\(x^{3} + 10 x^{2} + 28 x = -2 x^{2} - 7 x\)
\(x^{3} + 12 x^{2} + 35 x = 0\)
\(x (x^{2} + 12 x + 35) = 0\)

\(x (x + 7) (x + 5) = 0\)
\(x = -7 ∨ x = -5 ∨ x = 0\)

\(f(x) ≥ g(x)\) geeft \(-7 ≤ x ≤ -5 ∨ x ≥ 0 \text{.}\)

\(f(-6) = 10 \text{.}\)

\(-2 x + 4 ≥ 0\)
\(-2 x ≥ -4\)
\(x ≤ 2\)
Dus het randpunt is \((2 , -2) \text{.}\)

\(x > -6\) geeft \(-2 ≤ f(x) < 10 \text{.}\)

\(-4 - 5 \sqrt{2 x - 6} = -14\)
\(-5 \sqrt{2 x - 6} = -10\)
\(\sqrt{2 x - 6} = 2\)
\(2 x - 6 = 4\)
\(2 x = 10\)
\(x = 5 \text{.}\)

\(2 x - 6 ≥ 0\)
\(2 x ≥ 6\)
\(x ≥ 3\)
Dus het randpunt is \((3 , -4) \text{.}\)

\(f(x) ≥ -14\) geeft \(3 ≤ x ≤ 5 \text{.}\)

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

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

\(f(x) ≥ -1\) geeft \(0 < x ≤ 1 \text{.}\)

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

De \(a\kern{-.8pt}b\kern{-.8pt}c \text{-}\)formule geeft \(D = (-1)^{2} - 4 ⋅ 2 ⋅ -6 = 49\)
dus \(x = {1 + 7 \over 2 ⋅ 2} = 2\) en \(x = {1 - 7 \over 2 ⋅ 2} = -1\frac{1}{2} \text{.}\)

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

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