I Local inverse of non bijective functions

[S123456]

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!!

 

[FactChecker]

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 $x^2$.

 

[pasmith]

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 x for which f(x) = y_0. Saying that an inverse is local to y_0 is therefore ambiguous; saying that it is local to x_0 is not. (The domain of a local inverse of f at x_0 is f(D) for some open D \ni x_0.)

 

[S123456]

I see, thanks a lot for the help!!

 

[FactChecker]

Since the function is not actually invertible, there may be more than one x for which f(x) = y_0. Saying that an inverse is local to y_0 is therefore ambiguous; saying that it is local to x_0 is not. (The domain of a local inverse of f at x_0 is f(D) for some open D \ni x_0.)

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.

 

[WWGD]

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

 

[Bosko]

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 $f^{-1}(y)$ )

 

[SammyS]

Let's analyse this function ... $f(x)=x^4+4x^2$

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

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

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.)

 

[Bosko]

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

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 $1.414 \leq x$

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

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.

 

[FactChecker]

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.