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
Part I (spoiler)
