• Mar 22nd 2008, 04:08 PM
calcprincess88
Here's the question:

Find the composites for the relations defined.

R1={(x,y) member of real numbers x real numbers: y=x}
R2={(x,y) member of real numbers x real numbers: y=-5x+2}

a) R1compositeR1
b)R2compositeR2

a) I was thinking that R1compositeR1={(x,y) member of real numbers x real numbers: x=y}=R1
b) I'm not to sure how to do and thats the one I need help on.

Thank you!!
• Mar 22nd 2008, 04:48 PM
TheEmptySet
let
$f(x)=x$ and $g(x)=-5x+2$

Then

$(f \circ g)(x)=g(x)=-5x+2$ and

$(g \circ f)(x)=-5f(x)+2=-5x+2$

Since both f and g are defined on all of R and both map onto R

The domain and Range are all real numbers.

I hope this helps
good luck.
• Mar 22nd 2008, 05:08 PM
calcprincess88
Quote:

Originally Posted by TheEmptySet
let
$f(x)=x$ and $g(x)=-5x+2$

Then

$(f \circ g)(x)=g(x)=-5x+2$ and

$(g \circ f)(x)=-5f(x)+2=-5x+2$

Since both f and g are defined on all of R and both map onto R

The domain and Range are all real numbers.

I hope this helps
good luck.

I still don't understand. Does this mean that R2compositeR2={(x,y) member of real numbers x real numbers: y=-5x+2}=R2?
• Mar 22nd 2008, 05:17 PM
TheEmptySet
Quote:

Originally Posted by calcprincess88
I still don't understand. Does this mean that R2compositeR2={(x,y) member of real numbers x real numbers: y=-5x+2}=R2?

R1 and R2 are just fuctions.

R1= $f(x)=x$

R2= $g(x)=-5x+2$

I just called R1 f and R2 g

$R2 \circ R2 = g(x) \circ g(x)=g(g(x))=-5g(x)+2=-5(-5x+2)+2=25x-8$

and

$R1 \circ R1 = f(x) \circ f(x)=f(f(x))=f(x)=x$
• Mar 22nd 2008, 05:21 PM
calcprincess88
ohhhh I get it. I'm sorry. I didn't understand at first what you were saying but now I do! Thank you soooo much for your patience and help!! I really appreciate it!!!