Inverse of y=3x-(1/2x): Solve for y

[Nyasha]

Homework Statement

Find the inverse of y=3x-(1/2x)

The Attempt at a Solution

y=(6x^2-1)/(2x)


x=(6y^2-1)/(2y)

2yx=(6y^2-1)

2yx-6y^2=-1

2y(x-3y)=-1

2y(x-3y)=-1

( I am stuck here how do l solve for "y")

 

[Dick]

You've got 6y^2-2xy-1=0. That's a quadratic equation in y. Solve it the way you would any other quadratic equation ay^2+by+c=0. a=6, b=-2x c=-1.

 

[Mark44]

Homework Statement

Find the inverse of y=3x-(1/2x)

The Attempt at a Solution

y=(6x^2-1)/(2x)


x=(6y^2-1)/(2y)

I realize that this switching of x and y is how many (most?) books present this process,
but it's really not necessary and can lead to some confusion on the part of students. The only reason for doing this switching of variables is that both the original function and its inverse can be written as y = f(x) and y = f^-1(x). In other words, with y as the dependent variable and x as the independent variable.

That convenience gets in the way, IMO, in much of what you do with inverses later on. For a function that has an inverse, the most important relationship between the two functions is this:
y = f(x) \Leftrightarrow x = f^{-1}(y)

This says that if one function gives you a y value associated with an x input value, the other, the inverse, gives you the x value if you know the y value. Because there is no switching of variables, the graph of both functions is exactly the same -- there is no reflection across the line y = x.

A very commonly used example is:
y = e^x \Leftrightarrow x = ln(y)

I'm probably fighting a losing battle on this...

2yx=(6y^2-1)

2yx-6y^2=-1

2y(x-3y)=-1

2y(x-3y)=-1

( I am stuck here how do l solve for "y")

 

[Nyasha]

You've got 6y^2-2xy-1=0. That's a quadratic equation in y. Solve it the way you would any other quadratic equation ay^2+by+c=0. a=6, b=-2x c=-1.

Thanks for the help l figured out a way of solving the inverse of the function without using quadratic formula.

 

[HallsofIvy]

Then please tell us how you solved that quadratic equation without using the quadratic formula!

 

[Nyasha]

Then please tell us how you solved that quadratic equation without using the quadratic formula!

y=3x-(1/2x)

Multiply entire equation by 2x to get 2xy = 3x – 1.

Add 1 both sides 2xy + 1 = 3x.

Subtract 2xy from both sides, giving 1 = 3x - 2xy

Factor out x giving 1=x(3-2y)


Divided both sides by 3-2y giving x=1/(3-2y)

Interchange x and y giving y= 1/(3-2x)

 

[Dick]

2x*(3x-1/(2x))=6x^2-1, not 3x-1. Or was your problem really y=(3x-1)/(2x)?

 

[Nyasha]

2x*(3x-1/(2x))=6x^2-1, not 3x-1. Or was your problem really y=(3x-1)/(2x)?

Ummmmm, thanks for point out that error. This means one way or the other l will have to use quadratic formula

 

[Nyasha]

After l solve the quadratic equation l get: x= (y±sqrt(y²+6))/6To get the inverse do l now need to interchange "x" and " y" ?

 

[Dick]

Sure. Or you could have interchanged them first and solved for y. Either way.

 

[Unco]

Homework Statement

Find the inverse of y=3x-(1/2x)

Are you given some restriction on x (i.e., the domain of f(x) = 3x - 1/(2x)) ? Otherwise, as you have discovered, f(x) does not have an inverse (an inverse function must be single-valued!).

 

[Nyasha]

Are you given some restriction on x (i.e., the domain of f(x) = 3x - 1/(2x)) ? Otherwise, as you have discovered, f(x) does not have an inverse (an inverse function must be single-valued!).

The domain of the original function is : x is an element of real # except zero and y is an element of real #. To get the inverse l am supposed to change the domain and the range, which l have done.

 

[Nyasha]

Sure. Or you could have interchanged them first and solved for y. Either way.

x= (y±sqrt(y²+6))/6


So if my restricted domain is D={x|x<0} do l choose the negative function ?