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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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    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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    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