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 January 27th, 2017, 02:27 PM #1 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 Check if $y = x^3 - 3x$ is injective. Hi, I have this function: $f(x) = x^3 - 3x$ To check if it is injective, following the definition of injective, I tried this: $a^3 - 3a = b^3 - 3b \\ a^3 - 3a -b^3 + 3b = 0 \\ (a-b)(a^2+ab+b^2)-3(a-b) = 0 \\ (a-b)(a^2 + ab+b^2-3) = 0$ but, I don't know how to continue. Please, can you help me? Many thanks!
 January 27th, 2017, 02:47 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,533 Thanks: 1390 $y = x^3 -3x$ $y = x(x^2-3)$ $y = x(x-\sqrt{3})(x+\sqrt{3})$ thus there are 3 values of $x$ for which $y=0$ Therefore $f(x)$ is not injective. Thanks from beesee Last edited by romsek; January 27th, 2017 at 03:44 PM.
 January 27th, 2017, 03:03 PM #3 Senior Member   Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 yes, thanks, your way is very clear! But, considering what I have done in my first post, I don't know how to resolve $a^2 + ab + b^2 - 3$. Consering the classic $ax^2 + bx +c$ the x in my equation is equal to a, so there would be this: the coefficient of a^2 is 1, the coefficient of a is b, and where is c?
January 27th, 2017, 03:37 PM   #4
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 Originally Posted by beesee yes, thanks, your way is very clear! But, considering what I have done in my first post, I don't know how to resolve $a^2 + ab + b^2 - 3$. Consering the classic $ax^2 + bx +c$ the x in my equation is equal to a, so there would be this: the coefficient of a^2 is 1, the coefficient of a is b, and where is c?
$c = b^2 -3$

 January 27th, 2017, 03:38 PM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 $c$ is $(b^2-3)$ $a = \dfrac{-b \pm \sqrt{b^2 - 4(1)(b^2-3)}}{2}$ Thanks from beesee
 January 27th, 2017, 09:52 PM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics An alternative method: Compute $$\frac{d}{dx} (x^3 - 3x) = 3x^2 - 3$$ and note that $3x^2 - 3$ has 2 real roots, each with multiplicity of 1. Thus, these must correspond to local extrema and therefore $f$ is not injective. Thanks from beesee
January 28th, 2017, 03:21 AM   #7
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 Originally Posted by skeeter $c$ is $(b^2-3)$ $a = \dfrac{-b \pm \sqrt{b^2 - 4(1)(b^2-3)}}{2}$
if I understand, we want to know if there are any real solutions.
so checking the delta:
\begin{align}\Delta &= b^2 - 4(b^2 - 3) \\ &= b^2 - 4b^2 + 12 \\ &= -3b^2+12\end{align}
I find the solutions setting the inequality $\ge 0$, otherwise we have complex solutions:
$-3b^2 + 12 \ge 0 \\ 3b^2 - 12 \le 0 \\ -2 \le b \le 2$
so $(a^2+ab+b^2−3) = 0$ only if b is within and equal to -2 and 2,
i.e. there are multiple values for b and not only one,
and the function is not injective. - What do you think?

And sorry, I have another trivial and simple and banal question:
for checking if a function is injective I have folowed the rule $f(a) = f(b) \iff a = b$,
but, considering this:
Quote:
 Originally Posted by romsek ...thus there are 3 values of $x$ for which $y=0$ Therefore $f(x)$ is not injective.
why consider $y=0$?
maybe depends from the passages $f(a) = f(b) \Rightarrow f(a) - f(b) = 0$?

thanks!

Last edited by beesee; January 28th, 2017 at 03:23 AM.

January 28th, 2017, 05:27 AM   #8
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 Originally Posted by beesee why consider $y=0$? maybe depends from the passages $f(a) = f(b) \Rightarrow f(a) - f(b) = 0$? thanks!
If you can show that any point is many to one, then the function isn't an injection. I chose 0 because it was convenient.

January 28th, 2017, 10:55 AM   #9
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 Originally Posted by romsek If you can show that any point is many to one, then the function isn't an injection. I chose 0 because it was convenient.
the intersection with x axis.

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