\(x^3+17x^2-16x=7x^2+8x\)
\(x^3+10x^2-24x=0\)
\(x(x^2+10x-24)=0\)

\(x(x+12)(x-2)=0\)
\(x=-12∨x=0∨x=2\)

\(f(x)≥g(x)\) geeft \(-12≤x≤0∨x≥2\text{.}\)

\(f(-5)=-5\text{.}\)

\(-4x-4≥0\)
\(-4x≥4\)
\(x≤-1\)
Dus het randpunt is \((-1, 3)\text{.}\)

\(x>-5\) geeft \(-5<f(x)≤3\text{.}\)

\(4-5\sqrt{-2x-2}=-6\)
\(-5\sqrt{-2x-2}=-10\)
\(\sqrt{-2x-2}=2\)
\(-2x-2=4\)
\(-2x=6\)
\(x=-3\text{.}\)

\(-2x-2≥0\)
\(-2x≥2\)
\(x≤-1\)
Dus het randpunt is \((-1, 4)\text{.}\)

\(f(x)>-6\) geeft \(-3<x≤-1\text{.}\)

\(f(x)=6\)
\(5⋅{}^{4}\!\log(x+1)+6=6\)
\(5⋅{}^{4}\!\log(x+1)=0\)
\({}^{4}\!\log(x+1)=0\)
\(x+1=4^0=1\)
\(x=0\)

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

\(f(x)≤6\) geeft \(-1<x≤0\text{.}\)