Matematik opgave 244 249-250 253-256

Indholdsfortegnelse

1 Opgave 244
2 Opgave 249
3 Opgave 250
4 Opgave 253
5 Opgave 254
6 Opgave 255
7 Opgave 256

Opgave 244

$LaTex: \begin{eqnarray} f(x) &=& \sqrt{x}\\ f'(x) &=& \frac{1}{2\cdot\sqrt{x}}\\ p(x) &=& f(16)+f'(16)\cdot(x-16)\\ p(13) &=& \frac{29}{8}\\ p(14) &=& \frac{15}{4}\\ p(15) &=& \frac{31}{8}\\ p(16) &=& 4\\ p(17) &=& \frac{33}{8}\\ \end{eqnarray}$

Opgave 249

$LaTex: f'(x)=\frac{log(1+10^{-4})-log(1-10^{-4})}{2\cdot 10^{-4}} = 0,43429$
$LaTex: f'(x)=\frac{\sqrt{2\cdot (2-10^{-2})^2+3}-\sqrt{2\cdot (2-10^{-2})^2+3}}{2\cdot 10^{-2}}=1,206$
$LaTex: f'(x)=\frac{\frac{2}{3\cdot (-3+10^{-2})-1}-\frac{2}{3\cdot (-3-10^{-2})-1}}{2\cdot 10^{-2}} = -0,06$

Opgave 250

Se Martin

Opgave 253

f(0)=f'(0): Er selvfølgelig det samme fordi toppunktet ligger i 0,0.

se evt: Matematik Bevis differential 2, for at finde f'(x).

$LaTex: \begin{eqnarray} f(x) &=& a\cdot x^2 \\ f'(x) &=& 2\cdot a\cdot x\\ a\cdot x^2 &=& 2\cdot a\cdot x\\ \frac{a\cdot x^2}{x} &=& 2\cdot a\\ x &=& 2\\ \end{eqnarray}$

Opgave 254

$LaTex: \begin{eqnarray} f(x) &=& \sqrt[4]{\frac{1}{8}}\cdot x^2\\ g(x) &=& \frac{1}{x}\\ f'(x) &=& 2\cdot \sqrt[4]{\frac{1}{8}} \cdot x\\ g'(x) &=& \frac{-1}{x^2}\\ x_0 &=& \sqrt[4]{2}\\ f'(x_0) &=& 2\cdot \sqrt[4]{\frac{1}{8}} \cdot \sqrt[4]{2} = 1,4241\\ g'(x_0) &=& \frac{-1}{(\sqrt[4]{2})^2} = \\ f'(x_0) \cdot g'(x_0) &=& -1\\ \end{eqnarray}$

Derfor er linjerne ortogonale.

Opgave 255

$LaTex: \begin{eqnarray} f(x) &=& \sqrt{x}\\ f'(x) &=& \frac{1}{2\cdot \sqrt{x}}\\ a\cdot y +4\cdot x &=& 7\\ y &=& \frac{7-4\cdot x}{a}\\ a_h &=& \frac{-1}{f'(16)} = -8\\ a &=& \frac{-4}{a_h} = \frac{1}{2} \end{eqnarray}$