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

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Manager
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Function f(x) satisfies f(x) = f(x^2) for all x. Which of [#permalink]

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03 Sep 2007, 18:44
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Function f(x) satisfies f(x) = f(x^2) for all x. Which of the following must be true?

A) f(4) = f(2)*f(2)

B) f(16) - f(-2) = 0

C) f(-2) + f(4) = 0

D) f(3) = 3*f(3)

E) f(0) = 0

Help
:lol:

P.S. - I know the answer. Tell me how you got it :D
Manager
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03 Sep 2007, 19:19
Huh huh :) I think you misread the stem. It's f(x)=f(x^2), and not f(x) = x^2.

I am not sure how to interpret f(x)=f(x^2).

Any other takers?
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03 Sep 2007, 20:12
raptr wrote:
Function f(x) satisfies f(x) = f(x^2) for all x. Which of the following must be true?

A) f(4) = f(2)*f(2)

B) f(16) - f(-2) = 0

C) f(-2) + f(4) = 0

D) f(3) = 3*f(3)

E) f(0) = 0

Help

P.S. - I know the answer. Tell me how you got it :D

I got B by eliminating the choices that did not satisfy the equation.

A) f(4) = f(4^2) NOT f(2)*f(2)

B) f(16) - f(-2) = 0---> f(16)=f(-2)---->f(-2)=f((-2)^2)=f(4), thus
f(16)=f(4) or f(4)=f(16) or f(4)=f(4^2)

C) f(-2) + f(4) = 0---->f(-2)=- f(4), this equation would work if not for the minus sign before f(4).

D) f(3) = 3*f(3)----->f(3)=f(9) NOT 3f(3)

E) f(0) = 0------>f(0)=f(0^2) or f(0)=f(0) NOT just zero.

of course under time pressure this long calculation is not a good approach. I suggest you try to eliminate most obvious wrong choices in this case E, D, and then try the others.
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03 Sep 2007, 22:03
Oh my ...

You confused me with your "Huh huh I think you misread the stem. It's f(x)=f(x^2), and not f(x) = x^2."

My way was exactly as carpeD's, but I was to shy too post it again =)
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04 Sep 2007, 07:09
i feel vindicated
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04 Sep 2007, 21:53
IrinaOK wrote:
raptr wrote:
Function f(x) satisfies f(x) = f(x^2) for all x. Which of the following must be true?

A) f(4) = f(2)*f(2)

B) f(16) - f(-2) = 0

C) f(-2) + f(4) = 0

D) f(3) = 3*f(3)

E) f(0) = 0

Help

P.S. - I know the answer. Tell me how you got it :D

I got B by eliminating the choices that did not satisfy the equation.

A) f(4) = f(4^2) NOT f(2)*f(2)

B) f(16) - f(-2) = 0---> f(16)=f(-2)---->f(-2)=f((-2)^2)=f(4), thus
f(16)=f(4) or f(4)=f(16) or f(4)=f(4^2)

C) f(-2) + f(4) = 0---->f(-2)=- f(4), this equation would work if not for the minus sign before f(4).

D) f(3) = 3*f(3)----->f(3)=f(9) NOT 3f(3)

E) f(0) = 0------>f(0)=f(0^2) or f(0)=f(0) NOT just zero.

of course under time pressure this long calculation is not a good approach. I suggest you try to eliminate most obvious wrong choices in this case E, D, and then try the others.

Initially though of (E).........but after reading this post m convinced that it should be (B).

Can anyone explain how its it (E)???
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04 Sep 2007, 23:14
OK... I NEED TO APOLOGIZE. THE OFFICIAL ANSWER IS NOT E)

I got confused because I picked E and it was wrong.

I guess if you think of this as a sequence it makes more sense... I don;t have a clear explanation. That's what I was hoping to get out of this tread :)
VP
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05 Sep 2007, 02:06
f(x) = f(x^2)

f(-2) = f(4) = f(16) and so on...

We don't need and can't solve f(16) !!!

since f(-2) = f(16) then:

f(16) - f(-2) = 0

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05 Sep 2007, 06:11
raptr wrote:
Function f(x) satisfies f(x) = f(x^2) for all x. Which of the following must be true?

A) f(4) = f(2)*f(2)

B) f(16) - f(-2) = 0

C) f(-2) + f(4) = 0

D) f(3) = 3*f(3)

E) f(0) = 0

Help

P.S. - I know the answer. Tell me how you got it :D

Got B
Knowing f(x) = f(x^2)
f(-2) = f(4) = f(16)
So f(16) - f(-2) = 0
Re: DS Functions   [#permalink] 05 Sep 2007, 06:11
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