\(f'(x) = -6 x^{2} - 6 x + 36\)

\(f'(x) = 0\) geeft
\(-6 x^{2} - 6 x + 36 = 0\)
\(x^{2} + x - 6 = 0\)
\((x + 3) (x - 2) = 0\)
\(x = -3 ∨ x = 2\)

Schets:

min. is \(f(-3) = -69\) en max. is \(f(2) = 56 \text{.}\)

\(f'(x) = 12 x^{3} - 72 x^{2} + 96 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} - 72 x^{2} + 96 x = 0\)
\(x^{3} - 6 x^{2} + 8 x = 0\)
\(x (x - 2) (x - 4) = 0\)
\(x = 0 ∨ x = 2 ∨ x = 4\)

Schets:

min. is \(f(0) = 48 \text{,}\) max. is \(f(2) = 96\) en min. is \(f(4) = 48 \text{.}\)

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

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

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 \(x = \sqrt{2} \text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(2 \sqrt{x + 5} = 4\)
\(\sqrt{x + 5} = 2\)

Kwadrateren geeft
\(x + 5 = 4\)
\(x = -1\)

Schets:

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

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

max. is \(f(-1) = 2\frac{1}{4} \text{,}\) dus \(B_{f} = ⟨\leftarrow , 2\frac{1}{4}] \text{.}\)

Uitdelen geeft
\(f(x) = {5 x^{2} + 8 x + 45 \over x} = {5 x^{2} \over x} + {8 x \over x} + {45 \over x} = 5 x + 8 + 45 x^{-1}\)

De afgeleide is dan
\(f'(x) = 5 + 45 ⋅ -1 ⋅ x^{-2} = 5 - {45 \over x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(5 - {45 \over x^{2}} = 0\)
\(\frac{5}{1} = {45 \over x^{2}}\)

Kruislings vermenigvuldigen geeft
\(5 x^{2} = 45\)
\(x^{2} = 9\)
\(x = \sqrt{9} = 3 ∨ x = -\sqrt{9} = -3\)

Schets:

min. is \(f(-3) = -22\) en max. is \(f(3) = 38 \text{.}\)