\(x^{3} - 8 x^{2} + 37 x = 3 x^{2} + 7 x\)
\(x^{3} - 11 x^{2} + 30 x = 0\)
\(x (x^{2} - 11 x + 30) = 0\)

\(x (x - 5) (x - 6) = 0\)
\(x = 0 ∨ x = 5 ∨ x = 6\)

\(f(x) ≥ g(x)\) geeft \(0 ≤ x ≤ 5 ∨ x ≥ 6 \text{.}\)

\(f(-5) = -7 \text{.}\)

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

\(x ≥ -5\) geeft \(-7 ≤ f(x) ≤ -3 \text{.}\)

\(3 - 4 \sqrt{2 x - 4} = -5\)
\(-4 \sqrt{2 x - 4} = -8\)
\(\sqrt{2 x - 4} = 2\)
\(2 x - 4 = 4\)
\(2 x = 8\)
\(x = 4 \text{.}\)

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

\(f(x) > -5\) geeft \(2 ≤ x < 4 \text{.}\)

\(f(x) = -5\)
\(4 ⋅ {}^{\frac{1}{4}}\!\log(-3 x + 8) + 1 = -5\)
\(4 ⋅ {}^{\frac{1}{4}}\!\log(-3 x + 8) = -6\)
\({}^{\frac{1}{4}}\!\log(-3 x + 8) = -1\frac{1}{2}\)
\(-3 x + 8 = \frac{1}{4}^{-1\frac{1}{2}} = 8\)
\(-3 x = 0\)
\(x = 0\)

Bereking van het domein geeft
\(-3 x + 8 > 0\)
\(-3 x > -8\)
\(x < 2\frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x = 2\frac{2}{3} \text{.}\)

\(f(x) > -5\) geeft \(0 < x < 2\frac{2}{3} \text{.}\)