\(f(1)=5\text{,}\) dus \(A(1, 5)\text{.}\)

\(f(x)=-5x^3+6x^2+6x-2\) geeft \(f'(x)=-15x^2+12x+6\text{.}\)

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

\(\begin{rcases}y=3x+b \\ \text{door }A(1, 5)\end{rcases}\begin{matrix}3⋅1+b=5 \\ 3+b=5 \\ b=2\end{matrix}\)
Dus \(l{:}\,y=3x+2\text{.}\)

\(f(x)=\frac{1}{3}x^3+x^2-5x+3\frac{5}{6}\) geeft \(f'(x)=x^2+2x-5\text{.}\)

\(f'(x)=3\) geeft
\(x^2+2x-5=3\)
\(x^2+2x-8=0\)
\((x+4)(x-2)=0\)
\(x=-4∨x=2\text{.}\)

\(f(-4)=18\frac{1}{2}\text{,}\) dus \(A(-4, 18\frac{1}{2})\text{.}\)

\(f(2)=\frac{1}{2}\text{,}\) dus \(B(2, \frac{1}{2})\text{.}\)

\(f'(x)=6x^2-6x-72\)

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

Schets:

max. is \(f(-3)=146\) en min. is \(f(4)=-197\text{.}\)

\(f'(x)=-12x^3-12x^2+240x\)

\(f'(x)=0\) geeft
\(-12x^3-12x^2+240x=0\)
\(x^3+x^2-20x=0\)
\(x(x+5)(x-4)=0\)
\(x=0∨x=-5∨x=4\)

Schets:

max. is \(f(-5)=1\,589\text{,}\) min. is \(f(0)=-36\) en max. is \(f(4)=860\text{.}\)

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

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

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
\(f(x)={x^2+6x+9 \over 3x}={x^2 \over 3x}+{6x \over 3x}+{9 \over 3x}=\frac{1}{3}x+2+3x^{-1}\)

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

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

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

Schets:

min. is \(f(-3)=0\) en max. is \(f(3)=4\text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(2\sqrt{3x+3}=6\)
\(\sqrt{3x+3}=3\)

Kwadrateren geeft
\(3x+3=9\)
\(x=2\)

Schets:

max. is \(f(2)=2\text{.}\)

\(3x+3≥0\) geeft \(x≥-1\text{,}\) dus \(D_f=[-1, \rightarrow ⟩\text{.}\)

max. is \(f(2)=2\text{,}\) dus \(B_f=⟨\leftarrow , 2]\text{.}\)

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

\(f(x)={-6 \over 2x-7}=-6(2x-7)^{-1}\) geeft
\(f'(x)=-6⋅-1⋅(2x-7)^{-2}⋅2={12 \over (2x-7)^2}\)

\(\text{rc}_k=f'(5)=\frac{4}{3}\)

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

\(\begin{rcases}y=-\frac{3}{4}x+b \\ \text{door }A(5, -2)\end{rcases}\begin{matrix}-\frac{3}{4}⋅5+b=-2 \\ -3\frac{3}{4}+b=-2 \\ b=1\frac{3}{4}\end{matrix}\)
Dus \(l{:}\,y=-\frac{3}{4}x+1\frac{3}{4}\text{.}\)

\(B(0, 1\frac{3}{4})\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+3x-3=-x^2+2x+3\)
\(2x^2+x-6=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=1^2-4⋅2⋅-6=49\) geeft
\(x={-1-\sqrt{49} \over 2⋅2}=-2∨x={-1+\sqrt{49} \over 2⋅2}=1\frac{1}{2}\)

\(x_A=-2\text{,}\) dus \(y_A=g(-2)=-5\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-2)=6\text{.}\)

\(\begin{rcases}y=6x+b \\ \text{door }A(-2, -5)\end{rcases}\begin{matrix}6⋅-2+b=-5 \\ -12+b=-5 \\ b=7\end{matrix}\)
Dus \(l{:}\,y=6x+7\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+3x-3=6x+7\)
\(x^2-3x-10=0\)
\((x+2)(x-5)=0\)
\(x=-2∨x=5\)

\(x_C=5\text{,}\) dus \(y_C=f(5)=37\) en
\(C(5, 37)\text{.}\)