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The question basically is

For each of the following problems, identify the hypothesis (what you can assume is true) and the conclusion (what you are trying to show is true).

Let $f(x) = 2^{-x}$ for any real number $x$. Then $f(x) = x$ for some real number $x$ with $0 \leq x \leq 1$

My solution is

Hypothesis: $f(x) = 2^{-x}$ for any real number $x$ with $0 \leq x \leq 1$.

Conclusion: $f(x) = x$ for some real number $x$.

My answer matches the book's answer expect that in the book this "with $0 \leq x \leq 1$" is part of the conclusion not the hypothesis. My question is why the foregoing statement is part of the conclusion? I only suspect the word "some" in the conclusion. Any hints?

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The hypothesis is that the function $f(x)=2^{-x}$ is defined for any real number. This implies that it is defined also for $0\le x\le 1$, but this is a consequence of the hypothesis.

The thesis or conclusion is that the function has a fixed point $x=f(x)$ in the interval $0\le x\le 1$. In this statement the position of the fixed point is part of the conclusion that we want to prove.

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The hypothesis should be "Let $f(x)=2^{-x}$ for any real number $x$." This is what is stated in the original context. They did not restrict the domain of $f$, so the domain of $f$ is the entire real line.

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