\(x^{3} - 16 x^{2} + 33 x = -5 x^{2} + 5 x\)
\(x^{3} - 11 x^{2} + 28 x = 0\)
\(x (x^{2} - 11 x + 28) = 0\)

\(x (x - 4) (x - 7) = 0\)
\(x = 0 ∨ x = 4 ∨ x = 7\)

\(f(x) ≤ g(x)\) geeft \(x ≤ 0 ∨ 4 ≤ x ≤ 7 \text{.}\)

\(f(-2) = 7 \text{.}\)

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

\(x > -2\) geeft \(-2 ≤ f(x) < 7 \text{.}\)

\(5 + 4 \sqrt{2 x + 6} = 21\)
\(4 \sqrt{2 x + 6} = 16\)
\(\sqrt{2 x + 6} = 4\)
\(2 x + 6 = 16\)
\(2 x = 10\)
\(x = 5 \text{.}\)

\(2 x + 6 ≥ 0\)
\(2 x ≥ -6\)
\(x ≥ -3\)
Dus het randpunt is \((-3 , 5) \text{.}\)

\(f(x) < 21\) geeft \(-3 ≤ x < 5 \text{.}\)

\(f(x) = 5\)
\(4 ⋅ {}^{2}\!\log(-3 x + 13) + 5 = 5\)
\(4 ⋅ {}^{2}\!\log(-3 x + 13) = 0\)
\({}^{2}\!\log(-3 x + 13) = 0\)
\(-3 x + 13 = 2^{0} = 1\)
\(-3 x = -12\)
\(x = 4\)

Bereking van het domein geeft
\(-3 x + 13 > 0\)
\(-3 x > -13\)
\(x < 4\frac{1}{3}\)
Dus de verticale asymptoot is de lijn \(x = 4\frac{1}{3} \text{.}\)

\(f(x) < 5\) geeft \(4 < x < 4\frac{1}{3} \text{.}\)

Gelijkstellen geeft
\({2 x + 4 \over -2 x + 3} = -x + 3 \text{.}\)
Kruislings vermenigvuldigen geeft
\(2 x + 4 = (-2 x + 3) (-x + 3)\)
\(2 x + 4 = 2 x^{2} - 6 x - 3 x + 9\)
\(2 x^{2} - 11 x + 5 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-11)^{2} - 4 ⋅ 2 ⋅ 5 = 81\)
dus \(x = {11 + 9 \over 2 ⋅ 2} = 5\) en \(x = {11 - 9 \over 2 ⋅ 2} = \frac{1}{2} \text{.}\)

Noemer gelijkstellen aan \(0\) geeft \(-2 x + 3 = 0 \text{,}\) dus de verticale asymptoot is de lijn \(x = 1\frac{1}{2} \text{.}\)

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