\(f'(x) = -3 x^{2} + 24 x - 45\)

\(f'(x) = 0\) geeft
\(-3 x^{2} + 24 x - 45 = 0\)
\(x^{2} - 8 x + 15 = 0\)
\((x - 3) (x - 5) = 0\)
\(x = 3 ∨ x = 5\)

Schets:

min. is \(f(3) = -8\) en max. is \(f(5) = -4 \text{.}\)

\(f'(x) = -12 x^{3} + 48 x\)

\(f'(x) = 0\) geeft
\(-12 x^{3} + 48 x = 0\)
\(x^{3} - 4 x = 0\)
\(x (x + 2) (x - 2) = 0\)
\(x = 0 ∨ x = -2 ∨ x = 2\)

Schets:

max. is \(f(-2) = 59 \text{,}\) min. is \(f(0) = 11\) en max. is \(f(2) = 59 \text{.}\)

\(f'(x) = 5 x^{4} - 14 x^{2} + 8\)

\(f'(\sqrt{2}) = 5 (\sqrt{2})^{4} - 14 (\sqrt{2})^{2} + 8 = 0\)

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

Uitdelen geeft
\(f(x) = {x^{2} + 12 x + 4 \over 2 x} = {x^{2} \over 2 x} + {12 x \over 2 x} + {4 \over 2 x} = \frac{1}{2} x + 6 + 2 x^{-1}\)

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

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

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

Schets:

min. is \(f(-2) = 4\) en max. is \(f(2) = 8 \text{.}\)

\(f(x) = \sqrt{3 x + 5} - \frac{1}{2} x = (3 x + 5)^{\frac{1}{2}} - \frac{1}{2} x\) geeft
\(f'(x) = \frac{1}{2} ⋅ (3 x + 5)^{-\frac{1}{2}} ⋅ 3 - \frac{1}{2} = {3 \over 2 \sqrt{3 x + 5}} - \frac{1}{2} \text{.}\)

\(f'(x) = 0\) geeft
\({3 \over 2 \sqrt{3 x + 5}} - \frac{1}{2} = 0\)
\({3 \over 2 \sqrt{3 x + 5}} = \frac{1}{2}\)

Kruislings vermenigvuldigen geeft
\(2 \sqrt{3 x + 5} = 6\)
\(\sqrt{3 x + 5} = 3\)

Kwadrateren geeft
\(3 x + 5 = 9\)
\(x = 1\frac{1}{3}\)

Schets:

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

\(3 x + 5 ≥ 0\) geeft \(x ≥ -1\frac{2}{3} \text{,}\) dus \(D_{f} = [-1\frac{2}{3} , \rightarrow ⟩ \text{.}\)

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