A question about restrictions of inverse functions

In summary, to find the needed restrictions for the inverse of a function, you can look at the range and domain of the original function and reverse the order. In this specific case, the restriction is x <= 3. As for your special question, you cannot choose the restriction [x:x\geq3] as it would not satisfy the requirements for a one-to-one inverse function.
  • #1
mindauggas
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Homework Statement



Hi. I found in the answears that the inverse of function [itex]f(x)=3-\sqrt{x-2}[/itex] is [itex]f^{-1}(x)=(3-x)^{2}+2[/itex] only if we restrict it to [itex]{x:x\leq3}[/itex]. 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 [itex][x:x\geq3][/itex]?

The Attempt at a Solution



No attempt since I regard this as a general mathematical knowledge question.
 
Last edited:
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  • #2
mindauggas said:

Homework Statement



Hi. I found in the answears that the inverse of function [itex]f(x)=3-\sqrt{x-2}[/itex] is [itex]f^{-1}(x)=(3-x)^{2}+2[/itex] only if we restrict it to [itex]{x:x\leq3}[/itex]. 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.
mindauggas said:
My special question (for the above case) (2): can I chose the restriction [itex][x:x\geq3][/itex]?

The Attempt at a Solution



No attempt since I regard this as a general mathematical knowledge question.
 
  • #3
Thanks
 

FAQ: A question about restrictions of inverse functions

1. What are restrictions on inverse functions?

The restrictions on inverse functions depend on the original function. Generally, the inverse function must have a one-to-one mapping, meaning that each input has only one corresponding output. Additionally, the domain and range of the inverse function must be switched from the original function.

2. How do I find the restrictions on an inverse function?

To find the restrictions on an inverse function, you must first find the domain and range of the original function. Then, switch the domain and range to determine the domain and range of the inverse function. The restrictions will be any values that are not included in the new domain and range.

3. Can an inverse function have restrictions?

Yes, an inverse function can have restrictions. In fact, most inverse functions will have some restrictions since they must have a one-to-one mapping and the domain and range must be switched from the original function.

4. What happens if an inverse function has restrictions?

If an inverse function has restrictions, it means that certain inputs will not have a corresponding output. This can limit the domain and range of the inverse function and may affect its behavior and properties.

5. How do restrictions on inverse functions affect their graphs?

Restrictions on inverse functions can affect their graphs in a few ways. If the restrictions limit the domain and range, the graph may have gaps or breaks. Additionally, the restrictions may also affect the shape and symmetry of the graph.

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