# difficulty with one-to-ones

• Feb 3rd 2009, 04:45 PM
lsnyder
difficulty with one-to-ones
okay, my professor did teach me this but he was hard to understand (due to having a foreign accent) when he tried to explain.

so if someone could show me what to do with this problem, I would be so grateful.

Problem:

If g is one-to-one and g(12)=1, then g^(-1)=(12) and (g(12))^(-1) =___
• Feb 3rd 2009, 06:36 PM
mollymcf2009
Quote:

Originally Posted by lsnyder
okay, my professor did teach me this but he was hard to understand (due to having a foreign accent) when he tried to explain.

so if someone could show me what to do with this problem, I would be so grateful.

Problem:

If g is one-to-one and g(12)=1, then g^(-1)=(12) and (g(12))^(-1) =___

Problem:
If g is one to one,

$g(12) = 1$
$g^{-1} = 12$
$(g(12))^{-1} = ?$

Hello!
I assume that you have learned the properties of inverse functions? Each of these conditions given to you can be calculated if you know your inverse function rules. Here's a quickie review:
1) For a function to have an inverse, it must be a one-to-one function.
2) If g(x) is the inverse of f(x), then f(x) is the inverse of g(x).
3) The domain of f(x) is equal to the range of $f^{-1}(x)$; the range of f(x) is equal to the domain of $f^{-1}(x)$**Think about what this mean when you have given (x, y) values. If you are given an (x,y) coordinate for your graph of f(x), don't you also have a coordinate on your graph of $f^{-1}(x)$? Think about it!!

You can plot a few point from your given info right? See what you can come up with based on the above information! **Inverse functions will not go away, you will see them from now on in any college math class, gotta learn 'em**
Ok? Good luck!(Wink)
• Feb 3rd 2009, 06:38 PM
GaloisTheory1
Quote:

Originally Posted by lsnyder

Problem:

If g is one-to-one and g(12)=1, then g^(-1)=(12) and (g(12))^(-1) =___

$g^{-1}(12) =1$

because of this fact:

If g is one-to-one and g(12)=1

do you see why?