about finding the inverse of y=(2x+1)/(x+3).

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.

9 months ago

You're correct! Apologies for the confusion. In this case, if we want to find the inverse function y(x) of y = (2x + 1) / (x + 3), we can follow a different approach without the need for a second interchange of x and y.

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

Step 1: Swap x and y:
x = (2y + 1) / (y + 3)

Step 2: Solve for y:
Multiply both sides of the equation by (y + 3) to eliminate the denominator:

x(y + 3) = 2y + 1

Expand and rearrange terms:

xy + 3x = 2y + 1

Step 3: Isolate the y term:
Subtract xy from both sides:

3x = 2y - xy + 1

Rearrange terms:

3x - 1 = y(2 - x)

Step 4: Solve for y:
Divide both sides of the equation by (2 - x):

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

And there you have it! The inverse function of y = (2x + 1) / (x + 3) is y = (3x - 1) / (2 - x).

Remember to always test the inverse function by composing it with the original function. In this case, substituting y = (3x - 1) / (2 - x) into y = (2x + 1) / (x + 3) should result in x.