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

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

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

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

\(f(x)=\frac{1}{3}x^3-3x^2+13x+\frac{5}{6}\) geeft \(f'(x)=x^2-6x+13\text{.}\)

\(f'(x)=5\) geeft
\(x^2-6x+13=5\)
\(x^2-6x+8=0\)
\((x-2)(x-4)=0\)
\(x=2∨x=4\text{.}\)

\(f(2)=17\frac{1}{2}\text{,}\) dus \(A(2, 17\frac{1}{2})\text{.}\)

\(f(4)=26\frac{1}{6}\text{,}\) dus \(B(4, 26\frac{1}{6})\text{.}\)

\(f'(x)=-6x^2-6x+72\)

\(f'(x)=0\) geeft
\(-6x^2-6x+72=0\)
\(x^2+x-12=0\)
\((x+4)(x-3)=0\)
\(x=-4∨x=3\)

Schets:

min. is \(f(-4)=-218\) en max. is \(f(3)=125\text{.}\)

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

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

Schets:

min. is \(f(-1)=-14\text{,}\) max. is \(f(0)=-11\) en min. is \(f(1)=-14\text{.}\)

\(f'(x)=-2x^4+8x^2-6\)

\(f'(\sqrt{3})=-2(\sqrt{3})^4+8(\sqrt{3})^2-6=0\)

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

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

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

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

Kwadrateren geeft
\(2x+5=25\)
\(x=10\)

Schets:

max. is \(f(10)=3\text{.}\)

\(2x+5≥0\) geeft \(x≥-2\frac{1}{2}\text{,}\) dus \(D_f=[-2\frac{1}{2}, \rightarrow ⟩\text{.}\)

max. is \(f(10)=3\text{,}\) dus \(B_f=⟨\leftarrow , 3]\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+2x+25 \over 3x}={4x^2 \over 3x}+{2x \over 3x}+{25 \over 3x}=\frac{4}{3}x+\frac{2}{3}+\frac{25}{3}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{4}{3}+\frac{25}{3}⋅-1⋅x^{-2}=\frac{4}{3}-{25 \over 3x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{4}{3}-{25 \over 3x^2}=0\)
\(\frac{4}{3}={25 \over 3x^2}\)

Kruislings vermenigvuldigen geeft
\(12x^2=75\)
\(x^2=\frac{25}{4}\)
\(x=\sqrt{\frac{25}{4}}=2\frac{1}{2}∨x=-\sqrt{\frac{25}{4}}=-2\frac{1}{2}\)

Schets:

min. is \(f(-2\frac{1}{2})=-6\) en max. is \(f(2\frac{1}{2})=7\frac{1}{3}\text{.}\)

\(f(4)=2\text{,}\) dus \(A(4, 2)\)

\(f(x)={-2 \over 2x-9}=-2(2x-9)^{-1}\) geeft
\(f'(x)=-2⋅-1⋅(2x-9)^{-2}⋅2={4 \over (2x-9)^2}\)

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

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

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

\(B(0, 3)\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+3x-4=-x^2+2x-3\)
\(2x^2+x-1=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)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 \(y_A=g(-1)=-6\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-1)=4\text{.}\)

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

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

\(x_C=2\text{,}\) dus \(y_C=f(2)=6\) en
\(C(2, 6)\text{.}\)