Local inverse of 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!!

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##.

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].)

I see, thanks a lot for the help!!

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.

There are results, theorems like the inverse or implicit function theorem that tell you the local conditions when that's edit: possible.

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)## )

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.)

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)

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.

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.