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

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

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

\(\begin{rcases}y=7x+b \\ \text{door }A(-1, -4)\end{rcases}\begin{matrix}7⋅-1+b=-4 \\ -7+b=-4 \\ b=3\end{matrix}\)
Dus \(l{:}\,y=7x+3\text{.}\)

\(f(x)=\frac{1}{3}x^3-2\frac{1}{2}x^2+11x+1\frac{1}{2}\) geeft \(f'(x)=x^2-5x+11\text{.}\)

\(f'(x)=5\) geeft
\(x^2-5x+11=5\)
\(x^2-5x+6=0\)
\((x-2)(x-3)=0\)
\(x=2∨x=3\text{.}\)

\(f(2)=16\frac{1}{6}\text{,}\) dus \(A(2, 16\frac{1}{6})\text{.}\)

\(f(3)=21\text{,}\) dus \(B(3, 21)\text{.}\)

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

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

Schets:

max. is \(f(1)=-35\) en min. is \(f(5)=-67\text{.}\)

\(f'(x)=12x^3-72x^2+96x\)

\(f'(x)=0\) geeft
\(12x^3-72x^2+96x=0\)
\(x^3-6x^2+8x=0\)
\(x(x-2)(x-4)=0\)
\(x=0∨x=2∨x=4\)

Schets:

min. is \(f(0)=-26\text{,}\) max. is \(f(2)=22\) en min. is \(f(4)=-26\text{.}\)

\(f'(x)=-4x^4+8x^2+12\)

\(f'(\sqrt{3})=-4(\sqrt{3})^4+8(\sqrt{3})^2+12=0\)

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

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

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

Kruislings vermenigvuldigen geeft
\(\sqrt{4x+5}=10\)

Kwadrateren geeft
\(4x+5=100\)
\(x=23\frac{3}{4}\)

Schets:

max. is \(f(23\frac{3}{4})=5\frac{1}{4}\text{.}\)

\(4x+5≥0\) geeft \(x≥-1\frac{1}{4}\text{,}\) dus \(D_f=[-1\frac{1}{4}, \rightarrow ⟩\text{.}\)

max. is \(f(23\frac{3}{4})=5\frac{1}{4}\text{,}\) dus \(B_f=⟨\leftarrow , 5\frac{1}{4}]\text{.}\)

Uitdelen geeft
\(f(x)={2x^2+5x+18 \over 8x}={2x^2 \over 8x}+{5x \over 8x}+{18 \over 8x}=\frac{1}{4}x+\frac{5}{8}+\frac{9}{4}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{1}{4}+\frac{9}{4}⋅-1⋅x^{-2}=\frac{1}{4}-{9 \over 4x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{1}{4}-{9 \over 4x^2}=0\)
\(\frac{1}{4}={9 \over 4x^2}\)

Kruislings vermenigvuldigen geeft
\(4x^2=36\)
\(x^2=9\)
\(x=\sqrt{9}=3∨x=-\sqrt{9}=-3\)

Schets:

min. is \(f(-3)=-\frac{7}{8}\) en max. is \(f(3)=2\frac{1}{8}\text{.}\)

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

\(f(x)={5 \over 8x+6}=5(8x+6)^{-1}\) geeft
\(f'(x)=5⋅-1⋅(8x+6)^{-2}⋅8={-40 \over (8x+6)^2}\)

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

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

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

\(B(0, 4\frac{1}{2})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2-2x+2=-x^2+5x-3\)
\(2x^2-7x+5=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=(-7)^2-4⋅2⋅5=9\) geeft
\(x={7-\sqrt{9} \over 2⋅2}=1∨x={7+\sqrt{9} \over 2⋅2}=2\frac{1}{2}\)

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

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

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

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

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

\(x_C=4\text{,}\) dus \(y_C=f(4)=10\) en
\(C(4, 10)\text{.}\)