 My Math Forum > Math Check if $y = x^3 - 3x$ is injective.
 User Name Remember Me? Password

 Math General Math Forum - For general math related discussion and news

 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
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,533
Thanks: 1390

Quote:
 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
Senior Member

Joined: Jan 2013
From: Italy

Posts: 154
Thanks: 7

Quote:
 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
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,533
Thanks: 1390

Quote:
 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
Senior Member

Joined: Jan 2013
From: Italy

Posts: 154
Thanks: 7

Quote:
 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. Tags 3x\$, check, injective Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ach4124 Calculus 2 March 27th, 2015 05:37 PM Senna Real Analysis 2 June 27th, 2014 10:00 PM zaff9 Abstract Algebra 6 January 22nd, 2013 08:21 PM Zilee Applied Math 3 October 2nd, 2012 08:07 PM problem Calculus 0 October 1st, 2010 04:54 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      