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

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

Schets:

max. is \(f(-5) = 303\) en min. is \(f(2) = -40 \text{.}\)

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

\(f'(x) = 0\) geeft
\(12 x^{3} + 12 x^{2} - 24 x = 0\)
\(x^{3} + x^{2} - 2 x = 0\)
\(x (x + 2) (x - 1) = 0\)
\(x = 0 ∨ x = -2 ∨ x = 1\)

Schets:

min. is \(f(-2) = -19 \text{,}\) max. is \(f(0) = 13\) en min. is \(f(1) = 8 \text{.}\)

\(f'(x) = 2 x^{4} - 3 x^{2} - 9\)

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

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

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

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

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

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

Schets:

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

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

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

Uitdelen geeft
\(f(x) = {x^{2} + 11 x + 16 \over 4 x} = {x^{2} \over 4 x} + {11 x \over 4 x} + {16 \over 4 x} = \frac{1}{4} x + \frac{11}{4} + 4 x^{-1}\)

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

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

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

Schets:

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