\(f(-2)=-6\text{,}\) dus \(A(-2, -6)\text{.}\)

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

Stel \(l{:}\,y=ax+b\) met \(a=f'(-2)=6\text{.}\)

\(\begin{rcases}y=6x+b \\ \text{door }A(-2, -6)\end{rcases}\begin{matrix}6⋅-2+b=-6 \\ -12+b=-6 \\ b=6\end{matrix}\)
Dus \(l{:}\,y=6x+6\text{.}\)

\(f(x)=\frac{1}{3}x^3-\frac{1}{2}x^2-22x+2\frac{1}{2}\) geeft \(f'(x)=x^2-x-22\text{.}\)

\(f'(x)=-2\) geeft
\(x^2-x-22=-2\)
\(x^2-x-20=0\)
\((x+4)(x-5)=0\)
\(x=-4∨x=5\text{.}\)

\(f(-4)=61\frac{1}{6}\text{,}\) dus \(A(-4, 61\frac{1}{6})\text{.}\)

\(f(5)=-78\frac{1}{3}\text{,}\) dus \(B(5, -78\frac{1}{3})\text{.}\)

\(f'(x)=6x^2+18x-24\)

\(f'(x)=0\) geeft
\(6x^2+18x-24=0\)
\(x^2+3x-4=0\)
\((x+4)(x-1)=0\)
\(x=-4∨x=1\)

Schets:

max. is \(f(-4)=64\) en min. is \(f(1)=-61\text{.}\)

\(f'(x)=-12x^3-96x^2-180x\)

\(f'(x)=0\) geeft
\(-12x^3-96x^2-180x=0\)
\(x^3+8x^2+15x=0\)
\(x(x+5)(x+3)=0\)
\(x=0∨x=-5∨x=-3\)

Schets:

max. is \(f(-5)=-156\text{,}\) min. is \(f(-3)=-220\) en max. is \(f(0)=-31\text{.}\)

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

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

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

Uitdelen geeft
\(f(x)={4x^2+x+49 \over 8x}={4x^2 \over 8x}+{x \over 8x}+{49 \over 8x}=\frac{1}{2}x+\frac{1}{8}+\frac{49}{8}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{1}{2}+\frac{49}{8}⋅-1⋅x^{-2}=\frac{1}{2}-{49 \over 8x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{1}{2}-{49 \over 8x^2}=0\)
\(\frac{1}{2}={49 \over 8x^2}\)

Kruislings vermenigvuldigen geeft
\(8x^2=98\)
\(x^2=\frac{49}{4}\)
\(x=\sqrt{\frac{49}{4}}=3\frac{1}{2}∨x=-\sqrt{\frac{49}{4}}=-3\frac{1}{2}\)

Schets:

min. is \(f(-3\frac{1}{2})=-3\frac{3}{8}\) en max. is \(f(3\frac{1}{2})=3\frac{5}{8}\text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(\sqrt{4x}=8\)

Kwadrateren geeft
\(4x=64\)
\(x=16\)

Schets:

min. is \(f(16)=-4\text{.}\)

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

min. is \(f(16)=-4\text{,}\) dus \(B_f=[-4, \rightarrow ⟩\text{.}\)

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

\(f(x)={9 \over 6x+9}=9(6x+9)^{-1}\) geeft
\(f'(x)=9⋅-1⋅(6x+9)^{-2}⋅6={-54 \over (6x+9)^2}\)

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

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=\frac{3}{2}\text{,}\) dus \(y=1\frac{1}{2}x+b\)

\(\begin{rcases}y=1\frac{1}{2}x+b \\ \text{door }A(-3, -1)\end{rcases}\begin{matrix}1\frac{1}{2}⋅-3+b=-1 \\ -4\frac{1}{2}+b=-1 \\ b=3\frac{1}{2}\end{matrix}\)
Dus \(l{:}\,y=1\frac{1}{2}x+3\frac{1}{2}\text{.}\)

\(B(0, 3\frac{1}{2})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2-4x-2=-x^2+x+5\)
\(2x^2-5x-7=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=(-5)^2-4⋅2⋅-7=81\) geeft
\(x={5-\sqrt{81} \over 2⋅2}=-1∨x={5+\sqrt{81} \over 2⋅2}=3\frac{1}{2}\)

\(x_A=-1\text{,}\) dus \(y_A=g(-1)=3\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-1)=3\text{.}\)

\(\begin{rcases}y=3x+b \\ \text{door }A(-1, 3)\end{rcases}\begin{matrix}3⋅-1+b=3 \\ -3+b=3 \\ b=6\end{matrix}\)
Dus \(l{:}\,y=3x+6\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2-4x-2=3x+6\)
\(x^2-7x-8=0\)
\((x+1)(x-8)=0\)
\(x=-1∨x=8\)

\(x_C=8\text{,}\) dus \(y_C=f(8)=30\) en
\(C(8, 30)\text{.}\)