\(f(1)=3\text{,}\) dus \(A(1, 3)\text{.}\)

\(f(x)=-x^3-x^2+6x-1\) geeft \(f'(x)=-3x^2-2x+6\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(1)=1\text{.}\)

\(\begin{rcases}y=x+b \\ \text{door }A(1, 3)\end{rcases}\begin{matrix}1⋅1+b=3 \\ 1+b=3 \\ b=2\end{matrix}\)
Dus \(l{:}\,y=x+2\text{.}\)

\(f(x)=\frac{1}{3}x^3+3\frac{1}{2}x^2+14x+4\) geeft \(f'(x)=x^2+7x+14\text{.}\)

\(f'(x)=4\) geeft
\(x^2+7x+14=4\)
\(x^2+7x+10=0\)
\((x+5)(x+2)=0\)
\(x=-5∨x=-2\text{.}\)

\(f(-5)=-20\frac{1}{6}\text{,}\) dus \(A(-5, -20\frac{1}{6})\text{.}\)

\(f(-2)=-12\frac{2}{3}\text{,}\) dus \(B(-2, -12\frac{2}{3})\text{.}\)

\(f'(x)=3x^2-3\)

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

Schets:

max. is \(f(-1)=17\) en min. is \(f(1)=13\text{.}\)

\(f'(x)=-12x^3+192x\)

\(f'(x)=0\) geeft
\(-12x^3+192x=0\)
\(x^3-16x=0\)
\(x(x+4)(x-4)=0\)
\(x=0∨x=-4∨x=4\)

Schets:

max. is \(f(-4)=782\text{,}\) min. is \(f(0)=14\) en max. is \(f(4)=782\text{.}\)

\(f'(x)=4x^4-7x^2-15\)

\(f'(\sqrt{3})=4(\sqrt{3})^4-7(\sqrt{3})^2-15=0\)

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

\(f(x)=\frac{1}{5}x-\sqrt{2x-3}=\frac{1}{5}x-(2x-3)^{\frac{1}{2}}\) geeft
\(f'(x)=\frac{1}{5}-\frac{1}{2}⋅(2x-3)^{-\frac{1}{2}}⋅2=\frac{1}{5}-{1 \over \sqrt{2x-3}}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{1}{5}-{1 \over \sqrt{2x-3}}=0\)
\(-{1 \over \sqrt{2x-3}}=-\frac{1}{5}\)

Kruislings vermenigvuldigen geeft
\(\sqrt{2x-3}=5\)

Kwadrateren geeft
\(2x-3=25\)
\(x=14\)

Schets:

min. is \(f(14)=-2\frac{1}{5}\text{.}\)

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

min. is \(f(14)=-2\frac{1}{5}\text{,}\) dus \(B_f=[-2\frac{1}{5}, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+5x+49 \over 7x}={4x^2 \over 7x}+{5x \over 7x}+{49 \over 7x}=\frac{4}{7}x+\frac{5}{7}+7x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{4}{7}+7⋅-1⋅x^{-2}=\frac{4}{7}-{7 \over x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{4}{7}-{7 \over x^2}=0\)
\(\frac{4}{7}={7 \over x^2}\)

Kruislings vermenigvuldigen geeft
\(4x^2=49\)
\(x^2=\frac{49}{4}\)
\(x=\sqrt{\frac{49}{4}}=3\frac{1}{2}∨x=-\sqrt{\frac{49}{4}}=-3\frac{1}{2}\)

Schets:

min. is \(f(-3\frac{1}{2})=-3\frac{2}{7}\) en max. is \(f(3\frac{1}{2})=4\frac{5}{7}\text{.}\)

\(f(2)=-2\text{,}\) dus \(A(2, -2)\)

\(f(x)={-8 \over 6x-8}=-8(6x-8)^{-1}\) geeft
\(f'(x)=-8⋅-1⋅(6x-8)^{-2}⋅6={48 \over (6x-8)^2}\)

\(\text{rc}_k=f'(2)=3\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=-\frac{1}{3}\text{,}\) dus \(y=-\frac{1}{3}x+b\)

\(\begin{rcases}y=-\frac{1}{3}x+b \\ \text{door }A(2, -2)\end{rcases}\begin{matrix}-\frac{1}{3}⋅2+b=-2 \\ -\frac{2}{3}+b=-2 \\ b=-1\frac{1}{3}\end{matrix}\)
Dus \(l{:}\,y=-\frac{1}{3}x-1\frac{1}{3}\text{.}\)

\(B(0, -1\frac{1}{3})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+4x+1=-x^2-x+4\)
\(2x^2+5x-3=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=5^2-4⋅2⋅-3=49\) geeft
\(x={-5-\sqrt{49} \over 2⋅2}=-3∨x={-5+\sqrt{49} \over 2⋅2}=\frac{1}{2}\)

\(x_A=-3\text{,}\) dus \(y_A=g(-3)=-2\)

\(g'(x)=-2x-1\)

\(l{:}\,y=ax+b\) met \(a=g'(-3)=5\text{.}\)

\(\begin{rcases}y=5x+b \\ \text{door }A(-3, -2)\end{rcases}\begin{matrix}5⋅-3+b=-2 \\ -15+b=-2 \\ b=13\end{matrix}\)
Dus \(l{:}\,y=5x+13\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+4x+1=5x+13\)
\(x^2-x-12=0\)
\((x+3)(x-4)=0\)
\(x=-3∨x=4\)

\(x_C=4\text{,}\) dus \(y_C=f(4)=33\) en
\(C(4, 33)\text{.}\)