R1 2008 vår LØSNING: Forskjell mellom sideversjoner
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<tex>f''(x) = (x-2)e^{-x}\\f''(x)=0 \Rightarrow x=2 \\ \text{Vendepunkt} | <tex>f''(x) = (x-2)e^{-x}\\f''(x)=0 \Rightarrow x=2 \\ \text{Vendepunkt:}\quad(2,f(2)) = (2,2e{-2}) = (2, \frac{2}{e^2})</tex> | ||
== Oppgave 2 == | == Oppgave 2 == | ||
Sideversjonen fra 9. mar. 2012 kl. 07:58
Del 1
Oppgave 1
a)
<tex>f(x) = x^2 \cdot lnx \\ f'(x) = 2x \cdot lnx + \frac 1x \cdot x^2 = 2xlnx+x = (2lnx+1)x</tex>
b)
<tex>\quad(x^3-4x^2+x+6):(x-2) =x^2 -2x - 3 \\ -(x^3-2x^2)\\ \quad \quad \quad\quad \quad -2x^2+x \\ \quad \quad \quad -(-2x^2+4x) \\ \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad -3x+6 \\ \quad \quad \quad\quad \quad \quad\quad \quad \quad \quad -(-3x+6)\\ \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad\quad \quad \quad0</tex>
c)
<tex>\lim_{x\to 8} \frac{x^2-64}{2x+16} =\lim_{x\to 8} \frac{(x-8)(x+8)}{2(x-8)}= \lim_{x\to 8} \frac{(x+8)}{2}=8 </tex>
d)
<tex>lg(x \cdot y^2)-2lgy+ lg(\frac{x}{y^2}) = lgx + 2lgy - 2lgy +lgx - 2lgy = 2(lgx-lgy)= 2lg ( \frac xy)</tex>
e)
1)
<tex>f(x) = xe^{-x} \\ f'(x) = 1 \cdot e^{-x} + x \cdot e^{-x} \cdot (-1) = e^{-x}-xe^{-x} = (1-x)e^{-x}</tex>
2)
<tex>f(x) = (x-2)e^{-x}\\f(x)=0 \Rightarrow x=2 \\ \text{Vendepunkt:}\quad(2,f(2)) = (2,2e{-2}) = (2, \frac{2}{e^2})</tex>