\(x^{3} + 10 x^{2} - x = 4 x^{2} - 9 x\)
\(x^{3} + 6 x^{2} + 8 x = 0\)
\(x (x^{2} + 6 x + 8) = 0\)

\(x (x + 4) (x + 2) = 0\)
\(x = -4 ∨ x = -2 ∨ x = 0\)

\(f(x) ≤ g(x)\) geeft \(x ≤ -4 ∨ -2 ≤ x ≤ 0 \text{.}\)

\(f(6) = -13 \text{.}\)

\(5 x - 5 ≥ 0\)
\(5 x ≥ 5\)
\(x ≥ 1\)
Dus het randpunt is \((1 , -3) \text{.}\)

\(x ≤ 6\) geeft \(-13 ≤ f(x) ≤ -3 \text{.}\)

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

\(-2 x + 2 ≥ 0\)
\(-2 x ≥ -2\)
\(x ≤ 1\)
Dus het randpunt is \((1 , 3) \text{.}\)

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

\(f(x) = -7\)
\(-3 ⋅ {}^{2}\!\log(x + 16) + 5 = -7\)
\(-3 ⋅ {}^{2}\!\log(x + 16) = -12\)
\({}^{2}\!\log(x + 16) = 4\)
\(x + 16 = 2^{4} = 16\)
\(x = 0\)

Bereking van het domein geeft
\(x + 16 > 0\)
\(x > -16\)
Dus de verticale asymptoot is de lijn \(x = -16 \text{.}\)

\(f(x) > -7\) geeft \(-16 < x < 0 \text{.}\)