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Function f(x) satisfies f(x) = f(x^2) for all x. Which of

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CEO
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Function f(x) satisfies f(x) = f(x^2) for all x. Which of [#permalink] New post 16 Oct 2007, 10:10
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Function f(x) satisfies f(x) = f(x^2) for all x. Which of the following must be true?

f(4) = f(2)*f(2)
f(16) - f(-2) = 0
f(-2) + f(4) = 0
f(3) = 3*f(3)
f(0) = 0
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 [#permalink] New post 16 Oct 2007, 10:31
For the function f(x) = f(x^2), f(0) is always = 0. E.
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 [#permalink] New post 16 Oct 2007, 11:23
the second
f(-2) = f(4)=f(16) - >
f(16) - f(-2) = 0

btw
f(x) = f(x^2) means f(x) = const, but not only 0
so f(0) could be 0, but not must be 0.
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 [#permalink] New post 16 Oct 2007, 11:40
Djames wrote:
the second
f(-2) = f(4)=f(16) - >
f(16) - f(-2) = 0

btw
f(x) = f(x^2) means f(x) = const, but not only 0
so f(0) could be 0, but not must be 0.



I get the part up til f(16) -f(-2)=0.

f(-2)= f(4)=f(16) ok this makes sense, but why does it equate to

f(16) -f(-2)=0? Can someone explain this in detail.

Thanks.
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 [#permalink] New post 16 Oct 2007, 13:47
Are you saying that all of these are correct?
bkk145 wrote:
GMATBLACKBELT wrote:
Djames wrote:
the second
f(-2) = f(4)=f(16) - >
f(16) - f(-2) = 0

btw
f(x) = f(x^2) means f(x) = const, but not only 0
so f(0) could be 0, but not must be 0.



I get the part up til f(16) -f(-2)=0.

f(-2)= f(4)=f(16) ok this makes sense, but why does it equate to

f(16) -f(-2)=0? Can someone explain this in detail.

Thanks.


f(16) = f(-2) is same as
f(16) - f(-2) = 0
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 [#permalink] New post 16 Oct 2007, 14:18
ramubhaiya wrote:
Are you saying that all of these are correct?


Yes.

These are all correct
f(-2) = f(4)
f( 4) = f(16)
f(-4) = f(16)
f(-2) = f(4)= f(-4) = f(16)

This is correct too
f(x) = f(x^2) means f(x) = const, but not only 0
so f(0) could be 0, but not must be 0.

Given
f(x) = f(x^2)
These are NOT correct:
f(x^2) = x
f(x) = x^2
  [#permalink] 16 Oct 2007, 14:18
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