\(f(2) = 19 \text{.}\)

\(3 x + 3 ≥ 0\)
\(3 x ≥ -3\)
\(x ≥ -1\)
Dus het randpunt is \((-1 , 4) \text{.}\)

\(x < 2\) geeft \(4 ≤ f(x) < 19 \text{.}\)

\(5 + 3 \sqrt{2 x + 6} = 11\)
\(3 \sqrt{2 x + 6} = 6\)
\(\sqrt{2 x + 6} = 2\)
\(2 x + 6 = 4\)
\(2 x = -2\)
\(x = -1 \text{.}\)

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

\(f(x) ≤ 11\) geeft \(-3 ≤ x ≤ -1 \text{.}\)

\(f(x) = 3\)
\(-3 ⋅ {}^{4}\!\log(2 x + 4) + 6 = 3\)
\(-3 ⋅ {}^{4}\!\log(2 x + 4) = -3\)
\({}^{4}\!\log(2 x + 4) = 1\)
\(2 x + 4 = 4^{1} = 4\)
\(2 x = 0\)
\(x = 0\)

Bereking van het domein geeft
\(2 x + 4 > 0\)
\(2 x > -4\)
\(x > -2\)
Dus de verticale asymptoot is de lijn \(x = -2 \text{.}\)

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