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# m25 q22

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Intern
Joined: 17 Mar 2010
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m25 q22 [#permalink]  13 May 2010, 11:46
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 $$x$$ .
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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Current Student
Joined: 31 Mar 2010
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Schools: Tuck Class of 2013
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Re: m25 q22 [#permalink]  14 May 2010, 06:41
Very interesting question, I will wait for the specialists.
Manager
Joined: 02 May 2012
Posts: 109
Location: United Kingdom
WE: Account Management (Other)
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Re: m25 q22 [#permalink]  18 Sep 2012, 07:08
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]  18 Sep 2012, 07:21
Expert's post
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 $$x$$ .
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, 07:21
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# m25 q22

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