- #1
knowLittle
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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\}##
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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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