Toepassingen van de afgeleide functie

\(f'(x)=4x^4-7x^2-15\)

\(f'(\sqrt{3})=4(\sqrt{3})^4-7(\sqrt{3})^2-15=0\)

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

\(f'(x)=-6x^2-18x+24\)

\(f'(x)=0\) geeft
\(-6x^2-18x+24=0\)
\(x^2+3x-4=0\)
\((x+4)(x-1)=0\)
\(x=-4∨x=1\)

Schets:

min. is \(f(-4)=-62\) en max. is \(f(1)=63\text{.}\)

\(f'(x)=12x^3-12x^2-72x\)

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

Schets:

min. is \(f(-2)=-16\text{,}\) max. is \(f(0)=48\) en min. is \(f(3)=-141\text{.}\)

\(f(x)=\sqrt{3x+5}-\frac{3}{4}x=(3x+5)^{\frac{1}{2}}-\frac{3}{4}x\) geeft
\(f'(x)=\frac{1}{2}⋅(3x+5)^{-\frac{1}{2}}⋅3-\frac{3}{4}={3 \over 2\sqrt{3x+5}}-\frac{3}{4}\text{.}\)

\(f'(x)=0\) geeft
\({3 \over 2\sqrt{3x+5}}-\frac{3}{4}=0\)
\({3 \over 2\sqrt{3x+5}}=\frac{3}{4}\)

Kruislings vermenigvuldigen geeft
\(6\sqrt{3x+5}=12\)
\(\sqrt{3x+5}=2\)

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

Schets:

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

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

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

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

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

\(f'(x)=0\) geeft
\(4-{81 \over x^2}=0\)
\(\frac{4}{1}={81 \over x^2}\)

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

Schets:

min. is \(f(-4\frac{1}{2})=-35\) en max. is \(f(4\frac{1}{2})=37\text{.}\)

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

\(f(x)={-2 \over 3x+10}=-2(3x+10)^{-1}\) geeft
\(f'(x)=-2⋅-1⋅(3x+10)^{-2}⋅3={6 \over (3x+10)^2}\)

\(\text{rc}_k=f'(-4)=\frac{3}{2}\)

\(\text{rc}_k⋅\text{rc}_l=-1\) geeft \(\text{rc}_l=-\frac{2}{3}\text{,}\) dus \(y=-\frac{2}{3}x+b\)

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

\(B(0, -1\frac{2}{3})\)

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

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

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

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

De snijpunten \(A\) en \(B\) volgen uit
\(x^2-4x-5=-x^2-5x-4\)
\(2x^2+x-1=0\)
\(abc\text{-}\)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 \(y_A=g(-1)=0\)

\(g'(x)=-2x-5\)

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

\(\begin{rcases}y=-3x+b \\ \text{door }A(-1, 0)\end{rcases}\begin{matrix}-3⋅-1+b=0 \\ 3+b=0 \\ b=-3\end{matrix}\)
Dus \(l{:}\,y=-3x-3\text{.}\)

Snijpunt \(C\) volgt uit
\(x^2-4x-5=-3x-3\)
\(x^2-x-2=0\)
\((x+1)(x-2)=0\)
\(x=-1∨x=2\)

\(x_C=2\text{,}\) dus \(y_C=f(2)=-9\) en
\(C(2, -9)\text{.}\)

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

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

\(f(-4)=16\frac{1}{3}\text{,}\) dus \(A(-4, 16\frac{1}{3})\text{.}\)

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