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$$ f,g\colon \mathbb{R} \to \mathbb{R}, $$

$$ g(f(x)) = 2x^5 + e^f + 1 $$

I need to show that f(x) is 1-1

Also:

$$ g(x) = x^2 -xf(x) + 1 $$

show that f is not 1-1

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    Are these two independent exercises? If yes, what $g$ is doing in the second? (nowhere used). And, is the exponent in your 2nd line $f(x)$? Or $x$?2012-10-30
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    In your first question, you have $e^f$ in right side. Should that be $e^{f(x)}$?2012-10-30
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    They are two questions in the same exercise. I also have no clue what g(x) is for. I corrected the f(x) typo.2012-10-30
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    @coffeemath yes. I could not get x next to f with MathJax.2012-10-30
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    In the second question $f$ maps one to one. Isn't that enough?:)2012-10-30
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    I typed the 2nd question wrong, it's corrected now.2012-10-30
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    Nick, I think there is still a problem with the 2nd question. Here is a "counterexample": Let $f(x) = x$, this is 1-1. Let $g(x)$ be defined as you said. There is no reason for contradiction :)2012-10-30
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    If $g\circ f$ is one-to-one, then $f$ is one-to-one (not necessarily $g$ though). So for the first part it is enough to the show that $g(f(x))$ is one-to-one. Although I can't see how to do that myself without knowing anything about $f'(x)$.2012-10-30
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    Thanks. the only thing I know is that $$ f\colon \mathbb{R} \to \mathbb{R} $$ . I expressed $$ 2x^5 $$ as $$ a(x) $$ and $$ e^f(x) $$ as $$ b(x) $$ . I did indeed show that a(x) is one-to-one but i have no clue about b(x)2012-10-30
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    Are you sure there is no more information there? If $f(x)=x^2$ then $g(f(x))$ in the first part is not even one-to-one at all, so there will be no way to prove that it is for all $f$.2012-10-30
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    For the first question, there is no more information. I am stuck there. For the second question, i guess it's obvious that with x^2 it can't be one-to-one right?2012-10-30

2 Answers 2

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)assume that $f(x_1)=f(x_2)$, then $g(f(x_1))=g(f(x_2))$, i.e., $2x_{1}^{5}+e^{f(x_1)}+1=2x_{2}^{5}+e^{f(x_2)}+1$, then we have $2x_{1}^{5}=2x_{1}^{5}$ so $x_1=x_2$

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    I have done this step. I am stuck at $$ e^f $$2012-10-30
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    Nick: The assumption for showing one to one is (as Tao started with) that $f(x_1)=f(x_2)$. So the terms $e^{f(x_i)}$ ($i=1,2$) cancel out to arrive at Tao's step $2x_1^5=2x_2^5$.2012-10-30
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    Yeah, I feel silly. Nick, we missed the fact that in the $g(f(x))$, the $f(x)$ part must be already built into the $2x^5$ term, which means $f(x)$ can't be even because 5 is prime.2012-10-30
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    where did $$ e^{f(x_2)} $$ go?2012-10-30
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    We assumed that $f(x_1)=f(x_2)$, so that means that $e^{f(x_1)}=e^{f(x_2)}$, and then they subtract out.2012-10-30
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in the second question, f(x) can be 1-1 for example f(x)=x-1

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    Nice example! It makes $g(x)$ come out $x+1$, clearly one to one.2012-10-30