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January 27th, 2017, 03:27 PM   #1
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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!
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January 27th, 2017, 03:47 PM   #2
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$y = x^3 -3x$

$y = x(x^2-3)$

$y = x(x-\sqrt{3})(x+\sqrt{3})$

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 04:44 PM.
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January 27th, 2017, 04:03 PM   #3
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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?
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January 27th, 2017, 04:37 PM   #4
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Quote:
Originally Posted by beesee View Post
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?
$c = b^2 -3$
Thanks from beesee
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January 27th, 2017, 04:38 PM   #5
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$c$ is $(b^2-3)$

$a = \dfrac{-b \pm \sqrt{b^2 - 4(1)(b^2-3)}}{2}$
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January 27th, 2017, 10:52 PM   #6
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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.
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January 28th, 2017, 04:21 AM   #7
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Quote:
Originally Posted by skeeter View Post
$c$ 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 View Post
...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 04:23 AM.
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January 28th, 2017, 06:27 AM   #8
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Quote:
Originally Posted by beesee View Post
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.
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January 28th, 2017, 11:55 AM   #9
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Quote:
Originally Posted by romsek View Post
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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