A question about restrictions of inverse functions

[mindauggas]

Homework Statement

Hi. I found in the answears that the inverse of function $f(x)=3-\sqrt{x-2}$ is $f^{-1}(x)=(3-x)^{2}+2$ only if we restrict it to ${x:x\leq3}$. I understand that the restriction is needed because the found inverse is a parabola (and thus not one-to-one function).

My general question (1): how to know/find out algebraically (without drawing graphs) the needed restrictions? Is there a general way, or some intuition?

My special question (for the above case) (2): can I chose the restriction $[x:x\geq3]$?

The Attempt at a Solution

No attempt since I regard this as a general mathematical knowledge question.

 

[Mark44]

Homework Statement

Hi. I found in the answears that the inverse of function $f(x)=3-\sqrt{x-2}$ is $f^{-1}(x)=(3-x)^{2}+2$ only if we restrict it to ${x:x\leq3}$. I understand that the restriction is needed because the found inverse is a parabola (and thus not one-to-one function).

My general question (1): how to know/find out algebraically (without drawing graphs) the needed restrictions? Is there a general way, or some intuition?

To find the domain and range of the inverse, look at the range and domain of the original function. Notice that I reversed the order.

For your problem, the domain of f is x >= 2, and the range of f is y <= 3. The reason that y has to be <= 3 can be seen from the formula, f(x) = 3 - √(x - 2). Here, we are subtracting a positive number from 3, so the function value (y) can be no larger than 3.

Since the domain and range of f are, respectively, x >= 2 and y <= 3, the domain and range of f^-1 are, respectively, x <= 3 and y >= 2.

My special question (for the above case) (2): can I chose the restriction $[x:x\geq3]$?

The Attempt at a Solution

No attempt since I regard this as a general mathematical knowledge question.

 

[mindauggas]

Thanks