For \(0<x<1\) you have \(f(x)=f(x^{(2^n)})=f(0)\) by continuity at 0. For \(x>1\) you have \(f(x)=f(x^{(1/2^n)})=f(1)\) by continuity at 1 . For negative \(x\) you have \(f(x)=f(x^2)=f(-x)\). Finally you have \(f(0)=f(1-1/n)=f(1)\) by continuity at 1. (Here, \(n\) denotes a positive integer.) Note that all we need is continuity at 0 and 1.

If \(f(x)\) is continuous , we can simply take \(f(x) = kx\) where k is some real constant.
Given , \[\Rightarrow f(x) = f(x^2)\]
\[\Rightarrow kx = kx^2 \]
\[\Rightarrow x(x-1) = 0\]
\[ x = 0 \ \text{or} \ x = 1 \]
Therefore, \(f(x) = 0 \ \text{or} \ f(x) = 1 = \ \text{constant}\). Hence Proved

If questions states that \(f(x)\) is a continuous function, them we can take \(f(x)\) as any continuous function, to prove \(f(x)\) continuous..
If you choose any continuous function which is not constant, then \(f(x) \neq f(x^2)\)

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TopNewestFor \(0<x<1\) you have \(f(x)=f(x^{(2^n)})=f(0)\) by continuity at 0. For \(x>1\) you have \(f(x)=f(x^{(1/2^n)})=f(1)\) by continuity at 1 . For negative \(x\) you have \(f(x)=f(x^2)=f(-x)\). Finally you have \(f(0)=f(1-1/n)=f(1)\) by continuity at 1. (Here, \(n\) denotes a positive integer.) Note that all we need is continuity at 0 and 1.

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If \(f(x)\) is continuous , we can simply take \(f(x) = kx\) where k is some real constant.

Given , \[\Rightarrow f(x) = f(x^2)\] \[\Rightarrow kx = kx^2 \] \[\Rightarrow x(x-1) = 0\] \[ x = 0 \ \text{or} \ x = 1 \]

Therefore, \(f(x) = 0 \ \text{or} \ f(x) = 1 = \ \text{constant}\). Hence Proved

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How do we know \(f(x)\) is a linear function? How do we even know that it is a polynomial? (We don't.)

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If questions states that \(f(x)\) is a continuous function, them we can take \(f(x)\) as any continuous function, to prove \(f(x)\) continuous..

If you choose any continuous function which is not constant, then \(f(x) \neq f(x^2)\)

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