Asked by kristie
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.
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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