R1 2011 vår LØSNING: Forskjell mellom sideversjoner
Fra Matematikk.net
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<tex>f(x)= x^3-3x^2-13x+15 \\ f(1)= 1-3-13+15 = 0 \\(x^3-3x^2-13x+15):(x-1)= | <tex>f(x)= x^3-3x^2-13x+15 \\ f(1)= 1-3-13+15 = 0 \\(x^3-3x^2-13x+15):(x-1)= x^2-2x-15 \\-(x^3-x^2) \\-2x^2-13x \\-(-2x^2+2x)\\-15x+15 \\ -(-15x+15) \\ 0 </tex> | ||
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Sideversjonen fra 13. mar. 2012 kl. 07:49
DEL 1
Oppgave 1
a)
<tex>O(x)= \frac{500}{x} + 8x^2 \\ O(x) = 500x^{-1} + 8x^2 \\ O'(x) = -500x^{-2}+ 16x = \frac{-500}{x^2} + 16x = \frac{-500 +16x^3}{x^2}</tex>
b)
1)
<tex>f(x)= 3ln(2x) \\ f'(x) = 3 \cdot \frac{1}{(2x)}\cdot 2 = \frac {6}{2x} = \frac 3x</tex>
2)
<tex>g(x) = 3x \cdot e^{x^2} \\ g'(x) = 3e^{x^2}+3x \cdot 2x \cdot e^{x^2} = (3+6x^2)e^{x^2}</tex>
c)
1)
<tex>f(x)= x^3-3x^2-13x+15 \\ f(1)= 1-3-13+15 = 0 \\(x^3-3x^2-13x+15):(x-1)= x^2-2x-15 \\-(x^3-x^2) \\-2x^2-13x \\-(-2x^2+2x)\\-15x+15 \\ -(-15x+15) \\ 0 </tex>