\(x^{3} - 26 x = x^{2} + 4 x\)
\(x^{3} - x^{2} - 30 x = 0\)
\(x (x^{2} - x - 30) = 0\)

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

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

\(f(4) = -13 \text{.}\)

\(3 x - 3 ≥ 0\)
\(3 x ≥ 3\)
\(x ≥ 1\)
Dus het randpunt is \((1 , 2) \text{.}\)

\(x < 4\) geeft \(-13 < f(x) ≤ 2 \text{.}\)

\(4 + 5 \sqrt{-6 x + 3} = 19\)
\(5 \sqrt{-6 x + 3} = 15\)
\(\sqrt{-6 x + 3} = 3\)
\(-6 x + 3 = 9\)
\(-6 x = 6\)
\(x = -1 \text{.}\)

\(-6 x + 3 ≥ 0\)
\(-6 x ≥ -3\)
\(x ≤ \frac{1}{2}\)
Dus het randpunt is \((\frac{1}{2} , 4) \text{.}\)

\(f(x) < 19\) geeft \(-1 < x ≤ \frac{1}{2} \text{.}\)

\(f(x) = 9\)
\(4 ⋅ {}^{3}\!\log(2 x - 5) + 5 = 9\)
\(4 ⋅ {}^{3}\!\log(2 x - 5) = 4\)
\({}^{3}\!\log(2 x - 5) = 1\)
\(2 x - 5 = 3^{1} = 3\)
\(2 x = 8\)
\(x = 4\)

Bereking van het domein geeft
\(2 x - 5 > 0\)
\(2 x > 5\)
\(x > 2\frac{1}{2}\)
Dus de verticale asymptoot is de lijn \(x = 2\frac{1}{2} \text{.}\)

\(f(x) ≤ 9\) geeft \(2\frac{1}{2} < x ≤ 4 \text{.}\)

Gelijkstellen geeft
\({4 x - 1 \over 2 x - 3} = -x - 2 \text{.}\)
Kruislings vermenigvuldigen geeft
\(4 x - 1 = (2 x - 3) (-x - 2)\)
\(4 x - 1 = -2 x^{2} - 4 x + 3 x + 6\)
\(-2 x^{2} - 5 x + 7 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-5)^{2} - 4 ⋅ -2 ⋅ 7 = 81\)
dus \(x = {5 + 9 \over 2 ⋅ -2} = -3\frac{1}{2}\) en \(x = {5 - 9 \over 2 ⋅ -2} = 1 \text{.}\)

Noemer gelijkstellen aan \(0\) geeft \(2 x - 3 = 0 \text{,}\) dus de verticale asymptoot is de lijn \(x = 1\frac{1}{2} \text{.}\)

\(f(x) ≥ g(x)\) geeft \(-3\frac{1}{2} ≤ x ≤ 1 ∨ x > 1\frac{1}{2} \text{.}\)