Guys i need help with my homework. Here we go: f is deffirentiable

Let g(x) := f(x)/x. Then g(xy) = f(xy)/xy = f(x)/x + f(y)/y = g(x) + g(y). We also have g(1) = g(1) + g(1), so g(1) = 0 and f(1) = 0.

Plugging in y=1/x we get:
0 = f(x*1/x) = xf(1/x) + f(x)/x
So f(1/x) = -f(x)/x^2 and thus g(1/x) = -f(x)/x = -g(x).

We have g'(x)=f'(x)/x - f(x)/x^2 so g'(1) = 1 - 0 = 1.

g(x+h) - g(x) = g(x+h) + g(1/x) = g(1+h/x) = g(1+h/x)
lim_(h -> 0) (g(x+h) - g(x))/h = (g(1+h/x) - g(1))/h = (g(1+h/x) - g(1))/(h/x) * 1/x = g'(1)/x = 1/x.

Integrating from 1 to T yields g(T) = ln T, which proves that f(T) = T ln T.

>>283397
Thx m'lad

B. Clearly x=1 is a solution.
The derivative of x^2 - 1 - 2x*ln(x) is equal to 2x - 2 - 2ln(x) which is positive (I think) so

2x ln x > x^2 - 1 if x < 1
and
2x ln x < x^2 - 1 if x > 1

so there are no other solutions.

>>>/sci/

Go to the 'questions that don't deserve their own thread' thread

>>283424
>Homework threads will be deleted, and the poster banned.
https://www.4chan.org/rules#sci2

>>283426
Huh.

I just always rephrased the problem and posted it as a thing to be solved rather then 'homework'. I actually got a few replies too. Strange that.

>>283427
Mods depend on reports from users/janitors.
The rules are there for a reason. Just because you can get away with selfish behavior doesn't mean you aren't being cancer.

>>283415
Thx m8