\(x^{3} + 7 x^{2} - 14 x = 6 x^{2} - 8 x\)
\(x^{3} + x^{2} - 6 x = 0\)
\(x (x^{2} + x - 6) = 0\)

\(x (x + 3) (x - 2) = 0\)
\(x = -3 ∨ x = 0 ∨ x = 2\)

\(f(x) ≥ g(x)\) geeft \(-3 ≤ x ≤ 0 ∨ x ≥ 2 \text{.}\)

\(f(4) = 13 \text{.}\)

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

\(x < 4\) geeft \(3 ≤ f(x) < 13 \text{.}\)

\(5 - 4 \sqrt{-6 x + 3} = -7\)
\(-4 \sqrt{-6 x + 3} = -12\)
\(\sqrt{-6 x + 3} = 3\)
\(-6 x + 3 = 9\)
\(-6 x = 6\)
\(x = -1 \text{.}\)

\(-6 x + 3 ≥ 0\)
\(-6 x ≥ -3\)
\(x ≤ \frac{1}{2}\)
Dus het randpunt is \((\frac{1}{2} , 5) \text{.}\)

\(f(x) ≥ -7\) geeft \(-1 ≤ x ≤ \frac{1}{2} \text{.}\)

\(f(x) = 1\)
\(-5 ⋅ {}^{3}\!\log(2 x - 1) + 6 = 1\)
\(-5 ⋅ {}^{3}\!\log(2 x - 1) = -5\)
\({}^{3}\!\log(2 x - 1) = 1\)
\(2 x - 1 = 3^{1} = 3\)
\(2 x = 4\)
\(x = 2\)

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

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