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

\(f(x)=5x^3+3x^2-6x-6\) geeft \(f'(x)=15x^2+6x-6\text{.}\)

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

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

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

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

\(f(2)=3\frac{2}{3}\text{,}\) dus \(A(2, 3\frac{2}{3})\text{.}\)

\(f(5)=-9\frac{5}{6}\text{,}\) dus \(B(5, -9\frac{5}{6})\text{.}\)

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

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

Schets:

max. is \(f(1)=-37\) en min. is \(f(5)=-69\text{.}\)

\(f'(x)=-12x^3-72x^2-60x\)

\(f'(x)=0\) geeft
\(-12x^3-72x^2-60x=0\)
\(x^3+6x^2+5x=0\)
\(x(x+5)(x+1)=0\)
\(x=0∨x=-5∨x=-1\)

Schets:

max. is \(f(-5)=403\text{,}\) min. is \(f(-1)=19\) en max. is \(f(0)=28\text{.}\)

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

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

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
\(f(x)={4x^2+7x+25 \over 2x}={4x^2 \over 2x}+{7x \over 2x}+{25 \over 2x}=2x+\frac{7}{2}+\frac{25}{2}x^{-1}\)

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

\(f'(x)=0\) geeft
\(2-{25 \over 2x^2}=0\)
\(\frac{2}{1}={25 \over 2x^2}\)

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

Schets:

min. is \(f(-2\frac{1}{2})=-6\frac{1}{2}\) en max. is \(f(2\frac{1}{2})=13\frac{1}{2}\text{.}\)

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

\(f'(x)=0\) geeft
\(\frac{1}{3}-{2 \over \sqrt{4x}}=0\)
\(-{2 \over \sqrt{4x}}=-\frac{1}{3}\)

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

Kwadrateren geeft
\(4x=36\)
\(x=9\)

Schets:

min. is \(f(9)=-3\text{.}\)

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

min. is \(f(9)=-3\text{,}\) dus \(B_f=[-3, \rightarrow ⟩\text{.}\)

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

\(f(x)={-9 \over 2x+3}=-9(2x+3)^{-1}\) geeft
\(f'(x)=-9⋅-1⋅(2x+3)^{-2}⋅2={18 \over (2x+3)^2}\)

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

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

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

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

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+4x+1=-x^2-4x-5\)
\(2x^2+8x+6=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=8^2-4⋅2⋅6=16\) geeft
\(x={-8-\sqrt{16} \over 2⋅2}=-3∨x={-8+\sqrt{16} \over 2⋅2}=-1\)

\(x_A=-3\text{,}\) dus \(y_A=g(-3)=-2\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-3)=2\text{.}\)

\(\begin{rcases}y=2x+b \\ \text{door }A(-3, -2)\end{rcases}\begin{matrix}2⋅-3+b=-2 \\ -6+b=-2 \\ b=4\end{matrix}\)
Dus \(l{:}\,y=2x+4\text{.}\)

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

\(x_C=1\text{,}\) dus \(y_C=f(1)=6\) en
\(C(1, 6)\text{.}\)