\(x^3+2x^2-5x=6x^2-8x\)
\(x^3-4x^2+3x=0\)
\(x(x^2-4x+3)=0\)

\(x(x-1)(x-3)=0\)
\(x=0∨x=1∨x=3\)

\(f(x)≥g(x)\) geeft \(0≤x≤1∨x≥3\text{.}\)

\(f(3)=-17\text{.}\)

\(2x+3≥0\)
\(2x≥-3\)
\(x≥-1\frac{1}{2}\)
Dus het randpunt is \((-1\frac{1}{2}, -5)\text{.}\)

\(x<3\) geeft \(-17<f(x)≤-5\text{.}\)

\(3+4\sqrt{-6x-3}=15\)
\(4\sqrt{-6x-3}=12\)
\(\sqrt{-6x-3}=3\)
\(-6x-3=9\)
\(-6x=12\)
\(x=-2\text{.}\)

\(-6x-3≥0\)
\(-6x≥3\)
\(x≤-\frac{1}{2}\)
Dus het randpunt is \((-\frac{1}{2}, 3)\text{.}\)

\(f(x)<15\) geeft \(-2<x≤-\frac{1}{2}\text{.}\)

\(f(x)=7\)
\(-2⋅{}^{\frac{1}{3}}\!\log(-3x+15)+5=7\)
\(-2⋅{}^{\frac{1}{3}}\!\log(-3x+15)=2\)
\({}^{\frac{1}{3}}\!\log(-3x+15)=-1\)
\(-3x+15=\frac{1}{3}^{-1}=3\)
\(-3x=-12\)
\(x=4\)

Bereking van het domein geeft
\(-3x+15>0\)
\(-3x>-15\)
\(x<5\)
Dus de verticale asymptoot is de lijn \(x=5\text{.}\)

\(f(x)≤7\) geeft \(4≤x<5\text{.}\)