\(x^{3} - 14 x^{2} + 2 x = -8 x^{2} - 3 x\)
\(x^{3} - 6 x^{2} + 5 x = 0\)
\(x (x^{2} - 6 x + 5) = 0\)

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

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

\(f(-2) = 1 \text{.}\)

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

\(x ≥ -2\) geeft \(1 ≤ f(x) ≤ 5 \text{.}\)

\(-4 - 5 \sqrt{-3 x - 3} = -19\)
\(-5 \sqrt{-3 x - 3} = -15\)
\(\sqrt{-3 x - 3} = 3\)
\(-3 x - 3 = 9\)
\(-3 x = 12\)
\(x = -4 \text{.}\)

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

\(f(x) > -19\) geeft \(-4 < x ≤ -1 \text{.}\)

\(f(x) = -9\)
\(5 ⋅ {}^{\frac{1}{2}}\!\log(3 x + 8) + 6 = -9\)
\(5 ⋅ {}^{\frac{1}{2}}\!\log(3 x + 8) = -15\)
\({}^{\frac{1}{2}}\!\log(3 x + 8) = -3\)
\(3 x + 8 = \frac{1}{2}^{-3} = 8\)
\(3 x = 0\)
\(x = 0\)

Bereking van het domein geeft
\(3 x + 8 > 0\)
\(3 x > -8\)
\(x > -2\frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x = -2\frac{2}{3} \text{.}\)

\(f(x) > -9\) geeft \(-2\frac{2}{3} < x < 0 \text{.}\)