Toepassingen van de afgeleide functie

\(f'(x)=3x^4-12x^2+9\)

\(f'(\sqrt{3})=3(\sqrt{3})^4-12(\sqrt{3})^2+9=0\)

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

\(f'(x)=3x^2-75\)

\(f'(x)=0\) geeft
\(3x^2-75=0\)
\(x^2-25=0\)
\((x+5)(x-5)=0\)
\(x=-5∨x=5\)

Schets:

max. is \(f(-5)=209\) en min. is \(f(5)=-291\text{.}\)

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

\(f'(x)=0\) 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:

min. is \(f(-5)=-1\,648\text{,}\) max. is \(f(0)=-23\) en min. is \(f(4)=-919\text{.}\)

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

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

Kruislings vermenigvuldigen geeft
\(2\sqrt{3x}=9\)
\(\sqrt{3x}=4\frac{1}{2}\)

Kwadrateren geeft
\(3x=20\frac{1}{4}\)
\(x=6\frac{3}{4}\)

Schets:

max. is \(f(6\frac{3}{4})=2\frac{1}{4}\text{.}\)

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

max. is \(f(6\frac{3}{4})=2\frac{1}{4}\text{,}\) dus \(B_f=⟨\leftarrow , 2\frac{1}{4}]\text{.}\)

Uitdelen geeft
\(f(x)={4x^2+81 \over 5x}={4x^2 \over 5x}+{81 \over 5x}=\frac{4}{5}x+\frac{81}{5}x^{-1}\)

De afgeleide is dan
\(f'(x)=\frac{4}{5}+\frac{81}{5}⋅-1⋅x^{-2}=\frac{4}{5}-{81 \over 5x^2}\text{.}\)

\(f'(x)=0\) geeft
\(\frac{4}{5}-{81 \over 5x^2}=0\)
\(\frac{4}{5}={81 \over 5x^2}\)

Kruislings vermenigvuldigen geeft
\(20x^2=405\)
\(x^2=\frac{81}{4}\)
\(x=\sqrt{\frac{81}{4}}=4\frac{1}{2}∨x=-\sqrt{\frac{81}{4}}=-4\frac{1}{2}\)

Schets:

min. is \(f(-4\frac{1}{2})=-7\frac{1}{5}\) en max. is \(f(4\frac{1}{2})=7\frac{1}{5}\text{.}\)

\(f(4)=2\text{,}\) dus \(A(4, 2)\)

\(f(x)={10 \over 2x-3}=10(2x-3)^{-1}\) geeft
\(f'(x)=10⋅-1⋅(2x-3)^{-2}⋅2={-20 \over (2x-3)^2}\)

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

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=\frac{5}{4}\text{,}\) dus \(y=1\frac{1}{4}x+b\)

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

\(B(0, -3)\)

\(f(1)=5\text{,}\) 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 \(l{:}\,y=3x+2\text{.}\)

De snijpunten \(A\) en \(B\) volgen uit
\(x^2+5x+5=-x^2-4x+1\)
\(2x^2+9x+4=0\)
\(a\kern{-.8pt}b\kern{-.8pt}c\text{-}\)formule met \(D=9^2-4⋅2⋅4=49\) geeft
\(x={-9-\sqrt{49} \over 2⋅2}=-4∨x={-9+\sqrt{49} \over 2⋅2}=-\frac{1}{2}\)

\(x_A=-4\text{,}\) dus \(y_A=g(-4)=1\)

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

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

\(\begin{rcases}y=4x+b \\ \text{door }A(-4, 1)\end{rcases}\begin{matrix}4⋅-4+b=1 \\ -16+b=1 \\ b=17\end{matrix}\)
Dus \(l{:}\,y=4x+17\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2+5x+5=4x+17\)
\(x^2+x-12=0\)
\((x+4)(x-3)=0\)
\(x=-4∨x=3\)

\(x_C=3\text{,}\) dus \(y_C=f(3)=29\) en
\(C(3, 29)\text{.}\)

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

\(f'(x)=4\) geeft
\(x^2+x-2=4\)
\(x^2+x-6=0\)
\((x+3)(x-2)=0\)
\(x=-3∨x=2\text{.}\)

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

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