Extreme waarden bepalen

\(f'(x) = 2 x^{4} - 2 x^{2} - 12\)

\(f'(\sqrt{3}) = 2 (\sqrt{3})^{4} - 2 (\sqrt{3})^{2} - 12 = 0\)

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

\(f'(x) = 6 x^{2} + 30 x + 36\)

\(f'(x) = 0\) geeft
\(6 x^{2} + 30 x + 36 = 0\)
\(x^{2} + 5 x + 6 = 0\)
\((x + 3) (x + 2) = 0\)
\(x = -3 ∨ x = -2\)

Schets:

max. is \(f(-3) = -63\) en min. is \(f(-2) = -64 \text{.}\)

\(f'(x) = -12 x^{3} + 12 x^{2} + 144 x\)

\(f'(x) = 0\) geeft
\(-12 x^{3} + 12 x^{2} + 144 x = 0\)
\(x^{3} - x^{2} - 12 x = 0\)
\(x (x + 3) (x - 4) = 0\)
\(x = 0 ∨ x = -3 ∨ x = 4\)

Schets:

max. is \(f(-3) = 272 \text{,}\) min. is \(f(0) = -25\) en max. is \(f(4) = 615 \text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(\sqrt{4 x + 3} = 6\)

Kwadrateren geeft
\(4 x + 3 = 36\)
\(x = 8\frac{1}{4}\)

Schets:

max. is \(f(8\frac{1}{4}) = 3\frac{1}{4} \text{.}\)

\(4 x + 3 ≥ 0\) geeft \(x ≥ -\frac{3}{4} \text{,}\) dus \(D_{f} = [-\frac{3}{4} , \rightarrow ⟩ \text{.}\)

max. is \(f(8\frac{1}{4}) = 3\frac{1}{4} \text{,}\) dus \(B_{f} = ⟨\leftarrow , 3\frac{1}{4}] \text{.}\)

Uitdelen geeft
\(f(x) = {5 x^{2} + 9 x + 80 \over x} = {5 x^{2} \over x} + {9 x \over x} + {80 \over x} = 5 x + 9 + 80 x^{-1}\)

De afgeleide is dan
\(f'(x) = 5 + 80 ⋅ -1 ⋅ x^{-2} = 5 - {80 \over x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(5 - {80 \over x^{2}} = 0\)
\(\frac{5}{1} = {80 \over x^{2}}\)

Kruislings vermenigvuldigen geeft
\(5 x^{2} = 80\)
\(x^{2} = 16\)
\(x = \sqrt{16} = 4 ∨ x = -\sqrt{16} = -4\)

Schets:

min. is \(f(-4) = -31\) en max. is \(f(4) = 49 \text{.}\)