Solving f(x) Inverse: x^2-4x, x∈R, |x|≤1

[thereddevils]

Given f(x)=x^2-4x , x\in R , |x|\leq 1

so f(x) is a one to one function , and i am supposed to find its inverse .

and i found :

f^{-1}(x)=2\pm \sqrt{4+x}

this is weird since a single input would give 2 different outputs and it can't be considered a function . But f(x) is a one-one function , so it should have an inverse ?

 

[Cyosis]

You have solved the equation y=x^2-4x[/tex] for all x in R. Yet the original function only exist for |x|<=1. This restricts the domain such that it is one to one (it is not one to one for all x in R). Use this domain to determine which branch you need.

 

[thereddevils]

You have solved the equation y=x^2-4x[/tex] for all x in R. Yet the original function only exist for |x|<=1. This restricts the domain such that it is one to one (it is not one to one for all x in R). Use this domain to determine which branch you need.

<br /> <br /> thanks ,i thought so too but i am not sure how to see from the domain that the inverse should be 2+root(4+x) OR 2-root(4+x)

 

[Mark44]

The domain of f is {x | |x| <= 1}. That domain can be written as an interval, which should help you figure out which branch to use for the inverse.

 

[Cyosis]

The range of f is the domain of f^-1. If f(a)=b then f^-1(b)=a.

 

[thereddevils]

The range of f is the domain of f^-1. If f(a)=b then f^-1(b)=a.

thanks again !

so the domain of f(x) is between 1 and -1 , and the range is between -3 and 5 , so the domain of f^(-1)(x) is also between -3 and 5 ? so both +ve and -ve would produce such range , err i am still confused ,

 

[Cyosis]

I suggest you plug in some numbers to see what happens.

 

[thereddevils]

I suggest you plug in some numbers to see what happens.

thanks !

 

[D H]

But f(x) is a one-one function , so it should have an inverse ?

Broadening things out, it depends on what you mean by one-to-one. The term one-to-one means injective to some, bijective to others. The function is a one-to-one function by both meanings of the term over the domain |x|≤1. It is not bijective over all of the reals, so it does not have an inverse function for this extended domain. (It does however have an inverse multivalued function.)