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

\(f(x)=3x^3-3x^2-6x-6\) geeft \(f'(x)=9x^2-6x-6\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(-1)=9\text{.}\)

\(\begin{rcases}y=9x+b \\ \text{door }A(-1, -6)\end{rcases}\begin{matrix}9⋅-1+b=-6 \\ -9+b=-6 \\ b=3\end{matrix}\)
Dus \(l{:}\,y=9x+3\text{.}\)

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

\(f'(x)=3\) geeft
\(x^2+3x-7=3\)
\(x^2+3x-10=0\)
\((x+5)(x-2)=0\)
\(x=-5∨x=2\text{.}\)

\(f(-5)=32\frac{1}{3}\text{,}\) dus \(A(-5, 32\frac{1}{3})\text{.}\)

\(f(2)=-3\frac{5}{6}\text{,}\) dus \(B(2, -3\frac{5}{6})\text{.}\)

\(f'(x)=6x^2-18x+12\)

\(f'(x)=0\) geeft
\(6x^2-18x+12=0\)
\(x^2-3x+2=0\)
\((x-1)(x-2)=0\)
\(x=1∨x=2\)

Schets:

max. is \(f(1)=38\) en min. is \(f(2)=37\text{.}\)

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

\(f'(x)=0\) 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 \(f(-5)=-149\text{,}\) min. is \(f(-3)=-213\) en max. is \(f(0)=-24\text{.}\)

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

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

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

Uitdelen geeft
\(f(x)={4x^2+9x+81 \over 6x}={4x^2 \over 6x}+{9x \over 6x}+{81 \over 6x}=\frac{2}{3}x+\frac{3}{2}+\frac{27}{2}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{2}{3}+\frac{27}{2}⋅-1⋅x^{-2}=\frac{2}{3}-{27 \over 2x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{2}{3}-{27 \over 2x^2}=0\)
\(\frac{2}{3}={27 \over 2x^2}\)

Kruislings vermenigvuldigen geeft
\(4x^2=81\)
\(x^2=\frac{81}{4}\)
\(x=\sqrt{\frac{81}{4}}=4\frac{1}{2}∨x=-\sqrt{\frac{81}{4}}=-4\frac{1}{2}\)

Schets:

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

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

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

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

Kwadrateren geeft
\(4x+4=9\)
\(x=1\frac{1}{4}\)

Schets:

min. is \(f(1\frac{1}{4})=-2\frac{1}{6}\text{.}\)

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

min. is \(f(1\frac{1}{4})=-2\frac{1}{6}\text{,}\) dus \(B_f=[-2\frac{1}{6}, \rightarrow ⟩\text{.}\)

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

\(f(x)={2 \over 4x+10}=2(4x+10)^{-1}\) geeft
\(f'(x)=2⋅-1⋅(4x+10)^{-2}⋅4={-8 \over (4x+10)^2}\)

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

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

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

\(B(0, 2)\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+5x-4=-x^2-4x+1\)
\(2x^2+9x-5=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=9^2-4⋅2⋅-5=121\) geeft
\(x={-9-\sqrt{121} \over 2⋅2}=-5∨x={-9+\sqrt{121} \over 2⋅2}=\frac{1}{2}\)

\(x_A=-5\text{,}\) dus \(y_A=g(-5)=-4\)

\(g'(x)=-2x-4\)

\(l{:}\,y=ax+b\) met \(a=g'(-5)=6\text{.}\)

\(\begin{rcases}y=6x+b \\ \text{door }A(-5, -4)\end{rcases}\begin{matrix}6⋅-5+b=-4 \\ -30+b=-4 \\ b=26\end{matrix}\)
Dus \(l{:}\,y=6x+26\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+5x-4=6x+26\)
\(x^2-x-30=0\)
\((x+5)(x-6)=0\)
\(x=-5∨x=6\)

\(x_C=6\text{,}\) dus \(y_C=f(6)=62\) en
\(C(6, 62)\text{.}\)