R1 2012 vår LØSNING
DEL EN
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 2:
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+ b</tex>