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If function f(x) satisfies f(x) = f(x^2) for all x [#permalink]
 04 Jan 2011, 15:21
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If 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) = 0C. f(-2) + f(4) = 0D. f(3) = 3f(3)E. f(0) = 0I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down.
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tonebeeze wrote: If 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) = 3f(3)
e. f(0) = 0
I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down. This function question is different from the problems you are referring to. We are told that some function f(x) has following property f(x) = f(x^2) for all values of x. Note that we don't know the actual function, just this one property of it. For example for this function f(3)=f(3^2) --> f(3)=f(9), similarly: f(9)=f(81), so f(3)=f(9)=f(81)=.... Now, the question asks: which of the following MUST be true? A. f(4)=f(2)*f(2): we know that f(2)=f(4), but it's not necessary f(2)=f(2)*f(2) to be true (it will be true if f(2)=1 or f(2)=0 but as we don't know the actual function we can not say for sure); B. f(16) - f(-2) = 0: again f(-2)=f(4) =f(16)=... so f(16)-f(-2)=f(16)-f(16)=0 and thus this option is always true; C. f(-2) + f(4) = 0: f(-2)=f(4), but it's not necessary f(4) + f(4)=2f(4)=0 to be true (it will be true only if f(4)=0, but again we don't know that for sure); D. f(3)=3*f(3): is 3*f(3)-f(3)=0? is 2*f(3)=0? is f(3)=0? As we don't know the actual function we can not say for sure; E. f(0)=0: And again as we don't know the actual function we can not say for sure. Answer: B. Hope it's clear.
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Re: If function f(x) satisfies f(x) = f(x^2) for all x [#permalink]
18 Oct 2012, 13:50
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tonebeeze wrote: If 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) = 3f(3)
E. f(0) = 0
I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down. Rather than analyzing each answer, I would like to point out how the correct answer should look like. From the equation f(x) = f(x^2) we can get a chain of equalities between the values of the function f at different points. So, we will be able to deduce different equalities of the type f(a)=f(b), but there is no way to find explicit values of the function in any specific point. The correct answer should be of this form, or its equivalent f(a)-f(b)=0. Only answer B is of this type. For any specific value of x, except 0 and 1, we can start an infinite chain of equalities. For example, start with x=2: f(2)=f(4)=f((-2)^2)=f(-2)=f(4^2)=f(16)=f((-4)^2)=f(-4)=f(16^2)=f(256)=f((-16)^2)=f(-16)=.... We can see that for a given x, the function f will have the same value at all the points x, x^2,x^4,x^8,..., -x,-x^2,-x^4,-x^8,...For 0, we just get f(0)=f(0^2), while for x=1, we have f(1)=f((-1)^2)=f(-1).
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f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..
as f(x) = f(x^2)
so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0
so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have
f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0
Please clarify
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Joy111 wrote: f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..
as f(x) = f(x^2)
so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0
so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have
f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0
Please clarify Can you please tell me what do you mean by the red part? Anyway: we are told that f(x) = f(x^2), so f(-2)=f(4) =f(16)=... --> f(16)-f(-2)=f(16)-f(16)=0. Thus option D is always true.
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Bunuel wrote: Joy111 wrote: f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..
as f(x) = f(x^2)
so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0
so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have
f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0
Please clarify Can you please tell me what do you mean by the red part? Anyway: we are told that f(x) = f(x^2), so f(-2)=f(4) =f(16)=... --> f(16)-f(-2)=f(16)-f(16)=0. Thus option D is always true. given statement = f(16)-f(-2)= 0 which can be written as 1) f(16)-f(4) lets say we stop here , how do we know that this equation will be 0 2) f(16)-f(16) this as we can see will of course yield a 0 3)f(16)-f(256) this is yet another way of writing f(16) - f(-2) , again how can we be sure that this will always be 0 without knowing the function . we are manipulating f(16)-f(-2) to become f(16)-f(16) this of course as we can see will be zero but what if we write f(16)-f(-2) as f(-2)-f(256 ) and stop here , without the function it is difficult to see how this will yield a 0 if (ii) is the answer then f(16)-f(-2) = 0 must be true for all functions for the value's of x=16 and x= -2 but since f(x) = f(x^2) then f(-2)- f( 256) = 0 must also be true for all functions for the value's of x=-2 and x= 256 I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero. is it possible to clearly prove this by taking the example of one actual function rather then dealing ambiguously. thank you.
Last edited by Joy111 on 24 May 2012, 03:13, edited 3 times in total.
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Joy111 wrote: Bunuel wrote: Joy111 wrote: f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..
as f(x) = f(x^2)
so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0
so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have
f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0
Please clarify Can you please tell me what do you mean by the red part? Anyway: we are told that f(x) = f(x^2), so f(-2)=f(4) =f(16)=... --> f(16)-f(-2)=f(16)-f(16)=0. Thus option D is always true. since f(x) = f(x^2) so we can write f(16) = f( 256) similarly f(-2)= f(4) = f(16)= f(256) =f( 256^2) hence f(16) - f(-2) can also be written as f( 256) - f( 256^2) so how is this statement f(16) -f(-2) = 0 = f( 256) - f( 256^2) always valid I don't understand your question at all. f(16)-f(-2) and f(16)-f(16^2) both equal to zero since f(x) = f(x^2) for all x, from which we have that f(16)=f(-2) and f(16)=f(16^2).
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I don't understand your question at all.
f(16)-f(-2) and f(16)-f(16^2) both equal to zero since f(x) = f(x^2) for all x, from which we have that f(16)=f(-2) and f(16)=f(16^2).[/quote]
I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero. f(-2) could equal = f(256) = f(256^2) etc
so f(16)- f( 256^2) , is this equal to 0 as well.
