A function f, defined for all non-zero real numbers x, satisfies:

f(x) + 4f(1/x) = 3x.

[1] Find all values of x for which f(x)=f(1/x);

[2] Find all values of x for which f(x)=f(-x).

Really? Nobody has done this yet?

Here's part 1, solving for x if f(x) = f(1/x)

Let x = y

Then 3y = f(y)+4f(1/y) = 5f(y)

Let x = 1/y

Then 3/y = f(1/y)+4f(y) = 5f(y)

So 3y = 3/y

So y*y = 1.

x = y = 1 or -1 (only two solutions).

In fact, if af(x) + bf(1/x) = cx (where c is non-zero), then 1 and -1 are the still the only x values for which f(x) = f(1/x).

I'll let somebody else solve part 2 before I present my solution, but I wanted to get the ball rolling.

*Edited on ***October 25, 2010, 10:24 am**

*Edited on ***October 25, 2010, 10:27 am**