\(f'(x) = -3 x^{2} + 6 x + 9\)

\(f'(x) = 0\) geeft
\(-3 x^{2} + 6 x + 9 = 0\)
\(x^{2} - 2 x - 3 = 0\)
\((x + 1) (x - 3) = 0\)
\(x = -1 ∨ x = 3\)

Schets:

min. is \(f(-1) = -46\) en max. is \(f(3) = -14 \text{.}\)

\(f'(x) = 12 x^{3} - 24 x^{2} - 180 x\)

\(f'(x) = 0\) geeft
\(12 x^{3} - 24 x^{2} - 180 x = 0\)
\(x^{3} - 2 x^{2} - 15 x = 0\)
\(x (x + 3) (x - 5) = 0\)
\(x = 0 ∨ x = -3 ∨ x = 5\)

Schets:

min. is \(f(-3) = -341 \text{,}\) max. is \(f(0) = 10\) en min. is \(f(5) = -1\,365 \text{.}\)

\(f'(x) = -5 x^{4} + 12 x^{2} + 9\)

\(f'(\sqrt{3}) = -5 (\sqrt{3})^{4} + 12 (\sqrt{3})^{2} + 9 = 0\)

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

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

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

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

Kwadrateren geeft
\(x = 4\)

Schets:

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

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

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

Uitdelen geeft
\(f(x) = {4 x^{2} + 8 x + 81 \over 2 x} = {4 x^{2} \over 2 x} + {8 x \over 2 x} + {81 \over 2 x} = 2 x + 4 + \frac{81}{2} x^{-1}\)

De afgeleide is dan
\(f'(x) = 2 + \frac{81}{2} ⋅ -1 ⋅ x^{-2} = 2 - {81 \over 2 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(2 - {81 \over 2 x^{2}} = 0\)
\(\frac{2}{1} = {81 \over 2 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(4 x^{2} = 81\)
\(x^{2} = \frac{81}{4}\)
\(x = \sqrt{\frac{81}{4}} = 4\frac{1}{2} ∨ x = -\sqrt{\frac{81}{4}} = -4\frac{1}{2}\)

Schets:

min. is \(f(-4\frac{1}{2}) = -14\) en max. is \(f(4\frac{1}{2}) = 22 \text{.}\)