How Do You Calculate Inverse Functions and Their Properties?

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The discussion centers on calculating inverse functions and their properties for the function f(x) = x^2, particularly concerning the sets C = {x ∈ R : x ≥ 1} and D = R+. Participants express confusion about the definitions and calculations of f(C), f^{-1}(C), f^{-1}(D), and f^{-1}({1}). It is clarified that f(C) includes all real numbers greater than or equal to 1, and f^{-1}(C) includes values that produce outputs in C, which leads to discussions about the inclusion of negative values. The conversation emphasizes understanding the pre-image concept and correctly applying the definitions of inverse functions.
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Homework Statement


Let ## C= \{ x \in R : x \geq 1 \} ## and ## D = R^+ ##
For each f defined below, determine ## f(C), f^{-1}(C), f^{-1}(D), f^{-1} (\{1\}) ##

a.) ## f: R -> R ## is defined by ## f(x) =x^2##
I have problems with the definitions

The Attempt at a Solution



a.)
## f(C)= { 1 , 4, 9, 16, ...} ## according to the definitions of x in C, x belongs to reals and it's greater or equal to 1. Then, 2 should be in f(C), but it's not. However, in the solutions I see this.

##f(C)= C, f^{-1}(C)=C \cup \{x \in R: -x \in C\} , f^{-1}(D) = R - \{0\} , f^{-1}(\{1\})= \{1, -1\}##

-----
About , ## f^{-1} (C) ## should we exclude C, since it is the inverse of this? How do they find this?
About the inverse of f(D), why do they exclude only 0, shouldn't they exclude all positive Reals?
The only one I understand is ## f^{-1} (\{1\}) = \{1, -1\} ## :/

The last one is true since we could do this:

f(x)= y =x^2
change vars.

## f^{-1} (x) =x = f(x)^2 ##
Then, x can be -1 or 1
 
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knowLittle said:

Homework Statement


Let ## C= \{ x \in R : x \geq 1 \} ## and ## D = R^+ ##
For each f defined below, determine ## f(C), f^{-1}(C), f^{-1}(D), f^{-1} (\{1\}) ##

I have problems with the definitions

The Attempt at a Solution



a.)
## f(C)= { 1 , 4, 9, 16, ...} ## according to the definitions of x in C, x belongs to reals and it's greater or equal to 1. Then, 2 should be in f(C), but it's not. However, in the solutions I see this.

##f(C)= C, f^{-1}(C)=C \cup \{x \in R: -x \in C\} , f^{-1}(D) = R - \{0\} , f^{-1}(\{1\})= \{1, -1\}##

-----
About , ## f^{-1} (C) ## should we exclude C, since it is the inverse of this? How do they find this?
About the inverse of f(D), why do they exclude only 0, shouldn't they exclude all positive Reals?
The only one I understand is ## f^{-1} (\{1\}) = \{1, -1\} ##

The last one is true since we could do this:

f(x)= y =x^2
change vars.

## f^{-1} (x) =x = f(x)^2 ##
Then, x can be -1 or 1

C isn't only integers {1,2,3,...}, C is all real numbers greater than or equal to 1. So that includes sqrt(2). f(sqrt(2))=2. So 2 is in f(C).
 
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Thank you, you are right.
How do I explain ## f^{-1} (C)## ? Why does it include all R+ too?
 
knowLittle said:
Thank you, you are right.
How do I explain ## f^{-1} (C)## ? Why does it include all R+ too?
What is the definition of ## f^{-1} (C)\ ?##


In more descriptive terms, it's the pre-image of set C .


So, what values of ##\ x\ ## will produce values of ##\ f(x)\ ## which are in C ?
 
knowLittle said:
Thank you, you are right.
How do I explain ## f^{-1} (C)## ? Why does it include all R+ too?
It doesn't. What is f(1/2)? Is that in C?
 
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