R1 2016 vår LØSNING

oppgaven som pdf

Diskusjon av denne oppgaven

a)

$f(x)=-3x^2+6x-4$

$f'(x)=-6x+6= -6(x-1)$

b)

$g(x)=5\ln(x^3-x)$

$g'(x)=\frac{5(3x^2-1)}{x^3-x}$

c)

$h(x)=\frac{x-1}{x+1}$

$h'(x)=\frac{x+1-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}$

a)

$p(x)=x^3-7x^2+14x+k$

$p(x)$ er delelig med $(x-2)$ hvis og bare hvis $p(2)=0$

$p(2)=8-7\cdot4+14\cdot2+k=8-28+28+k=8+k$

$8+k=0$

$k=-8$

a)

$f(x)=x^2e^{1-x^2}$

$f'(x)=2xe^{1-x^2}+x^2\cdot-2xe^{1-x^2}=2xe^{1-x^2}(1-x^2)$

a)

$AB=AC=BC=6 \ cm$

$HB=\frac{1}{2}AB=3 \ cm$

$CH=\sqrt{(BC)^2-(HB)^2}=\sqrt{6^2-3^2} \ cm=\sqrt{27}=\sqrt{3^3} \ cm=3\sqrt{3} \ cm$

$CF=CE=\sqrt{(BC)^2+(BE)^2}=\sqrt{6^2+6^2} \ cm=\sqrt{2\cdot6^2} \ cm=6\sqrt{2} \ cm$

$HF=\sqrt{(CF)^2-(CH)^2}=\sqrt{72-27} \ cm=\sqrt{45} \ cm=\sqrt{9\cdot5} \ cm=3\sqrt{5} \ cm$

b)

$\frac{AF}{AB}=\frac{3+3\sqrt{5}}{6}=\frac{3(1+\sqrt{5})}{2\cdot3}=\frac{1+\sqrt{5}}{2}=\phi$

b)

$f(x)=x^2+px+q$

$A=(0,1)$

$B=(-p,q)$

$\vec{OS}=\vec{OA}+\frac{1}{2}\vec{AB}=[0,1]+\frac{1}{2}[-p,q-1]=[\frac{-p}{2},1+\frac{q-1}{2}]=[\frac{-p}{2},\frac{q+1}{2}]$

$S=(\frac{-p}{2},\frac{q+1}{2})$

$r=|\vec{AS}|=\sqrt{(\frac{-p}{2})^2+(\frac{q-1}{2})^2}=\sqrt{\frac{p^2+(q-1)^2}{4}}=\frac{\sqrt{p^2+(q-1)^2}}{2}$