# Function question

• Feb 15th 2008, 05:48 AM
Tangera
Function question
Hello! My question is as follows:

f(x) = (x-3)^3 , x > 3

Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.

I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!
• Feb 15th 2008, 06:10 AM
mr fantastic
Quote:

Originally Posted by Tangera
Hello! My question is as follows:

f(x) = (x-3)^3 , x > 3

Find f^-1(f(x)) and f(f^-1(x)), stating clearly the rules and domains.

I know that the rules of both are the same: f^-1(f(x)) = x and f(f^-1(x)) = x. However, I don't understand why are the domains different? Please explain to me! Thank you!

You need reasoning similar to that used here.
• Feb 15th 2008, 06:56 AM
Tangera
I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?
• Feb 15th 2008, 01:34 PM
mr fantastic
Quote:

Originally Posted by Tangera
I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?

There is a rule that says:

$\text{dom} \, f^{-1} = \text{ran} \, f$

$\text{ran} \, f^{-1} = \text{dom} \, f$

These rules can be thought of as a consequence of swapping y and x around in y = f(x) to get the rule for the inverse function ..... y's and x's swap --> domain and range swap .....

$\text{dom}\, f = [3, \infty) \,$ and $\text{ran}\, f = [0, \infty) \,$.

Therefore:

$\text{dom} \, f^{-1} = \text{ran} \, f = [0, \infty)$.

$\text{ran} \, f^{-1} = \text{dom} \, f = [3, \infty)$.

The composite function g(f(x)) exists iff $\text{ran} f = \text{dom} \, g \,$. When that condition is met, the domain of g(f(x)) is $\text{dom} f$. In you present question, this condition is met for both $f(f^{-1}(x)) \,$ and $f^{-1}(f(x))$.

So:

The domain of $f(f^{-1}(x)) = x\,$ is $\text{dom} \, f^{-1} = [0, \infty)$.

The domain of $f^{-1}(f(x)) = x\,$ is $\text{dom} \, f = [3, \infty)$.