Local inverse of non bijective functions

  • #1
S123456
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TL;DR Summary
How do I find the local inverse of this non bijective functions
Hi,

I am having a hard time trying to solve this question. How do I find the local inverse at x0?

f (x) = x^4 − 4x^2
Find an expression for f^−1 for f at the point x = −2.
Thanks a lot! I would really appreciate any help!!
 
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  • #2
The inverse would not be "at ##x_0##", it would be defined locally at ##y_0 = f(x_0)##.
This is a homework-type of question for which we are only allowed to give hints and guidance to the work that you show us. So you need to show an attempt at a solution.

To get you started, write the equation as ##0 = x^4-4x^2-y## and apply the quadratic formula to get an equation for ##x^2##.
 
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  • #3
FactChecker said:
The inverse would not be "at ##x_0##", it would be defined locally at ##y_0 = f(x_0)##.

Since the function is not actually invertible, there may be more than one [itex]x[/itex] for which [itex]f(x) = y_0[/itex]. Saying that an inverse is local to [itex]y_0[/itex] is therefore ambiguous; saying that it is local to [itex]x_0[/itex] is not. (The domain of a local inverse of [itex]f[/itex] at [itex]x_0[/itex] is [itex]f(D)[/itex] for some open [itex]D \ni x_0[/itex].)
 
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  • #4
I see, thanks a lot for the help!!
 
  • #5
pasmith said:
Since the function is not actually invertible, there may be more than one [itex]x[/itex] for which [itex]f(x) = y_0[/itex]. Saying that an inverse is local to [itex]y_0[/itex] is therefore ambiguous; saying that it is local to [itex]x_0[/itex] is not. (The domain of a local inverse of [itex]f[/itex] at [itex]x_0[/itex] is [itex]f(D)[/itex] for some open [itex]D \ni x_0[/itex].)
Saying that a function is defined that is the local inverse at ##y_0## is valid. There might be multiple choices for the definition, but if there is a branch that gives an inverse, that can be the definition of an inverse function.
 
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  • #6
There are results, theorems like the inverse or implicit function theorem that tell you the local conditions when that's edit: possible.
 
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  • #7
S123456 said:
TL;DR Summary: How do I find the local inverse of this non bijective functions

Hi,

I am having a hard time trying to solve this question. How do I find the local inverse at x0?

f (x) = x^4 − 4x^2
Find an expression for f^−1 for f at the point x = −2.

Thanks a lot! I would really appreciate any help!!
Let's analyse this function ... ##f(x)=x^4+4x^2##

It can be written as ##f(x)=x^2(x^2+4)##.
It is always greater than or equal to zero because it is ##x^2 \geq 0## and ##x^2+4 \geq 4##
Only for ##x=0## is 0, means ##f(0)=0##.

It can be written as the square of a sum ##(x^2+b)^2##
##~(x^2+b)^2##
##=(x^2+b)(x^2+b)##
##=x^4+bx^2+bx^2+b^2##
##=x^4+2bx^2+b^2##

if b is 2 then above expression becomes ##=x^4+4x^2+4##

##f(x)=x^4+4x^2+4-4##

##f(x)=(x^2+2)^2-4##

Can you now from ##y=(x^2+2)^2-4##, find x=... as function of y ? ( That is ##f^{-1}(y)## )
 
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  • #8
Bosko said:
Let's analyse this function ... ##f(x)=x^4+4x^2##

Bosko said:
##f(x)=x^4+4x^2+4-4##

##f(x)=(x^2+2)^2-4##

Can you now from ##y=(x^2+2)^2-4##, find x=... as function of y ? ( That is ##f^{-1}(y)## )
I suppose you realize that you are working with a different function than given in the OP. What you have done is called "completing the square". The function given in the OP can be treated in a similar fashion. By the way, solving your final expression for ##x##, does not give ##x## as a function of ##y##. Rather it gives ##x## in terms of ##y## as what is sometimes referred to as a multi-valued function.

##\displaystyle \quad \quad f(x)=x^4-4x^2##

##\displaystyle \quad \quad \quad \quad =x^4-4x^2+4-4##

##\displaystyle \quad \quad \quad \quad =(x^2-2)^2-4##

Notice that this does not depend upon ##f(x)## being non-negative.
Setting ##y=f(x)##, we have.

##\displaystyle \quad \quad y=(x^2-2)^2-4##

Solving for ##x## gives the result:

##\displaystyle \quad \quad x=\pm\sqrt{2 \pm \sqrt{y+4\ } \ }##

Seeing as there are two ##\pm## symbols, we may expect as many as 4 distinct values of ##x## to result from a single value of ##y##.

(I expect to add a bit more to this Post as time permits.)

Added after @Bosko replied :

As mentioned in the OP, we are looking for a local inverse for ##f(x)## at ##x=-2=x_0##. So the corresponding ##y_0=f(x_0)=0##. The following function, ##g(y)## does the trick.

##\displaystyle \quad \quad x=g(y)=-\sqrt{2 + \sqrt{y+4\ } \ }##

##\displaystyle \quad \quad g(0)=-\sqrt{2 + \sqrt{0+4\ } \ }=-\sqrt{2 + 2 \ } = -2##

The domain of ##g## is ##[-4,\, \infty)##, the range is ##(-\infty,\,-\sqrt 2 \,] ##.
##g## is monotone decreasing and therefore is a bijection.

(I expect to add a bit more to this Post or make a new Post as time permits.)
 
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  • #9
SammyS said:
I suppose you realize that you are working with a different function than given in the OP. What you have done is called "completing the square".
Now I see. Thanks. I put + instead of - hahaha . All is wrong after that
SammyS said:
The function given in the OP can be treated in a similar fashion. By the way, solving your final expression for ##x##, does not give ##x## as a function of ##y##. Rather it gives ##x## in terms of ##y## as what is sometimes referred to as a multi-valued function.
That is true but, if we cut the original function in 4 parts : ( GeoGebra site drawing)
Screenshot 2024-02-05 at 23.41.57.png

1. for ##x \leq -1.414 ##
2. for ##-1.414 \leq x \leq 0 ##
3. for ##0 \leq x \leq 1.414 ##
4. for ##1.414 \leq x ##

SammyS said:
##\displaystyle \quad \quad x=\pm\sqrt{2 \pm \sqrt{y+4\ } \ }##

Seeing as there are two ##\pm## symbols, we may expect as many as 4 distinct values of ##x## to result from a single value of ##y##.

(I expect to add a bit more to this Post as time permits.)
By replacing ##\pm## with either + or - in your formula we can get 4 inverse functions that corresponds to 4 segments of the original function.
 
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  • #10
You have some options for the inverse function at ##y_0##. The choice of signs gives 4 options. You just need to pick the option that will give you ##x=-2##. At ##x=-2## you can calculate ##y=0##. So what does that tell you about the choice of signs in ##x=\pm \sqrt {2 \pm \sqrt {y+4}}##?
UPDATE: I see that @SammyS already said this in post #8.
 
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