\(x^{3} + 14 x^{2} + 34 x = -6 x^{2} - 2 x\)
\(x^{3} + 20 x^{2} + 36 x = 0\)
\(x (x^{2} + 20 x + 36) = 0\)

\(x (x + 18) (x + 2) = 0\)
\(x = -18 ∨ x = -2 ∨ x = 0\)

\(f(x) ≥ g(x)\) geeft \(-18 ≤ x ≤ -2 ∨ x ≥ 0 \text{.}\)

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

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

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

\(-5 + 3 \sqrt{4 x - 4} = 1\)
\(3 \sqrt{4 x - 4} = 6\)
\(\sqrt{4 x - 4} = 2\)
\(4 x - 4 = 4\)
\(4 x = 8\)
\(x = 2 \text{.}\)

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

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

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

Bereking van het domein geeft
\(3 x - 2 > 0\)
\(3 x > 2\)
\(x > \frac{2}{3}\)
Dus de verticale asymptoot is de lijn \(x = \frac{2}{3} \text{.}\)

\(f(x) > 2\) geeft \(\frac{2}{3} < x < 1 \text{.}\)

Gelijkstellen geeft
\({2 x - 2 \over x - 2} = 2 x - 5 \text{.}\)
Kruislings vermenigvuldigen geeft
\(2 x - 2 = (x - 2) (2 x - 5)\)
\(2 x - 2 = 2 x^{2} - 5 x - 4 x + 10\)
\(2 x^{2} - 11 x + 12 = 0 \text{.}\)

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-11)^{2} - 4 ⋅ 2 ⋅ 12 = 25\)
dus \(x = {11 + 5 \over 2 ⋅ 2} = 4\) en \(x = {11 - 5 \over 2 ⋅ 2} = 1\frac{1}{2} \text{.}\)

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

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