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

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

Schets:

max. is \(f(-2) = -23\) en min. is \(f(2) = -55 \text{.}\)

\(f'(x) = 12 x^{3} - 108 x^{2} + 240 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} - 108 x^{2} + 240 x = 0\)
\(x^{3} - 9 x^{2} + 20 x = 0\)
\(x (x - 4) (x - 5) = 0\)
\(x = 0 ∨ x = 4 ∨ x = 5\)

Schets:

min. is \(f(0) = -10 \text{,}\) max. is \(f(4) = 374\) en min. is \(f(5) = 365 \text{.}\)

\(f'(x) = -4 x^{4} + 3 x^{2} + 10\)

\(f'(\sqrt{2}) = -4 (\sqrt{2})^{4} + 3 (\sqrt{2})^{2} + 10 = 0\)

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
\(f(x) = {4 x^{2} + x + 25 \over 3 x} = {4 x^{2} \over 3 x} + {x \over 3 x} + {25 \over 3 x} = \frac{4}{3} x + \frac{1}{3} + \frac{25}{3} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{4}{3} + \frac{25}{3} ⋅ -1 ⋅ x^{-2} = \frac{4}{3} - {25 \over 3 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{4}{3} - {25 \over 3 x^{2}} = 0\)
\(\frac{4}{3} = {25 \over 3 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(12 x^{2} = 75\)
\(x^{2} = \frac{25}{4}\)
\(x = \sqrt{\frac{25}{4}} = 2\frac{1}{2} ∨ x = -\sqrt{\frac{25}{4}} = -2\frac{1}{2}\)

Schets:

min. is \(f(-2\frac{1}{2}) = -6\frac{1}{3}\) en max. is \(f(2\frac{1}{2}) = 7 \text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(\sqrt{2 x} = 4\)

Kwadrateren geeft
\(2 x = 16\)
\(x = 8\)

Schets:

max. is \(f(8) = 2 \text{.}\)

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

max. is \(f(8) = 2 \text{,}\) dus \(B_{f} = ⟨\leftarrow , 2] \text{.}\)