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I need an alternate proof for this problem.

Show that the function is one-one, provide a proof.

$f:x \rightarrow x^3 + x : x \in \mathbb{R}$

I needed to show that the function is a one-one function. I tried doing $f(x) = f(y) \Rightarrow x = y$, It ended up with $x(x^2 + 1) = y(y^2 + 1)$. I couldn't figure out how to solve this further.

Instead I had good idea of what the graph looked like, so I tried to show that it is a strictly increasing function.

$f(x_2) - f(x_1) = (x_2^3 - x_1^3) + (x_2 - x_1)$

Thus, $f(x_2) - f(x_1) > 0$

Hence, f(x) is strictly increasing. And hence is one-one.

Unfortunately this solution isn't acceptable. :( Can you guys help me work out how to do it the right way. Thanks again for your help.

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Why isn't your proof acceptable? It is a nice proof... btw, not sure what "right way" would mean... –  Aryabhata Jun 27 '11 at 11:44
    
Did you study about derivatives? –  Dennis Gulko Jun 27 '11 at 11:54
    
I have been doing some extra reading(monotonic functions, etc), I am supposed to solve it within covered course material. –  mathguy80 Jun 27 '11 at 12:09
    
@math: What has been covered? Has this been covered: If $a \gt b$ and $c \gt 0$ then $ac \gt bc$ and $-ac \lt -bc$? –  Aryabhata Jun 27 '11 at 12:27
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However there is a little piece missing: you need to say if $x_2 > x_1$ then $f(x_2) - f(x_1) > 0$ That would make the comment above a proof, but without the "if" part it doesn't work. –  Mark Bennet Jun 27 '11 at 14:58
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2 Answers

up vote 3 down vote accepted

Note that $f'(x) = 3x^{2} + 1 > 0$ for $x \in \mathbb{R}$. So $f$ is monotone.

Try doing this: $$x^{3} + x =y^{3}+y \Longrightarrow (x^{3}-y^{3})= -(x-y)$$

Then use $x^{3}-y^{3} = (x-y) \cdot (x^{2}+xy+y^{2})$. Using this you have $$(x-y) \cdot \Bigl[ x^{2}+xy+y^{2} +1\Bigr] =0$$

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Umm, $f'(x)=3x^2+1$. –  Thomas Andrews Jun 27 '11 at 12:00
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Question is tagged [algebra-precalculus] so talking of derivatives in the very first sentence might not be appropriate. –  Aryabhata Jun 27 '11 at 12:05
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Thanks, I had forgotten the a^3 - b^3 expansion, seems to have come in handy! I get that you then imply that either (x-y) = 0 or $(x^2 + xy + y^2 - 1) = 0$ But how do you show that $(x^2 + xy + y^2 - 1) \ne 0$.? –  mathguy80 Jun 27 '11 at 12:16
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btw, you have $x^2 + xy + y^2 -1$ instead of $x^2 + xy + y^2 + 1$. As stated, you can put $x=0, y=1$. –  Aryabhata Jun 27 '11 at 12:22
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@Chandru: The answer, though accepted, should be corrected. Note that in the first displayed line, you want something like $x^3-y^3=-(x-y)$, which, if $x \ne y$, leads to $x^2+xy+y^2=-1$. Then, for completeness, one shows that $x^2+xy+y^2 \ge 0$ always, maybe by completing the square. –  André Nicolas Jun 27 '11 at 13:16
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You need to show that if $x \neq y$ then $x^3 + x \neq y^3 + y$. If $x < y$, then also $x^3 < y^3$, hence $x+x^3 < y+y^3$; if $x > y$, then $x^3 > y^3$, hence $x+x^3 > y+y^3$. Hence $x \neq y$ implies $x^3 + x \neq y^3 + y$.

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Isn't this what OP has and does not want? I guess OP is missing some details... –  Aryabhata Jun 27 '11 at 11:55
    
@Aryabhata: Let's wait to OP's response... –  Shai Covo Jun 27 '11 at 12:01
    
I'm sorry I am not entirely sure I follow. To prove it's one-one, i need to show $f(x) = f(y) \Rightarrow x = y$, ie:- $x^3 + x = y^3 + y \Rightarrow x = y$. $$ $$ What you are showing is $x \ne y \Rightarrow x^3 + x \neq y^3 + y$. This then implies that $x = y \Rightarrow x^3 + x = y^3 + y$. Hence one-one. Do I interpret this correctly? –  mathguy80 Jun 27 '11 at 12:29
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@mathguy, no, it implies that if $x^3+x=y^3+y$, then $x=y$, and that's exactly what you need. –  Gerry Myerson Jun 27 '11 at 12:46
    
I really don't see how this is different from proving that the function is increasing. btw, your $x \gt y$ part of the proof is redundant, I suppose. –  Aryabhata Jun 27 '11 at 13:10
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