\(x^3-8x^2-50x=-7x^2+6x\)
\(x^3-x^2-56x=0\)
\(x(x^2-x-56)=0\)

\(x(x+7)(x-8)=0\)
\(x=-7∨x=0∨x=8\)

\(f(x)≤g(x)\) geeft \(x≤-7∨0≤x≤8\text{.}\)

\(f(-2)=-14\text{.}\)

\(-2x+5≥0\)
\(-2x≥-5\)
\(x≤2\frac{1}{2}\)
Dus het randpunt is \((2\frac{1}{2}, -2)\text{.}\)

\(x≥-2\) geeft \(-14≤f(x)≤-2\text{.}\)

\(-2-4\sqrt{-2x+3}=-14\)
\(-4\sqrt{-2x+3}=-12\)
\(\sqrt{-2x+3}=3\)
\(-2x+3=9\)
\(-2x=6\)
\(x=-3\text{.}\)

\(-2x+3≥0\)
\(-2x≥-3\)
\(x≤1\frac{1}{2}\)
Dus het randpunt is \((1\frac{1}{2}, -2)\text{.}\)

\(f(x)≥-14\) geeft \(-3≤x≤1\frac{1}{2}\text{.}\)

\(f(x)=5\)
\(-3⋅{}^{4}\!\log(-2x+9)+5=5\)
\(-3⋅{}^{4}\!\log(-2x+9)=0\)
\({}^{4}\!\log(-2x+9)=0\)
\(-2x+9=4^0=1\)
\(-2x=-8\)
\(x=4\)

Bereking van het domein geeft
\(-2x+9>0\)
\(-2x>-9\)
\(x<4\frac{1}{2}\)
Dus de verticale asymptoot is de lijn \(x=4\frac{1}{2}\text{.}\)

\(f(x)>5\) geeft \(4<x<4\frac{1}{2}\text{.}\)