\(x^3-6x^2+46x=7x^2+4x\)
\(x^3-13x^2+42x=0\)
\(x(x^2-13x+42)=0\)

\(x(x-6)(x-7)=0\)
\(x=0∨x=6∨x=7\)

\(f(x)>g(x)\) geeft \(0<x<6∨x>7\text{.}\)

\(f(-1)=-7\text{.}\)

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

\(x≤-1\) geeft \(-7≤f(x)≤3\text{.}\)

\(5+3\sqrt{2x-2}=11\)
\(3\sqrt{2x-2}=6\)
\(\sqrt{2x-2}=2\)
\(2x-2=4\)
\(2x=6\)
\(x=3\text{.}\)

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

\(f(x)≤11\) geeft \(1≤x≤3\text{.}\)

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

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

\(f(x)≥5\) geeft \(1<x≤2\text{.}\)