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

\(x(x+1)(x-3)=0\)
\(x=-1∨x=0∨x=3\)

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

\(f(-3)=13\text{.}\)

\(-4x+4≥0\)
\(-4x≥-4\)
\(x≤1\)
Dus het randpunt is \((1, 5)\text{.}\)

\(x≥-3\) geeft \(5≤f(x)≤13\text{.}\)

\(-4-3\sqrt{-2x+2}=-10\)
\(-3\sqrt{-2x+2}=-6\)
\(\sqrt{-2x+2}=2\)
\(-2x+2=4\)
\(-2x=2\)
\(x=-1\text{.}\)

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

\(f(x)≥-10\) geeft \(-1≤x≤1\text{.}\)

\(f(x)=12\)
\(5⋅{}^{4}\!\log(-x+19)+2=12\)
\(5⋅{}^{4}\!\log(-x+19)=10\)
\({}^{4}\!\log(-x+19)=2\)
\(-x+19=4^2=16\)
\(-x=-3\)
\(x=3\)

Bereking van het domein geeft
\(-x+19>0\)
\(-x>-19\)
\(x<19\)
Dus de verticale asymptoot is de lijn \(x=19\text{.}\)

\(f(x)<12\) geeft \(3<x<19\text{.}\)