\(x^{3} + 13 x^{2} - 30 x = 8 x^{2} + 6 x\)
\(x^{3} + 5 x^{2} - 36 x = 0\)
\(x (x^{2} + 5 x - 36) = 0\)

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

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

\(f(-6) = 14 \text{.}\)

\(-2 x - 3 ≥ 0\)
\(-2 x ≥ 3\)
\(x ≤ -1\frac{1}{2}\)
Dus het randpunt is \((-1\frac{1}{2} , 5) \text{.}\)

\(x > -6\) geeft \(5 ≤ f(x) < 14 \text{.}\)

\(3 + 5 \sqrt{-2 x - 2} = 13\)
\(5 \sqrt{-2 x - 2} = 10\)
\(\sqrt{-2 x - 2} = 2\)
\(-2 x - 2 = 4\)
\(-2 x = 6\)
\(x = -3 \text{.}\)

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

\(f(x) < 13\) geeft \(-3 < x ≤ -1 \text{.}\)

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

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

\(f(x) ≤ 2\frac{1}{2}\) geeft \(0 ≤ x < 1 \text{.}\)

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

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-5)^{2} - 4 ⋅ -2 ⋅ 3 = 49\)
dus \(x = {5 + 7 \over 2 ⋅ -2} = -3\) en \(x = {5 - 7 \over 2 ⋅ -2} = \frac{1}{2} \text{.}\)

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

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