Toepassingen van de afgeleide functie

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

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

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

\(f'(x)=6x^2-18x-60\)

\(f'(x)=0\) geeft
\(6x^2-18x-60=0\)
\(x^2-3x-10=0\)
\((x+2)(x-5)=0\)
\(x=-2∨x=5\)

Schets:

max. is \(f(-2)=111\) en min. is \(f(5)=-232\text{.}\)

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

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

Schets:

max. is \(f(-5)=-76\text{,}\) min. is \(f(-3)=-140\) en max. is \(f(0)=49\text{.}\)

\(f(x)=\frac{3}{5}x-\sqrt{3x-5}=\frac{3}{5}x-(3x-5)^{\frac{1}{2}}\) geeft
\(f'(x)=\frac{3}{5}-\frac{1}{2}⋅(3x-5)^{-\frac{1}{2}}⋅3=\frac{3}{5}-{3 \over 2\sqrt{3x-5}}\text{.}\)

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

Kruislings vermenigvuldigen geeft
\(6\sqrt{3x-5}=15\)
\(\sqrt{3x-5}=2\frac{1}{2}\)

Kwadrateren geeft
\(3x-5=6\frac{1}{4}\)
\(x=3\frac{3}{4}\)

Schets:

min. is \(f(3\frac{3}{4})=-\frac{1}{4}\text{.}\)

\(3x-5≥0\) geeft \(x≥1\frac{2}{3}\text{,}\) dus \(D_f=[1\frac{2}{3}, \rightarrow ⟩\text{.}\)

min. is \(f(3\frac{3}{4})=-\frac{1}{4}\text{,}\) dus \(B_f=[-\frac{1}{4}, \rightarrow ⟩\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+x+49 \over 8x}={4x^2 \over 8x}+{x \over 8x}+{49 \over 8x}=\frac{1}{2}x+\frac{1}{8}+\frac{49}{8}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{1}{2}+\frac{49}{8}⋅-1⋅x^{-2}=\frac{1}{2}-{49 \over 8x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{1}{2}-{49 \over 8x^2}=0\)
\(\frac{1}{2}={49 \over 8x^2}\)

Kruislings vermenigvuldigen geeft
\(8x^2=98\)
\(x^2=\frac{49}{4}\)
\(x=\sqrt{\frac{49}{4}}=3\frac{1}{2}∨x=-\sqrt{\frac{49}{4}}=-3\frac{1}{2}\)

Schets:

min. is \(f(-3\frac{1}{2})=-3\frac{3}{8}\) en max. is \(f(3\frac{1}{2})=3\frac{5}{8}\text{.}\)

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

\(f(x)={-1 \over 2x+9}=-1(2x+9)^{-1}\) geeft
\(f'(x)=-1⋅-1⋅(2x+9)^{-2}⋅2={2 \over (2x+9)^2}\)

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

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

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

\(B(0, -3)\)

\(f(1)=3\text{,}\) dus \(A(1, 3)\text{.}\)

\(f(x)=-4x^3+6x^2+2x-1\) geeft \(f'(x)=-12x^2+12x+2\text{.}\)

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

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

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

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

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

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

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

\(x_C=2\text{,}\) dus \(y_C=f(2)=-1\) en
\(C(2, -1)\text{.}\)

\(f(x)=\frac{1}{3}x^3+3x^2+13x+\frac{5}{6}\) geeft \(f'(x)=x^2+6x+13\text{.}\)

\(f'(x)=5\) geeft
\(x^2+6x+13=5\)
\(x^2+6x+8=0\)
\((x+4)(x+2)=0\)
\(x=-4∨x=-2\text{.}\)

\(f(-4)=-24\frac{1}{2}\text{,}\) dus \(A(-4, -24\frac{1}{2})\text{.}\)

\(f(-2)=-15\frac{5}{6}\text{,}\) dus \(B(-2, -15\frac{5}{6})\text{.}\)