f(1/x) + 2f(x) = x for all nonzero real x.Determine the function f(x)?

Question

I really need help on this problem. I need a proof but I have no idea how to do it.

Answer 1

1. f(1/x) + 2f(x) = x for all nonzero real x... therefore, the same should hold true if you substitute 1/x for x:

2. f(x) + 2f(1/x) = 1/x

Multiply the original equation through by 2:

3. 2f(1/x) + 4f(x) = 2x

Now substract equation #2 from equation #3, and solve for f(x):

( 2f(1/x) + 4f(x) ) - ( f(x) + 2f(1/x) ) = 2x - 1/x
2f(1/x) - 2f(1/x) + 4f(x) - f(x) = 2x^2/x - 1/x
3f(x) = (2x^2 - 1)/x
f(x) = (2x^2 - 1)/3x

Answer 2

f(x) = (2x)/3 - 1/(3x)

How I got it may not be the simplest way, but here goes...

The 1/x was bugging me, so I made a substitution to get something that would hopefully be a little easier to work with: x = e^u, so 1/x = e^-u. Then I defined g(u) = f(e^u), so then the problem is to find g so that

g(-u) + 2g(u) = e^u.

Then, I looked at power series expansions of both sides. Let g(x) = sum an*x^n, where the sum is from 0 to infinity. Then, you get

an((-1)^n + 2) = 1/n!

meaning that an = (1/3)(1/n!) when n is even, and 1/n! when n is odd.

Since the coefficient in front of the 1/n! alternates between 1/3 and 1, I felt it would be a combination of e^u and e^(-u), because the power series for e^-u has coefficients equal to those of e^u, but of alternating sign. Since 2/3 is halfway between 1/3 and 1, that told me the coefficient of e^u is 2/3, and the coefficients of the e^-u term would be added to 2/(3n!) when n is odd, and subtracted when n is even. I then assumed g(u) = (2/3)e^u + ae^-u and solved for a, getting a = -1/3.

I finished by replacing e^u with x to get f(x).

Answer 3

fairly f(x) and f(one million/x) are 2 distinctive applications, yet are related however the function f (think of of it as a mecanish). comparable as you could replace x with any allowed form, you could replace it with yet another variable or function. occasion g(x) = x^2 - 2x g(2) = 2^2 - 2 2 = 4 - 4 = 0 g(10) = 10^2 - 2 10 = a hundred - 20 = 80 g(y) = y^2 - 2y g(one million/y) = (one million/y)^2 - 2(one million/y) = one million/y^2 - 2/y g(x-one million) = (x-one million)^2 - 2(x-one million) additionally you could replace x or y with somthing else g(x-one million) = (x-one million)^2 - 2(x-one million) for x = 3 g(3-one million) = (3-one million)^2 - 2(3-one million) => g(2) = 2^2 - 2 2 = 0 (for sure g(2) can purely equivalent 0 as previously) +++++++++++++++++++++++++++++++++++++++... answer f(x) = 3f(one million/x) + 2x (one million) replace x with one million/x (meaning that one million/x is replaced via x) f(one million/x) = 3f(x) + 2(one million/x) => f(one million/x) = 3f(x) + 2/x (2) f(x) = 3f(one million/x) + 2x (one million) f(one million/x) = 3f(x) + 2/x (2) resolve the 2x2 gadget (the place unknowns are f(x) and f(one million/x)) f(x) - 3 f(one million/x) = 2x -3f(x) + f(one million/x)= 2/x multiply via 3 f(x) - 3 f(one million/x) = 2x -9f(x) + 3f(one million/x)= 6/x (+) upload the two equation ---------------------------- -8 f(x) + 0 f(one million/x) = 2x + 6/x => -8 f(x) = 2x + 6/x => f(x) = 2x / -8 + 6/x /-8 => f(x) = - x/4 - 3/4x ++++++++++++++++++ Amar Soni you're incorrect you replaces f(one million/x) with one million/y that's no longer perfect. for occasion if f(x) = x^2 -x^3 then f(one million/x) = one million/x^2 - one million/x^3 which isn't equivalent to one million/(x^2-x^3)

Answer 4

f(1/x) + 2f(x) = x

f(1/x) = x - 2f(x)

this should be true for all xbelongsto R

so if x=1/x,
f(x) = 1/x -2f(1/x)
or, f(1/x) = 1/(2x) - f(x) / 2

putting this value in f(1/x) = x - 2f(x)

1/(2x) - f(x) / 2 = x - 2f(x)
or, (2-1/2)f(x) = x - 1/(2x)
or (3/2)f(x) = x - 1/2/x
or f(x) = 2x/3 - 3/(2x)

Answer 5

f(x) as well as f(1/x) must have 2 components:
x & 1/x, thus assume f(x)=ax+b/x, a & b being unknown values.
if f(x)=ax+b/x then f(1/x)=a/x+bx.
thus a/x+bx+2ax+2b/x=x or =0/x+1*x
a+2b=0, b+2a=1 that is b=-1/3, a=2/3, & f(x)=2x/3-1/(3x)
CHECK!!!