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

\(f(x)=-4x^3+6x^2+2x-1\) geeft \(f'(x)=-12x^2+12x+2\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(1)=2\text{.}\)

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

\(f(x)=\frac{1}{3}x^3-x^2-11x+1\frac{1}{6}\) geeft \(f'(x)=x^2-2x-11\text{.}\)

\(f'(x)=-3\) geeft
\(x^2-2x-11=-3\)
\(x^2-2x-8=0\)
\((x+2)(x-4)=0\)
\(x=-2∨x=4\text{.}\)

\(f(-2)=16\frac{1}{2}\text{,}\) dus \(A(-2, 16\frac{1}{2})\text{.}\)

\(f(4)=-37\frac{1}{2}\text{,}\) dus \(B(4, -37\frac{1}{2})\text{.}\)

\(f'(x)=-3x^2-18x-15\)

\(f'(x)=0\) geeft
\(-3x^2-18x-15=0\)
\(x^2+6x+5=0\)
\((x+5)(x+1)=0\)
\(x=-5∨x=-1\)

Schets:

min. is \(f(-5)=23\) en max. is \(f(-1)=55\text{.}\)

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

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

Schets:

min. is \(f(-4)=-116\text{,}\) max. is \(f(-1)=19\) en min. is \(f(0)=12\text{.}\)

\(f'(x)=-2x^4-x^2+10\)

\(f'(\sqrt{2})=-2(\sqrt{2})^4-(\sqrt{2})^2+10=0\)

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

\(f(x)=\frac{2}{3}x-\sqrt{4x-3}=\frac{2}{3}x-(4x-3)^{\frac{1}{2}}\) geeft
\(f'(x)=\frac{2}{3}-\frac{1}{2}⋅(4x-3)^{-\frac{1}{2}}⋅4=\frac{2}{3}-{2 \over \sqrt{4x-3}}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{2}{3}-{2 \over \sqrt{4x-3}}=0\)
\(-{2 \over \sqrt{4x-3}}=-\frac{2}{3}\)

Kruislings vermenigvuldigen geeft
\(2\sqrt{4x-3}=6\)
\(\sqrt{4x-3}=3\)

Kwadrateren geeft
\(4x-3=9\)
\(x=3\)

Schets:

min. is \(f(3)=-1\text{.}\)

\(4x-3≥0\) geeft \(x≥\frac{3}{4}\text{,}\) dus \(D_f=[\frac{3}{4}, \rightarrow ⟩\text{.}\)

min. is \(f(3)=-1\text{,}\) dus \(B_f=[-1, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+9x+49 \over 3x}={4x^2 \over 3x}+{9x \over 3x}+{49 \over 3x}=\frac{4}{3}x+3+\frac{49}{3}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{4}{3}+\frac{49}{3}⋅-1⋅x^{-2}=\frac{4}{3}-{49 \over 3x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{4}{3}-{49 \over 3x^2}=0\)
\(\frac{4}{3}={49 \over 3x^2}\)

Kruislings vermenigvuldigen geeft
\(12x^2=147\)
\(x^2=\frac{49}{4}\)
\(x=\sqrt{\frac{49}{4}}=3\frac{1}{2}∨x=-\sqrt{\frac{49}{4}}=-3\frac{1}{2}\)

Schets:

min. is \(f(-3\frac{1}{2})=-6\frac{1}{3}\) en max. is \(f(3\frac{1}{2})=12\frac{1}{3}\text{.}\)

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

\(f(x)={4 \over 4x+10}=4(4x+10)^{-1}\) geeft
\(f'(x)=4⋅-1⋅(4x+10)^{-2}⋅4={-16 \over (4x+10)^2}\)

\(\text{rc}_k=f'(-2)=-4\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=\frac{1}{4}\text{,}\) dus \(y=\frac{1}{4}x+b\)

\(\begin{rcases}y=\frac{1}{4}x+b \\ \text{door }A(-2, 2)\end{rcases}\begin{matrix}\frac{1}{4}⋅-2+b=2 \\ -\frac{2}{4}+b=2 \\ b=2\frac{1}{2}\end{matrix}\)
Dus \(l{:}\,y=\frac{1}{4}x+2\frac{1}{2}\text{.}\)

\(B(0, 2\frac{1}{2})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+2x-4=-x^2+x+2\)
\(2x^2+x-6=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=1^2-4⋅2⋅-6=49\) geeft
\(x={-1-\sqrt{49} \over 2⋅2}=-2∨x={-1+\sqrt{49} \over 2⋅2}=1\frac{1}{2}\)

\(x_A=-2\text{,}\) dus \(y_A=g(-2)=-4\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-2)=5\text{.}\)

\(\begin{rcases}y=5x+b \\ \text{door }A(-2, -4)\end{rcases}\begin{matrix}5⋅-2+b=-4 \\ -10+b=-4 \\ b=6\end{matrix}\)
Dus \(l{:}\,y=5x+6\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+2x-4=5x+6\)
\(x^2-3x-10=0\)
\((x+2)(x-5)=0\)
\(x=-2∨x=5\)

\(x_C=5\text{,}\) dus \(y_C=f(5)=31\) en
\(C(5, 31)\text{.}\)