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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 $x$? –  Berci Oct 30 '12 at 14:42
    
In your first question, you have $e^f$ in right side. Should that be $e^{f(x)}$? –  coffeemath Oct 30 '12 at 14:43
    
They are two questions in the same exercise. I also have no clue what g(x) is for. I corrected the f(x) typo. –  Nickolas Oct 30 '12 at 14:43
    
@coffeemath yes. I could not get x next to f with MathJax. –  Nickolas Oct 30 '12 at 14:44
    
In the second question $f$ maps one to one. Isn't that enough?:) –  Berci Oct 30 '12 at 14:51
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2 Answers

up vote 2 down vote accepted

)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 $$ e^f $$ –  Nickolas Oct 30 '12 at 15:27
    
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 $2x_1^5=2x_2^5$. –  coffeemath Oct 30 '12 at 15:37
    
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. –  Todd Wilcox Oct 30 '12 at 15:41
    
where did $$ e^{f(x_2)} $$ go? –  Nickolas Oct 30 '12 at 15:51
    
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. –  Todd Wilcox Oct 30 '12 at 16:07
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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. –  coffeemath Oct 30 '12 at 15:54
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