\(f'(x) = 6 x^{2} - 6 x - 72\)

\(f'(x) = 0\) geeft
\(6 x^{2} - 6 x - 72 = 0\)
\(x^{2} - x - 12 = 0\)
\((x + 3) (x - 4) = 0\)
\(x = -3 ∨ x = 4\)

Schets:

max. is \(f(-3) = 90\) en min. is \(f(4) = -253 \text{.}\)

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

\(f'(x) = 0\) 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 \(f(-3) = -379 \text{,}\) max. is \(f(0) = -28\) en min. is \(f(5) = -1\,403 \text{.}\)

\(f'(x) = -3 x^{4} + 7 x^{2} + 6\)

\(f'(\sqrt{3}) = -3 (\sqrt{3})^{4} + 7 (\sqrt{3})^{2} + 6 = 0\)

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

Uitdelen geeft
\(f(x) = {3 x^{2} + 5 x + 12 \over x} = {3 x^{2} \over x} + {5 x \over x} + {12 \over x} = 3 x + 5 + 12 x^{-1}\)

De afgeleide is dan
\(f'(x) = 3 + 12 ⋅ -1 ⋅ x^{-2} = 3 - {12 \over x^{2}} \text{.}\)

\(f'(x) = 0\) geeft
\(3 - {12 \over x^{2}} = 0\)
\(\frac{3}{1} = {12 \over x^{2}}\)

Kruislings vermenigvuldigen geeft
\(3 x^{2} = 12\)
\(x^{2} = 4\)
\(x = \sqrt{4} = 2 ∨ x = -\sqrt{4} = -2\)

Schets:

min. is \(f(-2) = -7\) en max. is \(f(2) = 17 \text{.}\)

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

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

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

Kwadrateren geeft
\(x - 2 = 4\)
\(x = 6\)

Schets:

min. is \(f(6) = -\frac{1}{2} \text{.}\)

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

min. is \(f(6) = -\frac{1}{2} \text{,}\) dus \(B_{f} = [-\frac{1}{2} , \rightarrow ⟩ \text{.}\)