Bunuel wrote:

snkrhed wrote:

the key to solving this problem is that if not given a restriction as in b... the function can go on forever right?

Below is a detailed solution of this problem. Hope it helps.

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. \(f(0) = 0\)

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.

Hi Bunuel,

thanks for the explanation. I found this question to be very tough and I seem to be the only one not being able to understand this even after your explanation

I could understand why B is the answer and why not C and D. But I am still confused with A and E. I have tried to explain my thought process below -

We know that f(x) = f(x^2) means f(2) = f(4) = f(16) and so on.....

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

LHS is

f(4) which means (16)RHS is [color=#ff0000]f(2)*f(2) means 4*4 means (16)[/color]

Hence LHS = RHS?

As I am writing this, it occurred to me that here we are multiplying to functions {f(2)*f(2)} and we don't really know if multiplying of two functions will actually result in the multiplication of those two numbers/integers? it could result in some other function as well? Am I thinking in the right direction?

Coming to option E - it says f(0)=0

we know that f(x) = f(x^2)

if x is 0...its square (infact any exponent) or function of its square should always result in 0. I can't think through what's wrong with E?

Could you please help me understand where I am going wrong. Many thanks for your help.

Kind regards

f(4)=f(16), not 16 and f(2)*f(2)=f(4)*f(4), not 4*4.