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f(x+y)=f(x)+f(y)/1+f(x)f(y) what are even functions f(x),f(y)?

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lemong | Student, Undergraduate | (Level 1) Honors

Posted July 21, 2012 at 12:51 PM via web

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

what are even functions f(x),f(y)?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 21, 2012 at 3:31 PM (Answer #1)

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You should remember what an even function represents such that:

`f(-x) = f(x), for all x in R`

You need to make the following assumption such that:

`x = y = 0`

`f(0) = (f(0)+f(0))/(1+f(0)f(0))`

`f(0) = (2f(0))/(1+f^2(0)) =gt f(0) + f^3(0) = 2f(0) =gt f^3(0) = 2f(0) - f(0)`

`f^3(0) = f(0) =gt f(0) in {-1;0;1}`

You also need to make the following assumption such that:

`y = -x`

`f(x - x) = (f(x)+f(-x))/(1+f(x)f(-x))`

Since the function f(x) is even, hence `f(-x) = f(x)`  such that:

`f(0) = (2f(x))/(1 + f^2(x))`

Considering `f(0) = 0 =gt (2f(x))/(1 + f^2(x)) = 0`

Since the denominator `(1 + f^2(x)) != 0 =gt 2f(x) = 0 =gt f(x)=0`

Considering`f(0) =1 =gt (2f(x))/(1 + f^2(x)) = 1`

`f^2(x) - 2f(x) + 1 = 0`

`(f(x) - 1)^2 = 0 =gt f(x) - 1 = 0 =gt f(x) = 1`

Considering `f(0) = -1 =gt (2f(x))/(1 + f^2(x)) = -1`

`f^2(x)+ 2f(x) + 1 = 0`

`(f(x) + 1)^2 = 0 =gt f(x) + 1 = 0 =gt f(x) = -1`

Hence, evaluating the even functions under the given conditions yields `f(x) in {-1;0;1}`  for all `x in R` .

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