\(x^3+17x^2-31x=7x^2-7x\)
\(x^3+10x^2-24x=0\)
\(x(x^2+10x-24)=0\)

\(x(x+12)(x-2)=0\)
\(x=-12∨x=0∨x=2\)

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

\(f(-1)=10\text{.}\)

\(2x+6≥0\)
\(2x≥-6\)
\(x≥-3\)
Dus het randpunt is \((-3, 4)\text{.}\)

\(x≤-1\) geeft \(4≤f(x)≤10\text{.}\)

\(-4-3\sqrt{2x+6}=-16\)
\(-3\sqrt{2x+6}=-12\)
\(\sqrt{2x+6}=4\)
\(2x+6=16\)
\(2x=10\)
\(x=5\text{.}\)

\(2x+6≥0\)
\(2x≥-6\)
\(x≥-3\)
Dus het randpunt is \((-3, -4)\text{.}\)

\(f(x)>-16\) geeft \(-3≤x<5\text{.}\)

\(f(x)=6\)
\(5⋅{}^{2}\!\log(-3x+11)+1=6\)
\(5⋅{}^{2}\!\log(-3x+11)=5\)
\({}^{2}\!\log(-3x+11)=1\)
\(-3x+11=2^1=2\)
\(-3x=-9\)
\(x=3\)

Bereking van het domein geeft
\(-3x+11>0\)
\(-3x>-11\)
\(x<3\frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x=3\frac{2}{3}\text{.}\)

\(f(x)≤6\) geeft \(3≤x<3\frac{2}{3}\text{.}\)