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

\(f(x)=3x^3-x^2-4x-6\) geeft \(f'(x)=9x^2-2x-4\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(-1)=7\text{.}\)

\(\begin{rcases}y=7x+b \\ \text{door }A(-1, -6)\end{rcases}\begin{matrix}7⋅-1+b=-6 \\ -7+b=-6 \\ b=1\end{matrix}\)
Dus \(l{:}\,y=7x+1\text{.}\)

\(f(x)=\frac{1}{3}x^3+3\frac{1}{2}x^2+10x+1\frac{1}{3}\) geeft \(f'(x)=x^2+7x+10\text{.}\)

\(f'(x)=-2\) geeft
\(x^2+7x+10=-2\)
\(x^2+7x+12=0\)
\((x+4)(x+3)=0\)
\(x=-4∨x=-3\text{.}\)

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

\(f(-3)=-6\frac{1}{6}\text{,}\) dus \(B(-3, -6\frac{1}{6})\text{.}\)

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

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

Schets:

min. is \(f(1)=-50\) en max. is \(f(5)=-18\text{.}\)

\(f'(x)=12x^3+36x^2-120x\)

\(f'(x)=0\) geeft
\(12x^3+36x^2-120x=0\)
\(x^3+3x^2-10x=0\)
\(x(x+5)(x-2)=0\)
\(x=0∨x=-5∨x=2\)

Schets:

min. is \(f(-5)=-1\,088\text{,}\) max. is \(f(0)=37\) en min. is \(f(2)=-59\text{.}\)

\(f'(x)=5x^4-12x^2-9\)

\(f'(\sqrt{3})=5(\sqrt{3})^4-12(\sqrt{3})^2-9=0\)

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
\(f(x)={5x^2+45 \over 4x}={5x^2 \over 4x}+{45 \over 4x}=\frac{5}{4}x+\frac{45}{4}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{5}{4}+\frac{45}{4}⋅-1⋅x^{-2}=\frac{5}{4}-{45 \over 4x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{5}{4}-{45 \over 4x^2}=0\)
\(\frac{5}{4}={45 \over 4x^2}\)

Kruislings vermenigvuldigen geeft
\(20x^2=180\)
\(x^2=9\)
\(x=\sqrt{9}=3∨x=-\sqrt{9}=-3\)

Schets:

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

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

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

Kruislings vermenigvuldigen geeft
\(2\sqrt{5x}=25\)
\(\sqrt{5x}=12\frac{1}{2}\)

Kwadrateren geeft
\(5x=156\frac{1}{4}\)
\(x=31\frac{1}{4}\)

Schets:

max. is \(f(31\frac{1}{4})=6\frac{1}{4}\text{.}\)

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

max. is \(f(31\frac{1}{4})=6\frac{1}{4}\text{,}\) dus \(B_f=⟨\leftarrow , 6\frac{1}{4}]\text{.}\)

\(f(-5)=2\text{,}\) dus \(A(-5, 2)\)

\(f(x)={-10 \over 2x+5}=-10(2x+5)^{-1}\) geeft
\(f'(x)=-10⋅-1⋅(2x+5)^{-2}⋅2={20 \over (2x+5)^2}\)

\(\text{rc}_k=f'(-5)=\frac{4}{5}\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=-\frac{5}{4}\text{,}\) dus \(y=-1\frac{1}{4}x+b\)

\(\begin{rcases}y=-1\frac{1}{4}x+b \\ \text{door }A(-5, 2)\end{rcases}\begin{matrix}-1\frac{1}{4}⋅-5+b=2 \\ 6\frac{1}{4}+b=2 \\ b=-4\frac{1}{4}\end{matrix}\)
Dus \(l{:}\,y=-1\frac{1}{4}x-4\frac{1}{4}\text{.}\)

\(B(0, -4\frac{1}{4})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+2x-1=-x^2-3x+2\)
\(2x^2+5x-3=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=5^2-4⋅2⋅-3=49\) geeft
\(x={-5-\sqrt{49} \over 2⋅2}=-3∨x={-5+\sqrt{49} \over 2⋅2}=\frac{1}{2}\)

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

\(g'(x)=-2x-3\)

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

\(\begin{rcases}y=3x+b \\ \text{door }A(-3, 2)\end{rcases}\begin{matrix}3⋅-3+b=2 \\ -9+b=2 \\ b=11\end{matrix}\)
Dus \(l{:}\,y=3x+11\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+2x-1=3x+11\)
\(x^2-x-12=0\)
\((x+3)(x-4)=0\)
\(x=-3∨x=4\)

\(x_C=4\text{,}\) dus \(y_C=f(4)=23\) en
\(C(4, 23)\text{.}\)