ya think?
3f(1/x)+(2f(x)/x)=x^2
now, if x = -2,
3f(-1/2) + 2f(-2)/-2 = 4
3f(-1/2) - f(-2) = 4
if x = -1/2,
3f(-2)+2f(-1/2)/(-1/2) = 1/4
-4f(-1/2) + 3f(-2) = 1/4
f(-2) = 67/20
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or, we have
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3f(1/x)+2(1/x)f(x) = x^2
3f(x) + 2xf(1/x) = 1/x^2
2f(x)+3xf(1/x) = x^3
3f(x)+2xf(1/x) = 1/x^2
4f(x)+6xf(1/x) = 2x^3
9f(x)+6xf(1/x) = 3/x^2
5f(x) = 3/x^2 - 2x^3 = (3-2x^5)/x^2
f(x) = (3-2x^5)/(5x^2)
f(-2) = (3+64)/20 = 67/20
Suppose f(x) is a rational function such that 3f(1/x)+(2f(x)/x)=x^2 for all non-zero x. Find f(-2).
I noticed that f appears as f(x) and f(1/x). That's probably important...
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