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).

Well, I can't let a good problem go unsolved

(1) Let x = y

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

(2) Let x = 1/y

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

(3) Multiplying by 4, gives

4f(1/y) + 16f(y) = 12/y

(4) Subtracting equation 1 from equation 3 gives

15f(y) = 12/y - 3y

(5) Dividing by 15 gives f(y) as

** f(y) = 4/5y - y/5**

(6) It is interesting to me that we did not need to calculate the function to solve part 1, but we do need it for part 2.

(7) At any rate, if f(x) = f(-x),

then 4/5x - x/5 = -4/5x + x/5

8/5x = 2x/5

Multiplying by 5x gives 8 = 2x*x

**x = +2 or -2, as the only answer to part 2.**

(8) Checking the solution.

f(2) = 4/10 - 2/5 = 0

f(-2) = -4/10 + 2/5 = 0, so f(2) = f(-2)

(9) Still checking

F(1/2) = 8/5 - 1/10 = 15/10 = 3/2

F(2) + 4f(1/2) = 6 = 3*2, so our basic definition is satisfied.

Nice problem, Jer!