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

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CEO
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Joined: 21 Jan 2007
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Function F(x) satisfies F(x) = F(x^2) for all of x. Which [#permalink] New post 24 Oct 2007, 13:00
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B
C
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E

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

Can someone explain the approach to solving this Q?
VP
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 [#permalink] New post 24 Oct 2007, 13:11
You don't need to solve the given function, only to know that F(x) = F(x^2) so:

F(-2) = F(4) = F(16)

F(16) - F(-2) = 0

the answer is (B)

:)
Director
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Re: PS Functions ^2 [#permalink] New post 24 Oct 2007, 13:15
bmwhype2 wrote:
Function F(x) satisfies F(x) = F(x^2) for all of x. Which 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

Can someone explain the approach to solving this Q?


heh yeah this question was a toughie. i remember struggling with the solution. B is the right answer.

we know that F(x) = F(x^2), so anything in F(x) also going to hold true for F(x^2)

A. F(4) = F(2) * F(2) -> F(2) = F(4) so F(2) * F(2) is actually [F(x)]^2 NOT
F(x^2)

B. F(16) - F(-2) = 0 -> can also rewrite F(16) = F(-2), so since F(x) = F(x^2), then this must be true: F(-2) also equal to F(4)= F(16),

C. F(-2) + F(4) = 0 -> adding them together would not make it true. it would be 2F(x) => F(-2) - F(4) = 0 would be true (same logic as B)

D. F(3) = 3* F(3) -> this is false. the RHS will always be 3 times more LHS or 0 depending on the function of F(x)

E. F(0) = 0 -> F(0) = 0 not necessarily equal to F(0) = F(0)!!! We do not know what F(x) is. For instance, F(x) = x + 1 = F(0) = 0 + 1 = 1 (not 0 in this case)
Re: PS Functions ^2   [#permalink] 24 Oct 2007, 13:15
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