Extreme waarden bepalen

\(f'(x) = 2 x^{4} - 9 x^{2} + 9\)

\(f'(\sqrt{3}) = 2 (\sqrt{3})^{4} - 9 (\sqrt{3})^{2} + 9 = 0\)

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

\(f'(x) = -3 x^{2} - 18 x - 15\)

\(f'(x) = 0\) geeft
\(-3 x^{2} - 18 x - 15 = 0\)
\(x^{2} + 6 x + 5 = 0\)
\((x + 5) (x + 1) = 0\)
\(x = -5 ∨ x = -1\)

Schets:

min. is \(f(-5) = -57\) en max. is \(f(-1) = -25 \text{.}\)

\(f'(x) = -12 x^{3} + 300 x\)

\(f'(x) = 0\) geeft
\(-12 x^{3} + 300 x = 0\)
\(x^{3} - 25 x = 0\)
\(x (x + 5) (x - 5) = 0\)
\(x = 0 ∨ x = -5 ∨ x = 5\)

Schets:

max. is \(f(-5) = 1\,906 \text{,}\) min. is \(f(0) = 31\) en max. is \(f(5) = 1\,906 \text{.}\)

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

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

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

Kwadrateren geeft
\(3 x - 3 = 4\)
\(x = 2\frac{1}{3}\)

Schets:

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

\(3 x - 3 ≥ 0\) geeft \(x ≥ 1 \text{,}\) dus \(D_{f} = [1 , \rightarrow ⟩ \text{.}\)

max. is \(f(2\frac{1}{3}) = \frac{1}{4} \text{,}\) dus \(B_{f} = ⟨\leftarrow , \frac{1}{4}] \text{.}\)

Uitdelen geeft
\(f(x) = {4 x^{2} + 9 x + 25 \over 9 x} = {4 x^{2} \over 9 x} + {9 x \over 9 x} + {25 \over 9 x} = \frac{4}{9} x + 1 + \frac{25}{9} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{4}{9} + \frac{25}{9} ⋅ -1 ⋅ x^{-2} = \frac{4}{9} - {25 \over 9 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{4}{9} - {25 \over 9 x^{2}} = 0\)
\(\frac{4}{9} = {25 \over 9 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(36 x^{2} = 225\)
\(x^{2} = \frac{25}{4}\)
\(x = \sqrt{\frac{25}{4}} = 2\frac{1}{2} ∨ x = -\sqrt{\frac{25}{4}} = -2\frac{1}{2}\)

Schets:

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