Extreme waarden bepalen

\(f'(x) = 4 x^{4} - 13 x^{2} + 3\)

\(f'(\sqrt{3}) = 4 (\sqrt{3})^{4} - 13 (\sqrt{3})^{2} + 3 = 0\)

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

\(f'(x) = -6 x^{2} + 18 x + 60\)

\(f'(x) = 0\) geeft
\(-6 x^{2} + 18 x + 60 = 0\)
\(x^{2} - 3 x - 10 = 0\)
\((x + 2) (x - 5) = 0\)
\(x = -2 ∨ x = 5\)

Schets:

min. is \(f(-2) = -55\) en max. is \(f(5) = 288 \text{.}\)

\(f'(x) = -12 x^{3} - 72 x^{2} - 60 x\)

\(f'(x) = 0\) geeft
\(-12 x^{3} - 72 x^{2} - 60 x = 0\)
\(x^{3} + 6 x^{2} + 5 x = 0\)
\(x (x + 5) (x + 1) = 0\)
\(x = 0 ∨ x = -5 ∨ x = -1\)

Schets:

max. is \(f(-5) = 345 \text{,}\) min. is \(f(-1) = -39\) en max. is \(f(0) = -30 \text{.}\)

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

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

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

Kwadrateren geeft
\(4 x + 2 = 16\)
\(x = 3\frac{1}{2}\)

Schets:

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

\(4 x + 2 ≥ 0\) geeft \(x ≥ -\frac{1}{2} \text{,}\) dus \(D_{f} = [-\frac{1}{2} , \rightarrow ⟩ \text{.}\)

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

Uitdelen geeft
\(f(x) = {3 x^{2} + 5 x + 48 \over 9 x} = {3 x^{2} \over 9 x} + {5 x \over 9 x} + {48 \over 9 x} = \frac{1}{3} x + \frac{5}{9} + \frac{16}{3} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{1}{3} + \frac{16}{3} ⋅ -1 ⋅ x^{-2} = \frac{1}{3} - {16 \over 3 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{1}{3} - {16 \over 3 x^{2}} = 0\)
\(\frac{1}{3} = {16 \over 3 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(3 x^{2} = 48\)
\(x^{2} = 16\)
\(x = \sqrt{16} = 4 ∨ x = -\sqrt{16} = -4\)

Schets:

min. is \(f(-4) = -2\frac{1}{9}\) en max. is \(f(4) = 3\frac{2}{9} \text{.}\)