\(x^{3} - 10 x^{2} + 2 x = -6 x^{2} + 7 x\)
\(x^{3} - 4 x^{2} - 5 x = 0\)
\(x (x^{2} - 4 x - 5) = 0\)

\(x (x + 1) (x - 5) = 0\)
\(x = -1 ∨ x = 0 ∨ x = 5\)

\(f(x) > g(x)\) geeft \(-1 < x < 0 ∨ x > 5 \text{.}\)

\(f(3) = 11 \text{.}\)

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

\(x < 3\) geeft \(3 ≤ f(x) < 11 \text{.}\)

\(5 - 3 \sqrt{4 x - 4} = -1\)
\(-3 \sqrt{4 x - 4} = -6\)
\(\sqrt{4 x - 4} = 2\)
\(4 x - 4 = 4\)
\(4 x = 8\)
\(x = 2 \text{.}\)

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

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

\(f(x) = -11\)
\(-5 ⋅ {}^{2}\!\log(-3 x + 20) + 4 = -11\)
\(-5 ⋅ {}^{2}\!\log(-3 x + 20) = -15\)
\({}^{2}\!\log(-3 x + 20) = 3\)
\(-3 x + 20 = 2^{3} = 8\)
\(-3 x = -12\)
\(x = 4\)

Bereking van het domein geeft
\(-3 x + 20 > 0\)
\(-3 x > -20\)
\(x < 6\frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x = 6\frac{2}{3} \text{.}\)

\(f(x) > -11\) geeft \(4 < x < 6\frac{2}{3} \text{.}\)

Gelijkstellen geeft
\({2 x - 2 \over 2 x - 5} = x - 4 \text{.}\)
Kruislings vermenigvuldigen geeft
\(2 x - 2 = (2 x - 5) (x - 4)\)
\(2 x - 2 = 2 x^{2} - 8 x - 5 x + 20\)
\(2 x^{2} - 15 x + 22 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-15)^{2} - 4 ⋅ 2 ⋅ 22 = 49\)
dus \(x = {15 + 7 \over 2 ⋅ 2} = 5\frac{1}{2}\) en \(x = {15 - 7 \over 2 ⋅ 2} = 2 \text{.}\)

Noemer gelijkstellen aan \(0\) geeft \(2 x - 5 = 0 \text{,}\) dus de verticale asymptoot is de lijn \(x = 2\frac{1}{2} \text{.}\)

\(f(x) ≥ g(x)\) geeft \(x ≤ 2 ∨ 2\frac{1}{2} < x ≤ 5\frac{1}{2} \text{.}\)