Likning som suger ...

[nøden&nåden]

Hei!

Denna oppg. sliter eg litt med.

Bestem konstanten p slik at g'(4)=1 da g(x)=1/(px)+ [symbol:rot] x

Fasit: p=-(1/12)

g(x)=1/(px)+ [symbol:rot] x

g'(x)=((1*1*p)-(0*p*x)+(0.5x^-0.5)) / (px)^2

som gir,

(1/(px^2))+0.5x^-0.5

g'(4)=1 gir,

1=(1/(p*4^2))+0.5*4^-0.5

p=(1/(1*4^2))+2^-0.5

1/16+2^-0.5

Her tar det stopp ... )))

[Georgio]

[tex] g(x) = \frac{1}{px} + \sqrt{x} [/tex]

[tex] \frac{dg(x)}{dx} = -\frac{1}{px^2} + \frac{1}{2\sqrt{x}} [/tex]

[tex] \frac{dg(4)}{dx} = 1 = -\frac{1}{16p} + \frac{1}{4} [/tex]

[tex] p(1-\frac{1}{4}) = -\frac{1}{16} [/tex]

[tex] p = \frac{-\frac{1}{16}}{1-\frac{1}{4}} = -\frac{1}{12} [/tex]