Domain of a composite function

[songoku]

TL;DR Summary

Let say I have f(x) = ##-x^2 + 3 , x \leq 0##. I want to find the domain of composite function ##ff^{-1}## and ##f^{-1} f##

##f(x)=-x^2 + 3## so ##f^{-1} (x)=- \sqrt{3-x} ~, x \leq 3##

##ff^{-1} = - (- \sqrt{3-x})^2 + 3 = x## and the domain will be ##x \leq 3##

##f^{-1} f = - \sqrt{3-(-x^2+3)} = -x ## and the domain will be ##x \leq 0##

My question:
##ff^{-1} (x)## or ##f^{-1} f(x) ## is not always equal to ##x## and their domain can be different?

Thanks

 

[Mark44]

Summary:: Let say I have f(x) = ##-x^2 + 3 , x \leq 0##. I want to find the domain of composite function ##ff^{-1}## and ##f^{-1} f##

##f(x)=-x^2 + 3## so ##f^{-1} (x)=- \sqrt{3-x} ~, x \leq 3##

##ff^{-1} = - (- \sqrt{3-x})^2 + 3 = x## and the domain will be ##x \leq 3##

##f^{-1} f = - \sqrt{3-(-x^2+3)} = -x ## and the domain will be ##x \leq 0##

My question:
##ff^{-1} (x)## or ##f^{-1} f(x) ## is not always equal to ##x## and their domain can be different?

Thanks

Here ##ff^{-1} (x)## and ##f^{-1} f(x) ## are equal (= x) on their common domain, which is ##x \le 0##. Outside that common domain, one or both of ##ff^{-1} (x)## and ##f^{-1} f(x) ## don't necessarily result in x.
For example, f(x) is defined at x = 1, but outside of the restricted domain of this problem.
##f^{-1}(f(1)) = f^{-1}(2) = -\sqrt{3 - 2} = -1 \ne 1##

 

[songoku]

Here ##ff^{-1} (x)## and ##f^{-1} f(x) ## are equal (= x) on their common domain, which is ##x \le 0##.

I try using ##x=-6## which in their common domain.

##ff^{-1} (-6) = f(-3) = -6##

##f^{-1}f (-6) = f^{-1}(-33) = -6##

But from my working in post#1, ##f^{-1} f = - \sqrt{3-(-x^2+3)} = -x \rightarrow f^{-1} f(x) = -x##
So, ##f^{-1} f(-6) = - (-6) = 6## which is clearly wrong but I do not understand why. Is ##f^{-1} f(x) = -x## correct?

Outside that common domain, one or both of ##ff^{-1} (x)## and ##f^{-1} f(x) ## don't necessarily result in x.
For example, f(x) is defined at x = 1, but outside of the restricted domain of this problem.
##f^{-1}(f(1)) = f^{-1}(2) = -\sqrt{3 - 2} = -1 \ne 1##

I thought the domain of ##f^{-1} f(x)## is ##x \le 0## so we can not put ##x=1## to ##f^{-1} f(x)## because it is outside the allowed domain?

Thanks

 

[Mark44]

I try using ##x=-6## which in their common domain.

##ff^{-1} (-6) = f(-3) = -6##

##f^{-1}f (-6) = f^{-1}(-33) = -6##

But from my working in post#1, ##f^{-1} f = - \sqrt{3-(-x^2+3)} = -x \rightarrow f^{-1} f(x) = -x##

The above should be ##f^{-1}( f(x))##

So, ##f^{-1} f(-6) = - (-6) = 6## which is clearly wrong

No, the correct result is in the work near the top of this post; namely ##f^{-1}(f (-6)) = f^{-1}(-33) = -6##

but I do not understand why. Is ##f^{-1} f(x) = -x## correct?

No. Keep in mind that the domain of f is ##x \le 0##.
##f^{-1}( f(x)) = f^{-1}(-x^2 + 3) = -\sqrt{3 + x^2 - 3} = -\sqrt{x^2} = -|x|##
But since ##x \le 0##, then |x| = -x, so -|x| = x.
Hence ##f^{-1}( f(x)) = x, which is the right result for the composition of a function and its inverse.

I thought the domain of ##f^{-1} f(x)## is ##x \le 0## so we can not put ##x=1## to ##f^{-1} f(x)## because it is outside the allowed domain?

