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f(x) = x^2 / (x^4 - 1) What is f(1/x) in terms of f(x)? f -f

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f(x) = x^2 / (x^4 - 1) What is f(1/x) in terms of f(x)? f -f [#permalink] New post 16 Nov 2007, 07:16
f(x) = x^2 / (x^4 - 1)
What is f(1/x) in terms of f(x)?

f[x]
-f[x]
1/f[x]
-1/f[x]
2*f[x]
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Re: f(1/x) [#permalink] New post 16 Nov 2007, 07:21
bmwhype2 wrote:
f(x) = x^2 / (x^4 - 1)
What is f(1/x) in terms of f(x)?

f[x]
-f[x]
1/f[x]
-1/f[x]
2*f[x]


f(1/x) = (1/x)^2 / ((1/x)^4 - 1) = x^4/((x^2)* (1- x^4))
= x^2/(1-x^4)= - ( x^2/(x^4-1))
= -f(x)

I pick B.

What is OA?
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 [#permalink] New post 16 Nov 2007, 11:57
f(x) = x^2 / (x^4 - 1)
What is f(1/x) in terms of f(x)?

f[x]
-f[x]
1/f[x]
-1/f[x]
2*f[x]

f(1/x) = (1/x)^2 / [(1/x)^4 - 1]
= 1/x^2 / [1/x^4 - 1]

Clearly, the only answer that could make sense is C


what is the OA ?
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 [#permalink] New post 16 Nov 2007, 12:14
It is B f(x) = x^2/(X^4 - 1)
f(1/x) = (1/x)^2/((1/x)^4 -1)
= (1/x)^2/(-(x^4- 1)/x^4)
= x^4/x^2 * -(1- x^4)
= (x^2 * x^2)/ x^2 * -(1-x^4)
= f(x) * (x2/ -x2)
= -f(x)
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Re: f(1/x) [#permalink] New post 17 Nov 2007, 00:55
bmwhype2 wrote:
f(x) = x^2 / (x^4 - 1)
What is f(1/x) in terms of f(x)?

f[x]
-f[x]
1/f[x]
-1/f[x]
2*f[x]


If x = 2 then f(x) = 4/15 and f(1/x) = -4/15 which is equal to -f(x)

answer B.
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Re: f(1/x) [#permalink] New post 17 Nov 2007, 22:49
bmwhype2 wrote:
f(x) = x^2 / (x^4 - 1)
What is f(1/x) in terms of f(x)?

f[x]
-f[x]
1/f[x]
-1/f[x]
2*f[x]


(1/x^2)/(1/x^4-1) --> (1/x^2)/(1-x^4)/(x^4) --> just cancel out the x's. we get x^2/(1-x^4) --> x^2/-(x^4-1) ---> -(x^2)/(x^4-1)

so we get -f(x). So B



U can do it algebraically, which i suggest u do first, but if its not workin out for ya then just plug in a value for x. (obviously use easy numbers).

Just make X=2.

4/16-1 ---> 4/15

Then (1/4)/(1/16-1) ---> 1/4/-15/16 --> -4/15 ---> so its b b/c -(-4/15) ---> 4/15.
Re: f(1/x)   [#permalink] 17 Nov 2007, 22:49
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