\(x^{3} + x^{2} - 45 x = -4 x^{2} - 9 x\)
\(x^{3} + 5 x^{2} - 36 x = 0\)
\(x (x^{2} + 5 x - 36) = 0\)

\(x (x + 9) (x - 4) = 0\)
\(x = -9 ∨ x = 0 ∨ x = 4\)

\(f(x) ≤ g(x)\) geeft \(x ≤ -9 ∨ 0 ≤ x ≤ 4 \text{.}\)

\(f(5) = 10 \text{.}\)

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

\(x < 5\) geeft \(4 ≤ f(x) < 10 \text{.}\)

\(-5 - 3 \sqrt{-3 x - 6} = -14\)
\(-3 \sqrt{-3 x - 6} = -9\)
\(\sqrt{-3 x - 6} = 3\)
\(-3 x - 6 = 9\)
\(-3 x = 15\)
\(x = -5 \text{.}\)

\(-3 x - 6 ≥ 0\)
\(-3 x ≥ 6\)
\(x ≤ -2\)
Dus het randpunt is \((-2 , -5) \text{.}\)

\(f(x) > -14\) geeft \(-5 < x ≤ -2 \text{.}\)

\(f(x) = 6\)
\(-2 ⋅ {}^{\frac{1}{4}}\!\log(-3 x + 14) + 5 = 6\)
\(-2 ⋅ {}^{\frac{1}{4}}\!\log(-3 x + 14) = 1\)
\({}^{\frac{1}{4}}\!\log(-3 x + 14) = -\frac{1}{2}\)
\(-3 x + 14 = \frac{1}{4}^{-\frac{1}{2}} = 2\)
\(-3 x = -12\)
\(x = 4\)

Bereking van het domein geeft
\(-3 x + 14 > 0\)
\(-3 x > -14\)
\(x < 4\frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x = 4\frac{2}{3} \text{.}\)

\(f(x) < 6\) geeft \(4 < x < 4\frac{2}{3} \text{.}\)