R1 2008 vår LØSNING: Forskjell mellom sideversjoner
Fra Matematikk.net
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=== b) === | === 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 -3x+6 \quad \quad \quad\quad \quad \quad -(-3x+6)\\ \quad \quad \quad\quad \quad \quad 0</tex> | <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 -3x+6 \\ \quad \quad \quad\quad \quad \quad -(-3x+6)\\ \quad \quad \quad\quad \quad \quad 0</tex> | ||
=== c) === | === c) === | ||
Sideversjonen fra 9. mar. 2012 kl. 06:37
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 -3x+6 \\ \quad \quad \quad\quad \quad \quad -(-3x+6)\\ \quad \quad \quad\quad \quad \quad 0</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)
2)