\(x^{3} - 7 x^{2} - 69 x = -3 x^{2} - 9 x\)
\(x^{3} - 4 x^{2} - 60 x = 0\)
\(x (x^{2} - 4 x - 60) = 0\)

\(x (x + 6) (x - 10) = 0\)
\(x = -6 ∨ x = 0 ∨ x = 10\)

\(f(x) ≥ g(x)\) geeft \(-6 ≤ x ≤ 0 ∨ x ≥ 10 \text{.}\)

\(f(2) = 12 \text{.}\)

\(4 x - 4 ≥ 0\)
\(4 x ≥ 4\)
\(x ≥ 1\)
Dus het randpunt is \((1 , 2) \text{.}\)

\(x ≤ 2\) geeft \(2 ≤ f(x) ≤ 12 \text{.}\)

\(-5 - 3 \sqrt{-4 x - 4} = -11\)
\(-3 \sqrt{-4 x - 4} = -6\)
\(\sqrt{-4 x - 4} = 2\)
\(-4 x - 4 = 4\)
\(-4 x = 8\)
\(x = -2 \text{.}\)

\(-4 x - 4 ≥ 0\)
\(-4 x ≥ 4\)
\(x ≤ -1\)
Dus het randpunt is \((-1 , -5) \text{.}\)

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

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

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

\(f(x) > 6\) geeft \(\frac{2}{3} < x < 1 \text{.}\)

Gelijkstellen geeft
\({4 x - 2 \over -2 x + 3} = -x + 3 \text{.}\)
Kruislings vermenigvuldigen geeft
\(4 x - 2 = (-2 x + 3) (-x + 3)\)
\(4 x - 2 = 2 x^{2} - 6 x - 3 x + 9\)
\(2 x^{2} - 13 x + 11 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-13)^{2} - 4 ⋅ 2 ⋅ 11 = 81\)
dus \(x = {13 + 9 \over 2 ⋅ 2} = 5\frac{1}{2}\) en \(x = {13 - 9 \over 2 ⋅ 2} = 1 \text{.}\)

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 \(x < 1 ∨ 1\frac{1}{2} < x < 5\frac{1}{2} \text{.}\)