\(x^{3} + 5 x^{2} + 15 x = -6 x^{2} - 9 x\)
\(x^{3} + 11 x^{2} + 24 x = 0\)
\(x (x^{2} + 11 x + 24) = 0\)

\(x (x + 8) (x + 3) = 0\)
\(x = -8 ∨ x = -3 ∨ x = 0\)

\(f(x) > g(x)\) geeft \(-8 < x < -3 ∨ x > 0 \text{.}\)

\(f(-1) = -14 \text{.}\)

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

\(x ≤ -1\) geeft \(-14 ≤ f(x) ≤ -4 \text{.}\)

\(-2 - 5 \sqrt{-3 x + 6} = -17\)
\(-5 \sqrt{-3 x + 6} = -15\)
\(\sqrt{-3 x + 6} = 3\)
\(-3 x + 6 = 9\)
\(-3 x = 3\)
\(x = -1 \text{.}\)

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

\(f(x) > -17\) geeft \(-1 < x ≤ 2 \text{.}\)

\(f(x) = 2\)
\(-2 ⋅ {}^{4}\!\log(-3 x + 28) + 6 = 2\)
\(-2 ⋅ {}^{4}\!\log(-3 x + 28) = -4\)
\({}^{4}\!\log(-3 x + 28) = 2\)
\(-3 x + 28 = 4^{2} = 16\)
\(-3 x = -12\)
\(x = 4\)

Bereking van het domein geeft
\(-3 x + 28 > 0\)
\(-3 x > -28\)
\(x < 9\frac{1}{3}\)
Dus de verticale asymptoot is de lijn \(x = 9\frac{1}{3} \text{.}\)

\(f(x) ≥ 2\) geeft \(4 ≤ x < 9\frac{1}{3} \text{.}\)