Wednesday

July 30, 2014

July 30, 2014

Posted by **kristie** on Thursday, July 20, 2006 at 1:31pm.

should you multiply (x+3) on both sides and get y(x+3)=2x+1?

then what would be your next step?

This one is a little tricky because we we have a rational function.

What we want to end up with is x(y)

To do this you can interchange x and y and try to solve for x. That function is the inverse. Thus:

y=(2x+1)/(x+3)

x=(2y+1)/(y+3), then solve for y

(y+3)x= (2y+1)

yx+3x = 2y+1

3x-1=2y-yx =

3x-1= y(2-x) =

y= (3x-1)/(2-x)

Now interchange the x and y again. (after isolating y)

y = (3x-1)/(2-x)

And if we did things correctly, we should have the inverse function. Test it by composing functions.

(3((2x+1)/(x+3) -1))/(2-(2x+1)/(x+3))=

The numerator is:((6x+3)-(x+3))/(x+3))=

5x/(x+3)

The denominator is:(2(x+3)-(2x+1))/(x+3)=

5/(x+3)

Putting them together we have 5x/(x+3) times (x+3)/5 = x.

Thus they are inverses. Test it by graphing too.

That makes sense! Thank you so much!!!

comment:I don't think the second interchange is necessary if we want y(x). It's been awhile since I worked this, so check my work and ask questions if anything isn't clear.

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