\(x^{3} - 2 x^{2} + 11 x = -6 x^{2} + 8 x\)
\(x^{3} + 4 x^{2} + 3 x = 0\)
\(x (x^{2} + 4 x + 3) = 0\)

\(x (x + 3) (x + 1) = 0\)
\(x = -3 ∨ x = -1 ∨ x = 0\)

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

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

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

\(x ≤ 3\) geeft \(5 ≤ f(x) ≤ 13 \text{.}\)

\(-3 - 4 \sqrt{-4 x - 4} = -11\)
\(-4 \sqrt{-4 x - 4} = -8\)
\(\sqrt{-4 x - 4} = 2\)
\(-4 x - 4 = 4\)
\(-4 x = 8\)
\(x = -2 \text{.}\)

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

\(f(x) ≥ -11\) geeft \(-2 ≤ x ≤ -1 \text{.}\)

\(f(x) = 6\)
\(5 ⋅ {}^{2}\!\log(x - 1) + 6 = 6\)
\(5 ⋅ {}^{2}\!\log(x - 1) = 0\)
\({}^{2}\!\log(x - 1) = 0\)
\(x - 1 = 2^{0} = 1\)
\(x = 2\)

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

\(f(x) < 6\) geeft \(1 < x < 2 \text{.}\)