\(f'(x) = -3 x^{2} + 48\)

\(f'(x) = 0\) geeft
\(-3 x^{2} + 48 = 0\)
\(x^{2} - 16 = 0\)
\((x + 4) (x - 4) = 0\)
\(x = -4 ∨ x = 4\)

Schets:

min. is \(f(-4) = -176\) en max. is \(f(4) = 80 \text{.}\)

\(f'(x) = 12 x^{3} + 36 x^{2} + 24 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} + 36 x^{2} + 24 x = 0\)
\(x^{3} + 3 x^{2} + 2 x = 0\)
\(x (x + 2) (x + 1) = 0\)
\(x = 0 ∨ x = -2 ∨ x = -1\)

Schets:

min. is \(f(-2) = -44 \text{,}\) max. is \(f(-1) = -41\) en min. is \(f(0) = -44 \text{.}\)

\(f'(x) = x^{4} + 2 x^{2} - 15\)

\(f'(\sqrt{3}) = (\sqrt{3})^{4} + 2 (\sqrt{3})^{2} - 15 = 0\)

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 \(x = \sqrt{3} \text{.}\)

Uitdelen geeft
\(f(x) = {5 x^{2} + x + 20 \over 4 x} = {5 x^{2} \over 4 x} + {x \over 4 x} + {20 \over 4 x} = \frac{5}{4} x + \frac{1}{4} + 5 x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{5}{4} + 5 ⋅ -1 ⋅ x^{-2} = \frac{5}{4} - {5 \over x^{2}} \text{.}\)

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

Kruislings vermenigvuldigen geeft
\(5 x^{2} = 20\)
\(x^{2} = 4\)
\(x = \sqrt{4} = 2 ∨ x = -\sqrt{4} = -2\)

Schets:

min. is \(f(-2) = -4\frac{3}{4}\) en max. is \(f(2) = 5\frac{1}{4} \text{.}\)

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

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

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

Kwadrateren geeft
\(4 x = 100\)
\(x = 25\)

Schets:

min. is \(f(25) = -5 \text{.}\)

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

min. is \(f(25) = -5 \text{,}\) dus \(B_{f} = [-5 , \rightarrow ⟩ \text{.}\)