.
is it possible to clearly prove this by taking the example of one actual function .Thank you
every thing will fall into place , if this question was COULD be true rather than MUST be true , could you please check this once more .
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Joy111 wrote: I think we are only considering one condition, the condition when f(-2) = f(16) then of course it is 0, but what if f(-2) is not equal to f(16) , in that case how can we say that f(16)- f( -2) will always be zero. f(-2) could equal = f(256) = f(256^2) etc
so f(16)- f( 256^2) , is this equal to 0 as well.
.
is it possible to clearly prove this by taking the example of one actual function .Thank you
every thing will fall into place , if this question was COULD be true rather than MUST be true , could you please check this once more . I think that you just have some problems understanding the question. See the red part " what if f(-2) is not equal to f(16) ". That's the point: since f(x) = f(x^2) for all x then f(16)=f(-2) without any "what if".
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Joy111 wrote: f(16) =f(16^2) = f( 16^4) .....and f(-2)= f(4) =f(16)=f(256)=f( 256^2)..
as f(x) = f(x^2)
so we can write f(16)-f(-2) as= f(16^2) -f(256^2) != 0
so f(16) = f( 16^2) = f(256)
if f(-2) can be written as f(16)
then f(-2) can also be written as f( 256^2) as shown above then when we have
f(256)-f(256^2) then for this particular condition I believe that f(x) cannot be true , so how are we confirming that f(16)-f(-2) MUST always be equal to 0
Please clarify Let us assume f(x) = y. f(x) = f(x^2) means that for all values of x, f(x) takes the exact same value as f(x^2) takes. And vice versa f(x^2) is always equal to f(x). so, f(x^2) also equals y. Now irrespective of how far you go on squaring x, the value f(x) or f(x^2) or f(x^4) etc will always be y.
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Bunuel wrote: tonebeeze wrote: If 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) = 3f(3)
e. f(0) = 0
I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down. This function question is different from the problems you are referring to. We are told that some function f(x) has following property f(x) = f(x^2) for all values of x. Note that we don't know the actual function, just this one property of it. For example for this function f(3)=f(3^2) --> f(3)=f(9), similarly: f(9)=f(81), so f(3)=f(9)=f(81)=.... Now, the question asks: which of the following MUST be true? A. f(4)=f(2)*f(2): we know that f(2)=f(4), but it's not necessary f(2)=f(2)*f(2) to be true (it will be true if f(2)=1 or f(2)=0 but as we don't know the actual function we can not say for sure); B. f(16) - f(-2) = 0: again f(-2)=f(4) =f(16)=... so f(16)-f(-2)=f(16)-f(16)=0 and thus this option is always true; C. f(-2) + f(4) = 0: f(-2)=f(4), but it's not necessary f(4) + f(4)=2f(4)=0 to be true (it will be true only if f(4)=0, but again we don't know that for sure); D. f(3)=3*f(3): is 3*f(3)-f(3)=0? is 2*f(3)=0? is f(3)=0? As we don't know the actual function we can not say for sure; E. f(0)=0: And again as we don't know the actual function we can not say for sure. Answer: B. Hope it's clear. who created this question, the solution is trivial has no end, how can you stop f(16) just as it is, f(16) can be f(16 sqaured) so on and so forth, there is no end to it unless some boundary conditions mentioned.
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koro12 wrote: Bunuel wrote: tonebeeze wrote: If 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) = 3f(3)
e. f(0) = 0
I reviewed the previous function posts that Bunuel referred me to and seem to understand those problems decently. But I don't clearly understand the process of substitution and POE on this problem. I would appreciate if someone could help me with understand how to break the answer choices down. This function question is different from the problems you are referring to. We are told that some function f(x) has following property f(x) = f(x^2) for all values of x. Note that we don't know the actual function, just this one property of it. For example for this function f(3)=f(3^2) --> f(3)=f(9), similarly: f(9)=f(81), so f(3)=f(9)=f(81)=.... Now, the question asks: which of the following MUST be true? A. f(4)=f(2)*f(2): we know that f(2)=f(4), but it's not necessary f(2)=f(2)*f(2) to be true (it will be true if f(2)=1 or f(2)=0 but as we don't know the actual function we can not say for sure); B. f(16) - f(-2) = 0: again f(-2)=f(4) =f(16)=... so f(16)-f(-2)=f(16)-f(16)=0 and thus this option is always true; C. f(-2) + f(4) = 0: f(-2)=f(4), but it's not necessary f(4) + f(4)=2f(4)=0 to be true (it will be true only if f(4)=0, but again we don't know that for sure); D. f(3)=3*f(3): is 3*f(3)-f(3)=0? is 2*f(3)=0? is f(3)=0? As we don't know the actual function we can not say for sure; E. f(0)=0: And again as we don't know the actual function we can not say for sure. Answer: B. Hope it's clear. who created this question, the solution is trivial has no end, how can you stop f(16) just as it is, f(16) can be f(16 sqaured) so on and so forth, there is no end to it unless some boundary conditions mentioned. The question asks which of the options presented must be true. Now, since f(x) = f(x^2) then f(-2)=f(16) so option B. f(16) - f(-2) = 0 is always true. Next, the fact that f(16)=f(16^2) does not make option B any less correct. So there is nothing wrong with the question above. Hope it's clear.
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Algebra (Functions) [#permalink]
18 Oct 2012, 09:17
Please can someone the solution for the appended problem. Dint quite get the explaination provided. -Source - Gmat Club Test
if 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)=3f(3)
E.f(0)=f(0)
Ans is B
Thank you.
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Re: Algebra (Functions) [#permalink]
18 Oct 2012, 10:00
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Re: Algebra (Functions)
[#permalink]
18 Oct 2012, 10:00
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