\(f'(x) = 6 x^{2} - 18 x + 12\)

\(f'(x) = 0\) geeft
\(6 x^{2} - 18 x + 12 = 0\)
\(x^{2} - 3 x + 2 = 0\)
\((x - 1) (x - 2) = 0\)
\(x = 1 ∨ x = 2\)

Schets:

max. is \(f(1) = -20\) en min. is \(f(2) = -21 \text{.}\)

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

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

Schets:

max. is \(f(-3) = -8 \text{,}\) min. is \(f(-2) = -13\) en max. is \(f(0) = 19 \text{.}\)

\(f'(x) = -x^{4} + 6 x^{2} - 9\)

\(f'(\sqrt{3}) = -(\sqrt{3})^{4} + 6 (\sqrt{3})^{2} - 9 = 0\)

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

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

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

Kruislings vermenigvuldigen geeft
\(2 \sqrt{x + 3} = 2\)
\(\sqrt{x + 3} = 1\)

Kwadrateren geeft
\(x + 3 = 1\)
\(x = -2\)

Schets:

max. is \(f(-2) = 2 \text{.}\)

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

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

Uitdelen geeft
\(f(x) = {3 x^{2} + 4 x + 27 \over 6 x} = {3 x^{2} \over 6 x} + {4 x \over 6 x} + {27 \over 6 x} = \frac{1}{2} x + \frac{2}{3} + \frac{9}{2} x^{-1}\)

De afgeleide is dan
\(f'(x) = \frac{1}{2} + \frac{9}{2} ⋅ -1 ⋅ x^{-2} = \frac{1}{2} - {9 \over 2 x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(\frac{1}{2} - {9 \over 2 x^{2}} = 0\)
\(\frac{1}{2} = {9 \over 2 x^{2}}\)

Kruislings vermenigvuldigen geeft
\(2 x^{2} = 18\)
\(x^{2} = 9\)
\(x = \sqrt{9} = 3 ∨ x = -\sqrt{9} = -3\)

Schets:

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