Definition of Funtion

1. May 7, 2014

knowLittle

1. The problem statement, all variables and given/known data
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

3. 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

Last edited: May 7, 2014
2. May 7, 2014

Dick

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).

3. May 7, 2014

knowLittle

Thank you, you are right.
How do I explain $f^{-1} (C)$ ? Why does it include all R+ too?

4. May 7, 2014

SammyS

Staff Emeritus
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 ?

5. May 8, 2014

HallsofIvy

Staff Emeritus
It doesn't. What is f(1/2)? Is that in C?