\(f(5) = -11 \text{.}\)

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

\(x ≤ 5\) geeft \(-11 ≤ f(x) ≤ 4 \text{.}\)

\(3 - 4 \sqrt{2 x - 6} = -5\)
\(-4 \sqrt{2 x - 6} = -8\)
\(\sqrt{2 x - 6} = 2\)
\(2 x - 6 = 4\)
\(2 x = 10\)
\(x = 5 \text{.}\)

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

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

\(f(x) = -19\)
\(-5 ⋅ {}^{2}\!\log(3 x + 4) + 1 = -19\)
\(-5 ⋅ {}^{2}\!\log(3 x + 4) = -20\)
\({}^{2}\!\log(3 x + 4) = 4\)
\(3 x + 4 = 2^{4} = 16\)
\(3 x = 12\)
\(x = 4\)

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

\(f(x) > -19\) geeft \(-1\frac{1}{3} < x < 4 \text{.}\)