Hi guys,

I'm a little bit confused over how to apply the Mean Value Theorem to this particular problem.

Suppose f is a function such that f'(x)=1/x for all x>0. Prove that if f(1)=0, then f(xy)=f(x)+f(y) for all x,y >0.

Thanks in advance.

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- Sep 18th 2008, 07:01 AMHweengeeProof using Mean Value Theorem
Hi guys,

I'm a little bit confused over how to apply the Mean Value Theorem to this particular problem.

Suppose f is a function such that f'(x)=1/x for all x>0. Prove that if f(1)=0, then f(xy)=f(x)+f(y) for all x,y >0.

Thanks in advance. - Sep 18th 2008, 08:17 AMThePerfectHacker
For $\displaystyle y>0$ define $\displaystyle F_y(x) = f(xy)-f(x)-f(y)$ for $\displaystyle x>0$.

Now $\displaystyle F_y$ is differenciable on this integral and $\displaystyle F'_y(x) = \frac{y}{yx} - \frac{1}{x} = 0$.

Thus, $\displaystyle F'_y$ is konstant on $\displaystyle (0,\infty)$ so $\displaystyle F_y(x) = k$ for some $\displaystyle k\in \mathbb{R}$.

But $\displaystyle F_y(1) = 0$ and so $\displaystyle F_y(x) = 0$.

Thus, $\displaystyle f(xy) = f(x)+f(y)$. - Sep 18th 2008, 09:00 AMHweengee
"http://www.mathhelpforum.com/math-he...09c4391d-1.gif is konstant on http://www.mathhelpforum.com/math-he...cc47912d-1.gif so http://www.mathhelpforum.com/math-he...a6f6f078-1.gif for some http://www.mathhelpforum.com/math-he...b010bac8-1.gif. "

Do you mean F'_y(x) = 0 --> F_y(x) = k for some k in R? Sorry but I'm still a little confused. - Sep 18th 2008, 09:10 AMThePerfectHacker
- Sep 19th 2008, 04:16 AMHweengee
yes i get it now. thank you very much.