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!
Part 2: Spoiler
