R1 2012 vår LØSNING

Oppgave 1:

a)

1)

<tex>f(x) = 5x^3+x-4 \\ f'(x) = 3 \cdot 5x^2 + 1 \\ f'(x) = 15x^2 + 1 </tex>

2)

<tex>g(x) = 5e^{3x} \\ u = 3x \wedge u' = 3 \\ g'(x) = 5e^u \cdot u' \\ g'(x) = 15e^{3x}</tex>

b)

<tex> 2ln(\frac{a^2}{b}) + ln (a \cdot b) - 3ln a = \\ 2ln a^2 - 2ln b + ln a + lnb - 3 lna = \\4ln a - 2ln b + ln a + lnb - 3 lna = \\ 2lna - lnb </tex>

c)

<tex> f(x)= x^3-3x</tex>

1)

Nullpunkter:

<tex>x^3-3x = x(x^2-3)= x(x- \sqrt 3 )(x + \sqrt 3) \\x = - \sqrt3 \quad \vee \quad x = 0 \quad \vee \quad x= \sqrt3</tex>

2)

<tex>f'(x) = 3x^2-3 \\f'(x) = 0 \\ 3(x^2-1) = 0 \\ x = -1 \quad \vee \quad x = 1 \\ f(-1)= 2 \quad \vee \quad f(1) = -2</tex>

Toppunkt (-1,2)

Bunnpunkt (1,-2)

3)

d)

<tex>P(x) = x^3-3x^2-x+3 \\ P(3) = 27-27-3+3 =0 \\ \\ P(x):(x-3) \\ (x^3-3x^2-x+3): (x-3) =x^2-1

\\-(x^3-3x^2)\\ \quad \quad \quad \quad\quad \quad \quad -(-x+3) \\ \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad\quad \quad \quad 0</tex>

Dette gir følgende løsninger:

x = - 1 eller x = 1 eller x = 3.

e)

<tex>\vec r(t) = [3,0t ,-4,9t^2] \\ \vec v(t) = \vec r'(t) = [3,0 , -9,8t] \\ \vec a(t) = \vec v'(t) = \vec r(t) = [0 , -9,8] </tex>

Oppgave 3:

a)

<tex> f(x)= \frac1x \\ f'(x) = - \frac {1}{x^2} \\ f'(a) = - \frac {1}{a^2} \\ Rett \quad linje: \quad y=ax+b \\ y= - \frac{1}{a^2}x+ b </tex>

Finner b ved å bruke punktet (a, f(a)):

<tex>y = - \frac{1}{a^2}x+ b \\ \frac 1a = - \frac{1}{a^2}a+ b \\ b= \frac 2a </tex>

Som gir likningen

<tex>y = - \frac{1}{a^2}x+ \frac 2a</tex>