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

In summary, the conversation discusses finding the inverse of a function and the confusion that can arise from switching variables. The conversation also addresses solving a quadratic equation without using the quadratic formula and the importance of having a single-valued inverse function. It concludes with a discussion on choosing the correct inverse function for a given restricted domain.
  • #1
Nyasha
127
0

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")
 
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  • #2
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.
 
  • #3
Nyasha said:

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:
[itex]y = f(x) \Leftrightarrow x = f^{-1}(y)[/itex]

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:
[itex]y = e^x \Leftrightarrow x = ln(y)[/itex]

I'm probably fighting a losing battle on this...
Nyasha said:
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")
 
  • #4
Dick said:
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.
 
  • #5
Then please tell us how you solved that quadratic equation without using the quadratic formula!
 
  • #6
HallsofIvy said:
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)
 
  • #7
2x*(3x-1/(2x))=6x^2-1, not 3x-1. Or was your problem really y=(3x-1)/(2x)?
 
  • #8
Dick said:
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
 
  • #9
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" ?
 
  • #10
Sure. Or you could have interchanged them first and solved for y. Either way.
 
  • #11
Nyasha said:

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!).
 
  • #12
Unco said:
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.
 
  • #13
Dick said:
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 ?
 

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

What is the inverse of y=3x-(1/2x)?

The inverse of a function is a function that "undoes" the original function. In other words, if you input the output of the original function into the inverse function, you will get the original input.

How do you solve for y in y=3x-(1/2x)?

To solve for y, we need to isolate it on one side of the equation. First, we can combine like terms on the right side to get y=2.5x. Then, we can divide both sides by 2.5 to get y=0.4x.

What is the domain of the inverse function of y=3x-(1/2x)?

The domain of the inverse function will be the range of the original function. In this case, the range of y=3x-(1/2x) is all real numbers, so the domain of its inverse will also be all real numbers.

Is y=3x-(1/2x) a one-to-one function?

Yes, this function is one-to-one because for every input (x value), there is only one output (y value). This means there are no repeated outputs for different inputs, which is a requirement for a function to be one-to-one.

How can we graph the inverse of y=3x-(1/2x)?

To graph the inverse, we can switch the x and y values of the original function and plot the points. This will result in a reflection of the original graph over the line y=x. Alternatively, we can use the inverse function to find the points of the inverse and plot them on the same graph as the original function.

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