Toepassingen van de afgeleide functie

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

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

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

\(f'(x)=3x^2+6x-45\)

\(f'(x)=0\) geeft
\(3x^2+6x-45=0\)
\(x^2+2x-15=0\)
\((x+5)(x-3)=0\)
\(x=-5∨x=3\)

Schets:

max. is \(f(-5)=129\) en min. is \(f(3)=-127\text{.}\)

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

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

Schets:

max. is \(f(-2)=130\text{,}\) min. is \(f(0)=34\) en max. is \(f(5)=1\,159\text{.}\)

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

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

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

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

Schets:

min. is \(f(5)=-2\frac{1}{2}\text{.}\)

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

min. is \(f(5)=-2\frac{1}{2}\text{,}\) dus \(B_f=[-2\frac{1}{2}, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={3x^2+27 \over 5x}={3x^2 \over 5x}+{27 \over 5x}=\frac{3}{5}x+\frac{27}{5}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{3}{5}+\frac{27}{5}⋅-1⋅x^{-2}=\frac{3}{5}-{27 \over 5x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{3}{5}-{27 \over 5x^2}=0\)
\(\frac{3}{5}={27 \over 5x^2}\)

Kruislings vermenigvuldigen geeft
\(15x^2=135\)
\(x^2=9\)
\(x=\sqrt{9}=3∨x=-\sqrt{9}=-3\)

Schets:

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

\(f(-3)=2\text{,}\) dus \(A(-3, 2)\)

\(f(x)={-8 \over 4x+8}=-8(4x+8)^{-1}\) geeft
\(f'(x)=-8⋅-1⋅(4x+8)^{-2}⋅4={32 \over (4x+8)^2}\)

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

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

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

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

\(f(-2)=-8\text{,}\) dus \(A(-2, -8)\text{.}\)

\(f(x)=x^3+2x^2+x-6\) geeft \(f'(x)=3x^2+4x+1\text{.}\)

Stel \(l{:}\,y=ax+b\) met \(a=f'(-2)=5\text{.}\)

\(\begin{rcases}y=5x+b \\ \text{door }A(-2, -8)\end{rcases}\begin{matrix}5⋅-2+b=-8 \\ -10+b=-8 \\ b=2\end{matrix}\)
Dus \(l{:}\,y=5x+2\text{.}\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+5x-2=-x^2-2x-5\)
\(2x^2+7x+3=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=7^2-4⋅2⋅3=25\) geeft
\(x={-7-\sqrt{25} \over 2⋅2}=-3∨x={-7+\sqrt{25} \over 2⋅2}=-\frac{1}{2}\)

\(x_A=-3\text{,}\) dus \(y_A=g(-3)=-8\)

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

\(l{:}\,y=ax+b\) met \(a=g'(-3)=4\text{.}\)

\(\begin{rcases}y=4x+b \\ \text{door }A(-3, -8)\end{rcases}\begin{matrix}4⋅-3+b=-8 \\ -12+b=-8 \\ b=4\end{matrix}\)
Dus \(l{:}\,y=4x+4\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+5x-2=4x+4\)
\(x^2+x-6=0\)
\((x+3)(x-2)=0\)
\(x=-3∨x=2\)

\(x_C=2\text{,}\) dus \(y_C=f(2)=12\) en
\(C(2, 12)\text{.}\)

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

\(f'(x)=-5\) geeft
\(x^2+7x+7=-5\)
\(x^2+7x+12=0\)
\((x+4)(x+3)=0\)
\(x=-4∨x=-3\text{.}\)

\(f(-4)=7\frac{1}{3}\text{,}\) dus \(A(-4, 7\frac{1}{3})\text{.}\)

\(f(-3)=2\frac{1}{6}\text{,}\) dus \(B(-3, 2\frac{1}{6})\text{.}\)