R1 2016 vår LØSNING

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Diskusjon av denne oppgaven

a)

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

$f^{\prime }(x)=-6x+6=-6(x-1)$

b)

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

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

c)

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

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

a)

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

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

$p(2)=8-7\cdot 4+14\cdot 2+k=8-28+28+k=8+k$

$8+k=0$

$k=-8$

a)

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

$f^{\prime }(x)=2x{e}^{1-x^2}+x^2\cdot -2x{e}^{1-x^2}=2x{e}^{1-x^2}(1-x^2)$

a)

$AB=AC=BC=6cm$

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

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

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

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

b)

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

b)

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

$A=(0,1)$

$B=(-p,q)$

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

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

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

c)

Likning for sirkel:

$(x-x_1{)}^2+(y-y_1{)}^2=r^2$

$(x+\frac{p}{2}{)}^2+(y-\frac{q+1}{2}{)}^2=\frac{p^2+(q-1{)}^2}{4}$

Skjæring med y-aksen:

$y=0$

$(x+\frac{p}{2}{)}^2+(-\frac{q+1}{2}{)}^2=\frac{p^2+(q-1{)}^2}{4}$

$(x+\frac{p}{2}{)}^2=\frac{p^2+(q-1{)}^2}{4}-\frac{(q+1{)}^2}{4}$

$x+\frac{p}{2}=\frac{\pm \sqrt{p^2-4q}}{2}$

$x=\frac{-p\pm \sqrt{p^2-4q}}{2}$

Nullpunkter til $f(x)$:

$x^2+px+q=0$

$x=\frac{-p\pm \sqrt{p^2-4q}}{2}$

Sirkelen skjærer x-aksen i nullpunktene til $f(x)$.