\(f(1)=4\text{,}\) dus \(A(1, 4)\text{.}\)

\(f(x)=-3x^3+4x^2+4x-1\) geeft \(f'(x)=-9x^2+8x+4\text{.}\)

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

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

\(f(x)=\frac{1}{3}x^3+x^2-17x+1\frac{1}{3}\) geeft \(f'(x)=x^2+2x-17\text{.}\)

\(f'(x)=-2\) geeft
\(x^2+2x-17=-2\)
\(x^2+2x-15=0\)
\((x+5)(x-3)=0\)
\(x=-5∨x=3\text{.}\)

\(f(-5)=69\frac{2}{3}\text{,}\) dus \(A(-5, 69\frac{2}{3})\text{.}\)

\(f(3)=-31\frac{2}{3}\text{,}\) dus \(B(3, -31\frac{2}{3})\text{.}\)

\(f'(x)=3x^2+6x-24\)

\(f'(x)=0\) geeft
\(3x^2+6x-24=0\)
\(x^2+2x-8=0\)
\((x+4)(x-2)=0\)
\(x=-4∨x=2\)

Schets:

max. is \(f(-4)=48\) en min. is \(f(2)=-60\text{.}\)

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

\(f'(x)=0\) geeft
\(12x^3-96x^2+180x=0\)
\(x^3-8x^2+15x=0\)
\(x(x-3)(x-5)=0\)
\(x=0∨x=3∨x=5\)

Schets:

min. is \(f(0)=-24\text{,}\) max. is \(f(3)=165\) en min. is \(f(5)=101\text{.}\)

\(f'(x)=2x^4-7x^2+3\)

\(f'(\sqrt{3})=2(\sqrt{3})^4-7(\sqrt{3})^2+3=0\)

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
\(f(x)={4x^2+6x+25 \over 2x}={4x^2 \over 2x}+{6x \over 2x}+{25 \over 2x}=2x+3+\frac{25}{2}x^{-1}\)

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 \(f(-2\frac{1}{2})=-7\) en max. is \(f(2\frac{1}{2})=13\text{.}\)

\(f(x)=\frac{2}{3}x-\sqrt{4x+4}=\frac{2}{3}x-(4x+4)^{\frac{1}{2}}\) 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 \(B_f=[-2\frac{1}{6}, \rightarrow ⟩\text{.}\)

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

\(f(x)={-6 \over 2x+1}=-6(2x+1)^{-1}\) geeft
\(f'(x)=-6⋅-1⋅(2x+1)^{-2}⋅2={12 \over (2x+1)^2}\)

\(\text{rc}_k=f'(-2)=\frac{4}{3}\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=-\frac{3}{4}\text{,}\) dus \(y=-\frac{3}{4}x+b\)

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

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

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+x-5=-x^2-4x-2\)
\(2x^2+5x-3=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)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 \(y_A=g(-3)=1\)

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

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

\(\begin{rcases}y=2x+b \\ \text{door }A(-3, 1)\end{rcases}\begin{matrix}2⋅-3+b=1 \\ -6+b=1 \\ b=7\end{matrix}\)
Dus \(l{:}\,y=2x+7\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+x-5=2x+7\)
\(x^2-x-12=0\)
\((x+3)(x-4)=0\)
\(x=-3∨x=4\)

\(x_C=4\text{,}\) dus \(y_C=f(4)=15\) en
\(C(4, 15)\text{.}\)