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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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    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.2011-06-27

2 Answers 2

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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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    What does $f$ being monotone have to do with this solution?...2011-06-27
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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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    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.2011-06-27