1T 2011 vår LØSNING

Del 1

Oppgave 1

a)

1) <tex>36 200 000 = 3.62 \cdot 10^7</tex>

2) <tex>0.034 \cdot 10^{-2} = 3.4 \cdot 10^{-4}</tex>

b)

<tex>x^2 + 6x = 16 \quad \Leftrightarrow \quad x^2 + 6x - 16 = 0</tex>

Ved fullstendig kvadrat:

<tex>$$\begin{array}{rl}{x}^2+6x-16& =x^2+6x+(\frac{6}{2}{)}^2-16-(\frac{6}{2}{)}^2\\ & =x^2+6x+9-25\\ & =(x+3{)}^2-5^2\\ & =(x+3-5)(x+3+5)\\ & =(x-2)(x+8)\\ & =0\end{array}$$ </tex>

<tex>x = 2 \quad \vee \quad x = -8</tex>

Eller med abc-formelen:

<tex>x = \frac{-6 \pm \sqrt{6^2 - 4\cdot 1 \cdot (-16)} }{2 \cdot 1} = \frac{-6 \pm \sqrt{100}}{2} = -3 \pm 5</tex>

<tex>x = 2 \quad \vee \quad x = -8</tex>

d)

1) E

2) C

3) J

4) B

5) G

6) H

e)

<tex>\text{lg}(2x - 1) = 2</tex>

<tex>2x - 1 = 10^2</tex>

<tex>2x = 101</tex>

<tex>x = \frac{101}{2}</tex>

f)

1)

2)

Oppgave 2

b)

Sekanten er gitt ved

<tex>$$\begin{array}{rl}S(x)& =f(0)+\frac{f(2)-f(0)}{2-0}(x-0)\\ & =-2+\frac{(2^2-2)-(-2)}{2}x\\ & =2x-2\end{array}$$</tex>

c)

Tangenten er gitt ved

<tex>$$\begin{array}{rl}T(x)& =f(1)+f^{\prime }(1)(x-1)\\ & =1^2-2+2(x-1)\\ & =2x-3\end{array}$$ </tex>

Der det er brukt at <tex>f^{\prime}(x) = 2x</tex>.

[BILDE]