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

\(f'(x) = 0\) geeft
\(-3 x^{2} + 3 = 0\)
\(x^{2} - 1 = 0\)
\((x + 1) (x - 1) = 0\)
\(x = -1 ∨ x = 1\)

Schets:

min. is \(f(-1) = -37\) en max. is \(f(1) = -33 \text{.}\)

\(f'(x) = 12 x^{3} + 36 x^{2} - 120 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} + 36 x^{2} - 120 x = 0\)
\(x^{3} + 3 x^{2} - 10 x = 0\)
\(x (x + 5) (x - 2) = 0\)
\(x = 0 ∨ x = -5 ∨ x = 2\)

Schets:

min. is \(f(-5) = -1\,158 \text{,}\) max. is \(f(0) = -33\) en min. is \(f(2) = -129 \text{.}\)

\(f'(x) = 3 x^{4} - 5 x^{2} - 12\)

\(f'(\sqrt{3}) = 3 (\sqrt{3})^{4} - 5 (\sqrt{3})^{2} - 12 = 0\)

Schets:

\(f'(\sqrt{3}) = 0\) en in de schets is te zien dat de grafiek van \(f\) een top heeft voor \(x = \sqrt{3} \text{,}\) dus \(f\) heeft een extreme waarde voor \(x = \sqrt{3} \text{.}\)

\(f(x) = \frac{1}{5} x - \sqrt{4 x - 5} = \frac{1}{5} x - (4 x - 5)^{\frac{1}{2}}\) geeft
\(f'(x) = \frac{1}{5} - \frac{1}{2} ⋅ (4 x - 5)^{-\frac{1}{2}} ⋅ 4 = \frac{1}{5} - {2 \over \sqrt{4 x - 5}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{1}{5} - {2 \over \sqrt{4 x - 5}} = 0\)
\(-{2 \over \sqrt{4 x - 5}} = -\frac{1}{5}\)

Kruislings vermenigvuldigen geeft
\(\sqrt{4 x - 5} = 10\)

Kwadrateren geeft
\(4 x - 5 = 100\)
\(x = 26\frac{1}{4}\)

Schets:

min. is \(f(26\frac{1}{4}) = -4\frac{3}{4} \text{.}\)

\(4 x - 5 ≥ 0\) geeft \(x ≥ 1\frac{1}{4} \text{,}\) dus \(D_{f} = [1\frac{1}{4} , \rightarrow ⟩ \text{.}\)

min. is \(f(26\frac{1}{4}) = -4\frac{3}{4} \text{,}\) dus \(B_{f} = [-4\frac{3}{4} , \rightarrow ⟩ \text{.}\)

Uitdelen geeft
\(f(x) = {2 x^{2} + 8 \over 5 x} = {2 x^{2} \over 5 x} + {8 \over 5 x} = \frac{2}{5} x + \frac{8}{5} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{2}{5} + \frac{8}{5} ⋅ -1 ⋅ x^{-2} = \frac{2}{5} - {8 \over 5 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{2}{5} - {8 \over 5 x^{2}} = 0\)
\(\frac{2}{5} = {8 \over 5 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(10 x^{2} = 40\)
\(x^{2} = 4\)
\(x = \sqrt{4} = 2 ∨ x = -\sqrt{4} = -2\)

Schets:

min. is \(f(-2) = -1\frac{3}{5}\) en max. is \(f(2) = 1\frac{3}{5} \text{.}\)