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

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

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

\(\begin{rcases}y=x+b \\ \text{door }A(2, 6)\end{rcases}\begin{matrix}1⋅2+b=6 \\ 2+b=6 \\ b=4\end{matrix}\)
Dus \(l{:}\,y=x+4\text{.}\)

\(f(x)=\frac{1}{3}x^3+3x^2+3x+3\frac{5}{6}\) geeft \(f'(x)=x^2+6x+3\text{.}\)

\(f'(x)=-5\) geeft
\(x^2+6x+3=-5\)
\(x^2+6x+8=0\)
\((x+4)(x+2)=0\)
\(x=-4∨x=-2\text{.}\)

\(f(-4)=18\frac{1}{2}\text{,}\) dus \(A(-4, 18\frac{1}{2})\text{.}\)

\(f(-2)=7\frac{1}{6}\text{,}\) dus \(B(-2, 7\frac{1}{6})\text{.}\)

\(f'(x)=6x^2+6x-120\)

\(f'(x)=0\) geeft
\(6x^2+6x-120=0\)
\(x^2+x-20=0\)
\((x+5)(x-4)=0\)
\(x=-5∨x=4\)

Schets:

max. is \(f(-5)=441\) en min. is \(f(4)=-288\text{.}\)

\(f'(x)=-12x^3+108x\)

\(f'(x)=0\) geeft
\(-12x^3+108x=0\)
\(x^3-9x=0\)
\(x(x+3)(x-3)=0\)
\(x=0∨x=-3∨x=3\)

Schets:

max. is \(f(-3)=227\text{,}\) min. is \(f(0)=-16\) en max. is \(f(3)=227\text{.}\)

\(f'(x)=-x^4-2x^2+8\)

\(f'(\sqrt{2})=-(\sqrt{2})^4-2(\sqrt{2})^2+8=0\)

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

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

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

Kruislings vermenigvuldigen geeft
\(2\sqrt{5x-5}=20\)
\(\sqrt{5x-5}=10\)

Kwadrateren geeft
\(5x-5=100\)
\(x=21\)

Schets:

max. is \(f(21)=4\frac{3}{4}\text{.}\)

\(5x-5≥0\) geeft \(x≥1\text{,}\) dus \(D_f=[1, \rightarrow ⟩\text{.}\)

max. is \(f(21)=4\frac{3}{4}\text{,}\) dus \(B_f=⟨\leftarrow , 4\frac{3}{4}]\text{.}\)

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

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

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

Kruislings vermenigvuldigen geeft
\(2x^2=8\)
\(x^2=4\)
\(x=\sqrt{4}=2∨x=-\sqrt{4}=-2\)

Schets:

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

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

\(f(x)={-10 \over 6x-2}=-10(6x-2)^{-1}\) geeft
\(f'(x)=-10⋅-1⋅(6x-2)^{-2}⋅6={60 \over (6x-2)^2}\)

\(\text{rc}_k=f'(2)=\frac{3}{5}\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=-\frac{5}{3}\text{,}\) dus \(y=-1\frac{2}{3}x+b\)

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

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

De snijpunten \(A\) en \(B\) volgen uit
\(x^2-4x+3=-x^2-5x+4\)
\(2x^2+x-1=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=1^2-4⋅2⋅-1=9\) geeft
\(x={-1-\sqrt{9} \over 2⋅2}=-1∨x={-1+\sqrt{9} \over 2⋅2}=\frac{1}{2}\)

\(x_A=-1\text{,}\) dus \(y_A=g(-1)=8\)

\(g'(x)=-2x-5\)

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

\(\begin{rcases}y=-3x+b \\ \text{door }A(-1, 8)\end{rcases}\begin{matrix}-3⋅-1+b=8 \\ 3+b=8 \\ b=5\end{matrix}\)
Dus \(l{:}\,y=-3x+5\text{.}\)

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

\(x_C=2\text{,}\) dus \(y_C=f(2)=-1\) en
\(C(2, -1)\text{.}\)