I mentioned in my previous post that x = 1 was not in the restricted domain of f, but we could still do the calculation of ##f^{-1}( f(x))##.
[/quote]The above

 

[songoku]

The above should be ##f^{-1}( f(x))##

You mean I should use bracket for "f(x)" part or maybe you have other purpose stating this part?

Hence ##f^{-1}( f(x)) = x##, which is the right result for the composition of a function and its inverse.

Actually I have note from teacher saying this exact thing: "the composition of a function and its inverse will always be equal to ##x##"

When I was practicing, I encounter problem on post#1 and I am confused why ##f(f^{-1}(x))## is ##x## but ##f^{-1}( f(x))## is ##-x##
You have explained that actually my working is wrong, both should be equal to ##x##

So the composition of a function and its inverse will always be equal to ##x##?

I mentioned in my previous post that x = 1 was not in the restricted domain of f, but we could still do the calculation of ##f^{-1}( f(x))##.

Oh, I thought we can only use value of ##x## in the restricted domain. Same as ##f(x)##, the domain is ##x \le 0## so I thought we can not put ##x > 0## into ##f(x)##

If I want to draw the graph of ##f(f^{-1}(x))##, should I just draw ##y=x## that extends from ##(-\infty , \infty)## or I draw ##y=x## that extends only from ##(-\infty , 3]## ?

Thanks

 

[Mark44]

You mean I should use bracket for "f(x)" part or maybe you have other purpose stating this part?

Yes, because ##f^{-1]f## could be mistaken for the product rather than the composition of the two functions.

Actually I have note from teacher saying this exact thing: "the composition of a function and its inverse will always be equal to ##x##"

When I was practicing, I encounter problem on post#1 and I am confused why ##f(f^{-1}(x))## is ##x## but ##f^{-1}( f(x))## is ##-x##
You have explained that actually my working is wrong, both should be equal to ##x##

So the composition of a function and its inverse will always be equal to ##x##?

Yes, but only on their shared domain.

Oh, I thought we can only use value of ##x## in the restricted domain. Same as ##f(x)##, the domain is ##x \le 0## so I thought we can not put ##x > 0## into ##f(x)##

If I want to draw the graph of ##f(f^{-1}(x))##, should I just draw ##y=x## that extends from ##(-\infty , \infty)## or I draw ##y=x## that extends only from ##(-\infty , 3]## ?

For the functions of this problem, only the part of the line y = x that's in the third quadrant; i.e., for ##x \in (-\infty, 0]##.

 

[songoku]

For the functions of this problem, only the part of the line y = x that's in the third quadrant; i.e., for ##x \in (-\infty, 0]##.

Sorry I don't understand why it should be only the 3rd quadrant part

Is ##x \le 3## the correct domain for ##f(f^{-1}(x))## ?

 

[Mark44]

Sorry I don't understand why it should be only the 3rd quadrant part

Is ##x \le 3## the correct domain for ##f(f^{-1}(x))## ?

For that composition, yes, but for ##f(f^{-1}(x))## to be equal to ##f^{-1}(f(x))##, we need the domain that both functions share: ##x \in (-\infty, 0]##. On that interval, the line y = x is in only the 3rd quadrant.

 

[songoku]

For that composition, yes, but for ##f(f^{-1}(x))## to be equal to ##f^{-1}(f(x))##, we need the domain that both functions share: ##x \in (-\infty, 0]##. On that interval, the line y = x is in only the 3rd quadrant.

So if I only want to draw graph of ##f(f^{-1}(x))##, I still have to take the domain of ##f^{-1}(f(x))## into consideration? I can not just consider only domain of ##f(f^{-1}(x))## eventhough that is the graph I want to draw?

Thanks

 

[Mark44]

So if I only want to draw graph of ##f(f^{-1}(x))##, I still have to take the domain of ##f^{-1}(f(x))## into consideration?

No, that would be the line y = x, for ##x \in (-\infty, 3]##. For the graph of the other composition, ##y =(f^{-1}(f(x))##, that would be the same line, but for ##x \in (-\infty, 0]##.
What I said was for both compositions to be equal, we have to be using the common domain, ##x \in (-\infty, 0]##.

 

[songoku]

I suppose all the ##\infty## is typo, should be ##-\infty## instead.

Thank you very much for all the explanation Mark44

 

[Mark44]

I suppose all the ##\infty## is typo, should be ##-\infty## instead.

Thank you very much for all the explanation Mark44

Yes, I neglected to add the minus sign. I've edited my post to correct them.