\(f'(x) = -6 x^{2} - 18 x + 60\)

\(f'(x) = 0\) geeft
\(-6 x^{2} - 18 x + 60 = 0\)
\(x^{2} + 3 x - 10 = 0\)
\((x + 5) (x - 2) = 0\)
\(x = -5 ∨ x = 2\)

Schets:

min. is \(f(-5) = -243\) en max. is \(f(2) = 100 \text{.}\)

\(f'(x) = -12 x^{3} + 12 x^{2} + 240 x\)

\(f'(x) = 0\) geeft
\(-12 x^{3} + 12 x^{2} + 240 x = 0\)
\(x^{3} - x^{2} - 20 x = 0\)
\(x (x + 4) (x - 5) = 0\)
\(x = 0 ∨ x = -4 ∨ x = 5\)

Schets:

max. is \(f(-4) = 849 \text{,}\) min. is \(f(0) = -47\) en max. is \(f(5) = 1\,578 \text{.}\)

\(f'(x) = -2 x^{4} - x^{2} + 10\)

\(f'(\sqrt{2}) = -2 (\sqrt{2})^{4} - (\sqrt{2})^{2} + 10 = 0\)

Schets:

\(f'(\sqrt{2}) = 0\) en in de schets is te zien dat de grafiek van \(f\) een top heeft voor \(x = \sqrt{2} \text{,}\) dus \(f\) heeft een extreme waarde voor \(x = \sqrt{2} \text{.}\)

Uitdelen geeft
\(f(x) = {x^{2} + 5 x + 4 \over 3 x} = {x^{2} \over 3 x} + {5 x \over 3 x} + {4 \over 3 x} = \frac{1}{3} x + \frac{5}{3} + \frac{4}{3} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{1}{3} + \frac{4}{3} ⋅ -1 ⋅ x^{-2} = \frac{1}{3} - {4 \over 3 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{1}{3} - {4 \over 3 x^{2}} = 0\)
\(\frac{1}{3} = {4 \over 3 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(3 x^{2} = 12\)
\(x^{2} = 4\)
\(x = \sqrt{4} = 2 ∨ x = -\sqrt{4} = -2\)

Schets:

min. is \(f(-2) = \frac{1}{3}\) en max. is \(f(2) = 3 \text{.}\)

\(f(x) = \sqrt{2 x + 3} - \frac{1}{5} x = (2 x + 3)^{\frac{1}{2}} - \frac{1}{5} x\) geeft
\(f'(x) = \frac{1}{2} ⋅ (2 x + 3)^{-\frac{1}{2}} ⋅ 2 - \frac{1}{5} = {1 \over \sqrt{2 x + 3}} - \frac{1}{5} \text{.}\)

\(f'(x) = 0\) geeft
\({1 \over \sqrt{2 x + 3}} - \frac{1}{5} = 0\)
\({1 \over \sqrt{2 x + 3}} = \frac{1}{5}\)

Kruislings vermenigvuldigen geeft
\(\sqrt{2 x + 3} = 5\)

Kwadrateren geeft
\(2 x + 3 = 25\)
\(x = 11\)

Schets:

max. is \(f(11) = 2\frac{4}{5} \text{.}\)

\(2 x + 3 ≥ 0\) geeft \(x ≥ -1\frac{1}{2} \text{,}\) dus \(D_{f} = [-1\frac{1}{2} , \rightarrow ⟩ \text{.}\)

max. is \(f(11) = 2\frac{4}{5} \text{,}\) dus \(B_{f} = ⟨\leftarrow , 2\frac{4}{5}] \text{.}\)