\(x^{3} + 10 x^{2} - 38 x = 4 x^{2} + 2 x\)
\(x^{3} + 6 x^{2} - 40 x = 0\)
\(x (x^{2} + 6 x - 40) = 0\)

\(x (x + 10) (x - 4) = 0\)
\(x = -10 ∨ x = 0 ∨ x = 4\)

\(f(x) > g(x)\) geeft \(-10 < x < 0 ∨ x > 4 \text{.}\)

\(f(5) = -7 \text{.}\)

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

\(x < 5\) geeft \(-7 < f(x) ≤ 5 \text{.}\)

\(5 + 3 \sqrt{2 x + 6} = 17\)
\(3 \sqrt{2 x + 6} = 12\)
\(\sqrt{2 x + 6} = 4\)
\(2 x + 6 = 16\)
\(2 x = 10\)
\(x = 5 \text{.}\)

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

\(f(x) < 17\) geeft \(-3 ≤ x < 5 \text{.}\)

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

Bereking van het domein geeft
\(2 x + 1 > 0\)
\(2 x > -1\)
\(x > -\frac{1}{2}\)
Dus de verticale asymptoot is de lijn \(x = -\frac{1}{2} \text{.}\)

\(f(x) < 4\) geeft \(-\frac{1}{2} < x < 0 \text{.}\)

Gelijkstellen geeft
\({2 x + 1 \over x - 1} = 2 x - 5 \text{.}\)
Kruislings vermenigvuldigen geeft
\(2 x + 1 = (x - 1) (2 x - 5)\)
\(2 x + 1 = 2 x^{2} - 5 x - 2 x + 5\)
\(2 x^{2} - 9 x + 4 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-9)^{2} - 4 ⋅ 2 ⋅ 4 = 49\)
dus \(x = {9 + 7 \over 2 ⋅ 2} = 4\) en \(x = {9 - 7 \over 2 ⋅ 2} = \frac{1}{2} \text{.}\)

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

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