculator: $x\ge \mathrm{1,92}$

$16x^{\frac{1}{4}}=\frac{1}{2}x$ gives you $x^{\frac{3}{4}}=32$ and $x={32}^{\frac{4}{3}}$.
Solution for the inequality: $x\ge {32}^{\frac{4}{3}}$.

${(2x-40)}^{\frac{1}{2}}=40$ gives you $2x-40=1600$ and $x=820$.
Solution for the inequality: $20\le x<820$.

$f\left(x\right)=10x^{-1\frac{1}{2}}+100$ and $g\left(x\right)=10x^{\frac{1}{2}}$

Only the graph of $f$ has asymptotes: $x=0$ and $y=100$.

Graphing calculator: $0<x<\mathrm{100,02}$