\(f'(x) = 3 x^{2} - 6 x - 9\)

\(f'(x) = 0\) geeft
\(3 x^{2} - 6 x - 9 = 0\)
\(x^{2} - 2 x - 3 = 0\)
\((x + 1) (x - 3) = 0\)
\(x = -1 ∨ x = 3\)

Schets:

max. is \(f(-1) = 18\) en min. is \(f(3) = -14 \text{.}\)

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

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

Schets:

max. is \(f(-5) = -418 \text{,}\) min. is \(f(-4) = -427\) en max. is \(f(0) = -43 \text{.}\)

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

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

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
\(f(x) = {x^{2} + 9 x + 4 \over 9 x} = {x^{2} \over 9 x} + {9 x \over 9 x} + {4 \over 9 x} = \frac{1}{9} x + 1 + \frac{4}{9} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{1}{9} + \frac{4}{9} ⋅ -1 ⋅ x^{-2} = \frac{1}{9} - {4 \over 9 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{1}{9} - {4 \over 9 x^{2}} = 0\)
\(\frac{1}{9} = {4 \over 9 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(9 x^{2} = 36\)
\(x^{2} = 4\)
\(x = \sqrt{4} = 2 ∨ x = -\sqrt{4} = -2\)

Schets:

min. is \(f(-2) = \frac{5}{9}\) en max. is \(f(2) = 1\frac{4}{9} \text{.}\)

\(f(x) = \frac{1}{4} x - \sqrt{3 x} = \frac{1}{4} x - (3 x)^{\frac{1}{2}}\) geeft
\(f'(x) = \frac{1}{4} - \frac{1}{2} ⋅ (3 x)^{-\frac{1}{2}} ⋅ 3 = \frac{1}{4} - {3 \over 2 \sqrt{3 x}} \text{.}\)

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

Kruislings vermenigvuldigen geeft
\(2 \sqrt{3 x} = 12\)
\(\sqrt{3 x} = 6\)

Kwadrateren geeft
\(3 x = 36\)
\(x = 12\)

Schets:

min. is \(f(12) = -3 \text{.}\)

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

min. is \(f(12) = -3 \text{,}\) dus \(B_{f} = [-3 , \rightarrow ⟩ \text{.}\)