Need help with finding an inverse function

[i077]

Hi, I'm having trouble finding the inverse for this function:

f(x) = sqrt(x)/(2x-3)

I know I need to swap f(x) for y and then solve for x, but I'm stuck on how to do that. Can anyone help me out?

 

[Deleted member 4993]

Hi, I'm having trouble finding the inverse for this function:

f(x) = sqrt(x)/(2x-3)

I know I need to swap f(x) for y and then solve for x, but I'm stuck on how to do that. Can anyone help me out?

What did you get after swapping?

 

[i077]

What did you get after swapping?

I got
x = (2x-3)²y²

Then, expanding that polynomial I got
x = y²(2x²-12x+9)

But now I'm not sure where to go from here.

 

[Deleted member 4993]

I got
x = (2x-3)²y²

Then, expanding that polynomial I got
x = y²(2x²-12x+9)

But now I'm not sure where to go from here.

f(x) = sqrt(x)/(2x-3)

y = √x/(2x-3)

swapping

x = √y/(2y-3)

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

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

4x^2y^2 - y(12x^2+1) + 9x^2 = 0

Now solve for 'y' (quadratic equation)

 

[Steven G]

I got
x = (2x-3)²y²

Then, expanding that polynomial I got
x = y²(2x²-12x+9)

But now I'm not sure where to go from here.

You have a quadratic equation in x.
Rewrite what you have as 2(y^2)x^2 -[12(y^2)+1]x + 9y^2 =0.
Now use the quadratic formula with a=2y^2, b =-[12(y^2)+1] and c= 9y^2 to solve for x which is really y inverse. Show us your work up to here.

 

[i077]

You have a quadratic equation in x.
Rewrite what you have as 2(y^2)x^2 -[12(y^2)+1]x + 9y^2 =0.
Now use the quadratic formula with a=2y^2, b =-[12(y^2)+1] and c= 9y^2 to solve for x which is really y inverse. Show us your work up to here.

Thanks. So now I have (after simplifying):
x=(12y²+1 ± sqrt(24y²(3y²+1)+1))/4y²

I think that's simplified enough, so now I just have to switch x and y then I'll have the inverse, right?

 

[Steven G]

Thanks. So now I have (after simplifying):
x=(12y²+1 ± sqrt(24y²(3y²+1)+1)/4y²

I think that's simplified enough, so now I just have to switch x and y then I'll have the inverse, right?

You opened 3 brackets but only closed 2! Just fix that.
I am not sure why you factored out the 24y^2.
Did you try to factor 72y^4 + 24y^2 + 1??
Now we know that x>0 (why?), so you did to check to make sure that your solutions for x is never 0 or negative. Let us know your results to these questions.

 

[i077]

You opened 3 brackets but only closed 2! Just fix that.
I am not sure why you factored out the 24y^2.
Did you try to factor 72y^4 + 24y^2 + 1??
Now we know that x>0 (why?), so you did to check to make sure that your solutions for x is never 0 or negative. Let us know your results to these questions.

I'm pretty sure it's not factorable. x>0 because if x were negative, it would generate an imaginary number not a real number.

Thanks for the help!

 

[Ishuda]

Hi, I'm having trouble finding the inverse for this function:

f(x) = sqrt(x)/(2x-3)

I know I need to swap f(x) for y and then solve for x, but I'm stuck on how to do that. Can anyone help me out?

Same sort of idea as already presented but another way
y = x^1/2/(2x-3)
Inverse is
x = y^1/2/(2y-3)
Let u = y^1/2
x = u/(2u²-3)
or
2x u² - u - 3 x = 0

EDIT: Fix dumb mistake [hope I haven't made more]