\(x^{3} - 3 x^{2} + 21 x = 8 x^{2} - 3 x\)
\(x^{3} - 11 x^{2} + 24 x = 0\)
\(x (x^{2} - 11 x + 24) = 0\)

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

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

\(f(1) = -13 \text{.}\)

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

\(x ≤ 1\) geeft \(-13 ≤ f(x) ≤ -5 \text{.}\)

\(-3 - 5 \sqrt{-4 x + 4} = -23\)
\(-5 \sqrt{-4 x + 4} = -20\)
\(\sqrt{-4 x + 4} = 4\)
\(-4 x + 4 = 16\)
\(-4 x = 12\)
\(x = -3 \text{.}\)

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

\(f(x) ≥ -23\) geeft \(-3 ≤ x ≤ 1 \text{.}\)

\(f(x) = 10\)
\(4 ⋅ {}^{3}\!\log(x + 2) + 6 = 10\)
\(4 ⋅ {}^{3}\!\log(x + 2) = 4\)
\({}^{3}\!\log(x + 2) = 1\)
\(x + 2 = 3^{1} = 3\)
\(x = 1\)

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

\(f(x) < 10\) geeft \(-2 < x < 1 \text{.}\)