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m25 q22

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m25 q22 [#permalink]

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New post 13 May 2010, 12:46
If function f(x) satisfiesf(x) = f(x^2) for all x , which of the following must be true?



* \(f(4) = f(2)f(2)\)
* \(f(16) - f(-2) = 0\)
* \(f(-2) + f(4) = 0\)
* \(f(3) = 3f(3)\)
* \(f(0) = 0\)

\(f(-2) = f(2) = f(4) = f(16)\) . The other choices are not necessarily true. Consider \(f(x) = 3\) for all \(x\) .
The correct answer is B.
I don't really understand why you are doing f(-2) = f(2) = f(4) = f(16). Where does it say in the problem that you can use the function on a previous function ?????
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Re: m25 q22 [#permalink]

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New post 14 May 2010, 07:41
Very interesting question, I will wait for the specialists. :|

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Re: m25 q22 [#permalink]

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New post 18 Sep 2012, 08:08
So the f(x)=f(x^2), so f(2)=f(2^2)=f(4), and f(4)=f(4^2)=f(16).... therefore f(2)=f(4)=f(16).... Cool?

So then f(-2)=f(-2^2)=f(4), and since f(4)=f(2) you can say f(-2)=f(2)=f(4)....... Cool?

Cool.

B.
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Re: m25 q22 [#permalink]

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New post 18 Sep 2012, 08:21
leilak wrote:
If function f(x) satisfiesf(x) = f(x^2) for all x , which of the following must be true?



* \(f(4) = f(2)f(2)\)
* \(f(16) - f(-2) = 0\)
* \(f(-2) + f(4) = 0\)
* \(f(3) = 3f(3)\)
* \(f(0) = 0\)

\(f(-2) = f(2) = f(4) = f(16)\) . The other choices are not necessarily true. Consider \(f(x) = 3\) for all \(x\) .
The correct answer is B.
I don't really understand why you are doing f(-2) = f(2) = f(4) = f(16). Where does it say in the problem that you can use the function on a previous function ?????


OPEN DISCUSSION OF THIS QUESTION IS HERE: if-function-f-x-satisfies-f-x-f-x-2-for-all-x-107351.html?hilit=function%20satisfies
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Re: m25 q22   [#permalink] 18 Sep 2012, 08:21
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