Maths help needed

[apiel06]

ok i hope somebody can help me here cause i'm not getting very far.

ok so i got f(x)= 2x + pln(x) and g(x)=px^2
and i gotta know for what p these two function TOUCH not cross over or w/e
so pasically for the same x and p -> f'(x)=g'(x) AND f(x)=g(x)

f'(x) = 2 + p/x
g'(x) = 2px

2x + p = 2px^2 and
2x + pln(x) = px^2

and thats where i got stuck :S both just fail hardcore lol...
hope you guys can help

thanks

 

[Koot]

Solve for p for both so you know the values when the 2 graphs intersect and when the 2 slopes are the same meaning they touch.

 

[apiel06]

yea <.< i got that far myself but how the **** do i solve p and x with two retarded formulas like this :S

 

[Determinate]

You could try a partial derivative?

I know they exist, and that they would probably solve your problem, but I don't know how you use them.

 

[video]

You could try a partial derivative?

I know they exist, and that they would probably solve your problem, but I don't know how you use them.

False. Keep your partial derivatives in your multivariable calculus, this is just a simple, 2-dimensional system of equations.

Take the first equation: 2x + p = 2px^2 and solve for p. You should get p=2x/(2x^2-1)

Take the second equation: 2x + pln(x) = px^2 and plug in what you got for p in the first equation. It's too messy to type out, so just use a calculator to solve for x. Then plug this value back into either equation in place of x and solve for p

 

[apiel06]

False. Keep your partial derivatives in your multivariable calculus, this is just a simple, 2-dimensional system of equations.

Take the first equation: 2x + p = 2px^2 and solve for p. You should get p=2x/(2x^2-1)

Take the second equation: 2x + pln(x) = px^2 and plug in what you got for p in the first equation. It's too messy to type out, so just use a calculator to solve for x. Then plug this value back into either equation in place of x and solve for p

aah thanks! for some reason i didn't get the p=2x/(2x^2 -1)
yess now it all fits, cheers!