Extreme waarden bepalen

\(f'(x) = -5 x^{4} + 6 x^{2} + 8\)

\(f'(\sqrt{2}) = -5 (\sqrt{2})^{4} + 6 (\sqrt{2})^{2} + 8 = 0\)

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

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

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

Schets:

min. is \(f(-2) = -27\) en max. is \(f(3) = 98 \text{.}\)

\(f'(x) = 12 x^{3} - 24 x^{2} - 180 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} - 24 x^{2} - 180 x = 0\)
\(x^{3} - 2 x^{2} - 15 x = 0\)
\(x (x + 3) (x - 5) = 0\)
\(x = 0 ∨ x = -3 ∨ x = 5\)

Schets:

min. is \(f(-3) = -307 \text{,}\) max. is \(f(0) = 44\) en min. is \(f(5) = -1\,331 \text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(\sqrt{4 x - 4} = 8\)

Kwadrateren geeft
\(4 x - 4 = 64\)
\(x = 17\)

Schets:

min. is \(f(17) = -3\frac{3}{4} \text{.}\)

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

min. is \(f(17) = -3\frac{3}{4} \text{,}\) dus \(B_{f} = [-3\frac{3}{4} , \rightarrow ⟩ \text{.}\)

Uitdelen geeft
\(f(x) = {4 x^{2} + 10 x + 1 \over 9 x} = {4 x^{2} \over 9 x} + {10 x \over 9 x} + {1 \over 9 x} = \frac{4}{9} x + \frac{10}{9} + \frac{1}{9} x^{-1}\)

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

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

Kruislings vermenigvuldigen geeft
\(36 x^{2} = 9\)
\(x^{2} = \frac{1}{4}\)
\(x = \sqrt{\frac{1}{4}} = \frac{1}{2} ∨ x = -\sqrt{\frac{1}{4}} = -\frac{1}{2}\)

Schets:

min. is \(f(-\frac{1}{2}) = \frac{2}{3}\) en max. is \(f(\frac{1}{2}) = 1\frac{5}{9} \text{.}\)