\(f'(x)=3x^2+6x-24\)

\(f'(x)=0\) geeft
\(3x^2+6x-24=0\)
\(x^2+2x-8=0\)
\((x+4)(x-2)=0\)
\(x=-4∨x=2\)

Schets:

max. is \(f(-4)=104\) en min. is \(f(2)=-4\text{.}\)

\(f'(x)=12x^3+48x^2-60x\)

\(f'(x)=0\) geeft
\(12x^3+48x^2-60x=0\)
\(x^3+4x^2-5x=0\)
\(x(x+5)(x-1)=0\)
\(x=0∨x=-5∨x=1\)

Schets:

min. is \(f(-5)=-889\text{,}\) max. is \(f(0)=-14\) en min. is \(f(1)=-25\text{.}\)

\(f'(x)=-5x^4+13x^2-6\)

\(f'(\sqrt{2})=-5(\sqrt{2})^4+13(\sqrt{2})^2-6=0\)

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

\(f(x)=\frac{2}{5}x-\sqrt{4x-1}=\frac{2}{5}x-(4x-1)^{\frac{1}{2}}\) geeft
\(f'(x)=\frac{2}{5}-\frac{1}{2}⋅(4x-1)^{-\frac{1}{2}}⋅4=\frac{2}{5}-{2 \over \sqrt{4x-1}}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{2}{5}-{2 \over \sqrt{4x-1}}=0\)
\(-{2 \over \sqrt{4x-1}}=-\frac{2}{5}\)

Kruislings vermenigvuldigen geeft
\(2\sqrt{4x-1}=10\)
\(\sqrt{4x-1}=5\)

Kwadrateren geeft
\(4x-1=25\)
\(x=6\frac{1}{2}\)

Schets:

min. is \(f(6\frac{1}{2})=-2\frac{2}{5}\text{.}\)

\(4x-1≥0\) geeft \(x≥\frac{1}{4}\text{,}\) dus \(D_f=[\frac{1}{4}, \rightarrow ⟩\text{.}\)

min. is \(f(6\frac{1}{2})=-2\frac{2}{5}\text{,}\) dus \(B_f=[-2\frac{2}{5}, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+12x+1 \over 9x}={4x^2 \over 9x}+{12x \over 9x}+{1 \over 9x}=\frac{4}{9}x+\frac{4}{3}+\frac{1}{9}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{4}{9}+\frac{1}{9}⋅-1⋅x^{-2}=\frac{4}{9}-{1 \over 9x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{4}{9}-{1 \over 9x^2}=0\)
\(\frac{4}{9}={1 \over 9x^2}\)

Kruislings vermenigvuldigen geeft
\(36x^2=9\)
\(x^2=\frac{1}{4}\)
\(x=\sqrt{\frac{1}{4}}=\frac{1}{2}∨x=-\sqrt{\frac{1}{4}}=-\frac{1}{2}\)

Schets:

min. is \(f(-\frac{1}{2})=\frac{8}{9}\) en max. is \(f(\frac{1}{2})=1\frac{7}{9}\text{.}\)