Domain of a composite function

[songoku]

TL;DR

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.