R1 2010 høst LØSNING

a)

1)

<tex>f(x)=2xe^x \\f'(x) = 2e^x+2xe^x = 2(1+x)e^x</tex>

2)

<tex>g(x) = 3\sqrt{x^2-1}\\ \text{setter u lik x i andre minus en og bruker kjerneregelen} \\ g'(x) = (\frac{3}{2\sqrt{x^2-1}}) \cdot (2x) = \frac {3x}{\sqrt{x^2-1}}</tex>

b)

<tex>P(x) = 2x^3-6x^2-2x+6 \\ P(1) = 2 \cdot 1^3- 6 \cdot 1^2 -2 \cdot1 + 6 = 0 \\ \quad \quad (2x^3-6x^2-2x+6 ):(x-1) =2x^2-4x-6 \\ -(2x^3-2x^2) \\ \quad \quad\quad \quad\quad \quad -4x^2-2x \\ \quad \quad\quad \quad-(-4x^2+4x) \\\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad -6x+6 \\\quad \quad\quad \quad\quad \quad\quad \quad\quad \quad -(-6x+6) \\\quad \quad \quad \quad\quad \quad\quad \quad\quad \quad \quad \quad\quad\quad0 </tex>

c)

1)

2)

<tex>\vec v (t) = \vec r'(t) = [6,-10t] \\\vec v (1) = [6,-10] </tex>

3)

<tex>\vec a (t) = \vec v'(t) = [0,-10] </tex> Akslerasjonen i x rettning er null. Akslerasjonen i y rettning er konstant lik 10, nedover.