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

\(f(x)=4x^3-x^2-5x-6\) geeft \(f'(x)=12x^2-2x-5\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(-1)=9\text{.}\)

\(\begin{rcases}y=9x+b \\ \text{door }A(-1, -6)\end{rcases}\begin{matrix}9⋅-1+b=-6 \\ -9+b=-6 \\ b=3\end{matrix}\)
Dus \(l{:}\,y=9x+3\text{.}\)

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

\(f'(x)=-3\) geeft
\(x^2+2x-11=-3\)
\(x^2+2x-8=0\)
\((x+4)(x-2)=0\)
\(x=-4∨x=2\text{.}\)

\(f(-4)=40\frac{1}{3}\text{,}\) dus \(A(-4, 40\frac{1}{3})\text{.}\)

\(f(2)=-13\frac{2}{3}\text{,}\) dus \(B(2, -13\frac{2}{3})\text{.}\)

\(f'(x)=-3x^2+48\)

\(f'(x)=0\) geeft
\(-3x^2+48=0\)
\(x^2-16=0\)
\((x+4)(x-4)=0\)
\(x=-4∨x=4\)

Schets:

min. is \(f(-4)=-148\) en max. is \(f(4)=108\text{.}\)

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

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

Schets:

max. is \(f(-1)=27\text{,}\) min. is \(f(0)=24\) en max. is \(f(1)=27\text{.}\)

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

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

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 \(x=\sqrt{2}\text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(2\sqrt{4x+5}=10\)
\(\sqrt{4x+5}=5\)

Kwadrateren geeft
\(4x+5=25\)
\(x=5\)

Schets:

min. is \(f(5)=-3\text{.}\)

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

min. is \(f(5)=-3\text{,}\) dus \(B_f=[-3, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+5x+1 \over 3x}={4x^2 \over 3x}+{5x \over 3x}+{1 \over 3x}=\frac{4}{3}x+\frac{5}{3}+\frac{1}{3}x^{-1}\)

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

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

Kruislings vermenigvuldigen geeft
\(12x^2=3\)
\(x^2=\frac{1}{4}\)
\(x=\sqrt{\frac{1}{4}}=\frac{1}{2}∨x=-\sqrt{\frac{1}{4}}=-\frac{1}{2}\)

Schets:

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

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

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

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

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

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

\(B(0, -2)\)

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

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

\(l{:}\,y=ax+b\) met \(a=g'(-1)=-2\text{.}\)

\(\begin{rcases}y=-2x+b \\ \text{door }A(-1, 0)\end{rcases}\begin{matrix}-2⋅-1+b=0 \\ 2+b=0 \\ b=-2\end{matrix}\)
Dus \(l{:}\,y=-2x-2\text{.}\)

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

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