S1 eksempeloppgave 2015 vår LØSNING

Oppgave 1

a)

$$\begin{gathered} f(x)=3x^2-4x+2 \\ f´(x)=6x-4 \end{gathered}$$

b)

$$\begin{gathered} g(x)=3x^3-3 \\ g´(x)=9x^2 \\ g´(2)=9\cdot 4=36 \end{gathered}$$

Oppgave 2

a)

$\frac{2^{-1}\cdot a\cdot b^{-1}}{4^{-1}\cdot a^{-2}\cdot b^2}=\frac{4a^3}{2b^3}=2(\frac{a}{b}{)}^3$

b)

$lg(a^{2}b)+lg(a{b}^2)+lb(\frac{a}{b^3})=2lga+lgb+lga+2lgb+lga-3lgb=4lga$

c)

$\frac{3a^2-75}{6a+30}=\frac{3(a+5)(a-5)}{6(a+5)}=\frac{a-5}{2}$

Oppgave 3

a)

$$\begin{gathered} \frac{{61}^2-{39}^2}{{51}^2-{49}^2}= \\ \frac{(61+39)(61-39)}{(51+49)(51-49)}= \\ \frac{100\cdot 22}{100\cdot 2}=11 \end{gathered}$$

b)

$$\begin{gathered} 1997\cdot 2003-1993\cdot 2007= \\ (2000-3)(2000+3)-(2000-7)(2000+7) \\ {2000}^2-3^2-{2000}^2+7^2 \\ -9+49=40 \end{gathered}$$

Oppgave 4

$$\begin{gathered} f(x)=g(x) \\ {x}^2-x-2=x+1 \\ {x}^2-2x-3=0 \\ x=\frac{2\pm \sqrt{4+12}}{2} \\ x=-1\vee x=3 \\ g(-1)=0\wedge g(3)=4 \end{gathered}$$

Skjæringspunktene mellom f og g er (-1, 0) og (3,4)

Oppgave 5

a)

$$\begin{gathered} 3x^2=18-3x \\ 3x^2+3x-18=0 \\ {x}^2+x-6=0 \\ x=\frac{-1\pm \sqrt{1+24}}{2} \\ x=\frac{-1\pm 5}{2} \\ x=-3\vee x=2 \end{gathered}$$

b)

$$\begin{gathered} 3\cdot 2^x=24 \\ {2}^x=8 \\ {2}^x=2^3 \\ x=3 \end{gathered}$$

c)

$$\begin{gathered} 3^8+3^8+3^8+3^8+3^8+3^8+3^8+3^8+3^8=3^x \\ 9\cdot 3^8=3^x \\ {3}^{10}=3^x \\ 10lg3=xlg3 \\ x=10 \end{gathered}$$

Oppgave 6

a)

$$\begin{gathered} y=a\cdot b^x \\ {b}^x=\frac{y}{a} \\ xlgb=lg(\frac{y}{a}) \\ x=\frac{lg(\frac{y}{a})}{lgb} \end{gathered}$$

Oppgave 8

$$\begin{gathered} f(x)=x^2+2x{D}_f=\text{R} \\ f´(x)=2x+2 \end{gathered}$$

Vi skal bruke definisjonen på den deriverte til å vise dette:

$$\begin{gathered} f´(x)=li{m}_{\mathrm{\Delta }x\to 0}\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x} \\ =li{m}_{\mathrm{\Delta }x\to 0}\frac{(x+\mathrm{\Delta }x{)}^2+2(x+\mathrm{\Delta }x)-x^2-2x}{\mathrm{\Delta }x} \\ =li{m}_{\mathrm{\Delta }x\to 0}\frac{x^2+2x\mathrm{\Delta }x+(\mathrm{\Delta }x{)}^2+2x+2\mathrm{\Delta }x-x^2-2x}{\mathrm{\Delta }x} \\ =li{m}_{\mathrm{\Delta }x\to 0}\frac{\mathrm{\Delta }x(2x+\mathrm{\Delta }x+2)}{\mathrm{\Delta }x} \\ =li{m}_{\mathrm{\Delta }x\to 0}2x+\mathrm{\Delta }x+2 \\ =2x+2 \end{gathered}$$

Oppgave 10

TREKANTTALL

a)

n	$a_n$	$a_n$	Total
Trafikalt	8	9	17
Ikke trafikalt	5	8	13
Total	13	17	30