
Math General Math Forum  For general math related discussion and news 
 LinkBack  Thread Tools  Display Modes 
January 27th, 2017, 02:27 PM  #1 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 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: $a^3  3a = b^3  3b \\ a^3  3a b^3 + 3b = 0 \\ (ab)(a^2+ab+b^2)3(ab) = 0 \\ (ab)(a^2 + ab+b^23) = 0$ 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: CA Posts: 1,296 Thanks: 664 
$y = x^3 3x$ $y = x(x^23)$ $y = x(x\sqrt{3})(x+\sqrt{3})$ thus there are 3 values of $x$ for which $y=0$ Therefore $f(x)$ is not injective. 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: 153 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: CA Posts: 1,296 Thanks: 664  Quote:
 
January 27th, 2017, 03:38 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,598 Thanks: 1287 
$c$ is $(b^23)$ $a = \dfrac{b \pm \sqrt{b^2  4(1)(b^23)}}{2}$ 
January 27th, 2017, 09:52 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 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.

January 28th, 2017, 03:21 AM  #7  
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  Quote:
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:
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: CA Posts: 1,296 Thanks: 664  
January 28th, 2017, 10:55 AM  #9 
Senior Member Joined: Jan 2013 From: Italy Posts: 153 Thanks: 7  

Tags 
3x$, check, injective 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Function  Disprove "Suppose g o f is injective, then f o g is injective."?  ach4124  Calculus  2  March 27th, 2015 05:37 PM 
Injective Function  Senna  Real Analysis  2  June 27th, 2014 10:00 PM 
Well Defined/Injective  zaff9  Abstract Algebra  6  January 22nd, 2013 08:21 PM 
Proving that "The sum of injective functions is injective."  Zilee  Applied Math  3  October 2nd, 2012 08:07 PM 
Injective  problem  Calculus  0  October 1st, 2010 04:54 AM 