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We have this relation: $$x=y^3+y+|y|$$ Problem: Determine if $y$ is a function of $x$.

If there were no $|y|$, I could prove that $y$ is a function of $x$: $$x_1=x_2 \Rightarrow y_1^3 + y_1 = y_2^3 + y_2 \Rightarrow y_1^3 - y_2^3 + y_1 - y_2 =0 $$ $$\Rightarrow (y_1 - y_2)(y_1^2 + y_1y_2 + y_2^2+1) = 0 \qquad (\mathcal{A})$$ And $y_1^2 + y_1y_2 + y_2^2 + 1$ is always greater than (or equal to) $1$ because: $$y_1^2 + y_1y_2 + y_2^2 + 1 \ge 1\Longleftrightarrow y_1^2 + y_1y_2 + y_2^2 \ge 0 \Longleftrightarrow 2y_1^2 + 2y_1y_2 + 2y_2^2 \ge 0 \Longleftrightarrow (y_1^2+2y_1y_2+y_2^2)+y_1^2+y_2^2 \ge 0 \Longleftrightarrow (y_1-y_2)^2+y_1^2+y_2^2 \ge 0$$ Based on $(\mathcal{A})$ we can deduct that: $$ y_1 - y_2 =0 \Rightarrow y_1 = y_2 $$ But this would be true if there were no $|y|$.

Any ideas to solve the problem would be appreciated.

Anonymous
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2 Answers2

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If $y \ge 0$ then $x = y^3 + 2y$ which is strictly increasing on $[0;\infty[$

Else $x = y^3 $ which is strictly increasing on $]-\infty;0[$

$f(y) = y^3+y+|y|$ is continuous and strictly increasing $]-\infty;0[$ and on $[0;\infty[$ therefore strictly increasing on $\mathbb{R}$

So $f$ is injective (and even bijective) and therefore $y$ is a function of $x$

stity
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Assume that $y>0$. Then it makes $x$ to be $>0$. Now let $x_1=x_2$ and we are on the way to show that the value $y_1$ respect to $x_1$ is the same as $y_2$ for $x_2$. If $y_1\neq y_2$ so $y_2-y_1\neq 0$ and so $$(y_1-y_2)(y_1^2+y_1y_2+y_2^2)=-2(y_1-y_2)$$ reaches us to have a contradiction! For $y<0$ is easy to see what we needs. enter image description here

Mikasa
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  • Nice idea. For $y < 0$ we have $x_1=x_2 \Rightarrow y_1^3=y_2^3 \Rightarrow y_1=y_2$. – Anonymous Dec 18 '15 at 13:44
  • yes that is right. If I were you I would ask Ron to undelete his nice answer. Indeed, we need all point of views as different answers. :-) – Mikasa Dec 18 '15 at 14:07