Inverse of absolute value function

[Coco12]

Homework Statement

Y=abs -4(x+3) +1
Note: the 1 is outside the absolute value

Homework Equations

Switch y and x

The Attempt at a Solution

So you subtract the 1 to bring it to the other side
After that do I put: x-1= -4(y+3)
And x-1= 4(y+3)

And solve for y?

Am I doing this right?? What should be the two equations that you get?

 

[Mark44]

Homework Statement

Y=abs -4(x+3) +1
Note: the 1 is outside the absolute value

Why not just use the | character? It should be on your keyboard. Also, note that |-4(x + 3)| + 1 = |4(x + 3)| + 1.

Homework Equations

Switch y and x

The Attempt at a Solution

So you subtract the 1 to bring it to the other side
After that do I put: x+1= -4(x+3)
And x+1= 4(x+3)

Am I doing this right??

No, you're way off. What happened to the absolute values? When you switch x and y, you should still have both x and y, but in different places. What happened to y here?

What should be the two equations that u get?

The only functions that have inverses are those that are one-to-one. Is your function one-to-one?

Textspeak - forum rules do not allow "textspeak," such as "u" for you and so on. Fair warning...

 

[Coco12]

Opps.. I meant to put x+1=-4(y+3) and so on for the other equation

 

[Coco12]

I took the positive and negative version for the absolute function.. Is that right?

 

[Mark44]

You didn't answer my question. Is your function one-to-one?

Opps.. I meant to put x+1=-4(y+3) and so on for the other equation

Why is +1 now appearing on the left side?

 

[Coco12]

The function is not one to one. I'm just trying to determine what the equations for the inverse of the absolute function is. I brought the 1 over from the other side now is just wondering what to do for the stuff inside the absolute value . Do I just take the positive version of it then solve and then the negative version and solve? Is my work above correct?

 

[Mark44]

The function is not one to one. I'm just trying to determine what the equations for the inverse of the absolute function is. I brought the 1 over from the other side now is just wondering what to do for the stuff inside the absolute value . Do I just take the positive version of it then solve and then the negative version and solve? Is my work above correct?

No it's not. "Bringing the 1 over" is not a valid operation. When you started, there was a +1 term on the right side. How do you get rid of it?

Since your function isn't one-to-one, it doesn't have an inverse. However, if you split the domain in the right way, then each part of the function becomes one-to-one, and so has an inverse.

It would be a useful exercise to sketch a graph of y = -4|x + 3| + 1. This might help you figure out how the domain (which is all real numbers) should be split up.

 

[Coco12]

I meant to say that it would be x-1=|-4(x+3)|
Where do I go from here assuming that it is one to one?

From my understanding a function does not have to be to one to be an inverse.
One to one simply means that both the function and it's inverse are functions.

I'm trying to find the equation for the inverse

 

[Mark44]

I meant to say that it would be x-1=|-4(x+3)|

If you mean y - 1 = |-4(x + 3)|, then yes.

Where do I go from here assuming that it is one to one?

Since it isn't one-to-one, you can't just assume that it is.

From my understanding a function does not have to be to one to be an inverse.

No, this is incorrect. For example, y = x² is not one-to-one, so it does not have an inverse that is a function.

One to one simply means that both the function and it's inverse are functions.

That's an outcome of a function being one-to-one, but it isn't the definition. Your book should have a definition of what it means for a function to be one-to-one, and several examples, including at least one where the function is not one-to-one, and what you need to do to find an inverse.

I'm trying to find the equation for the inverse

 

[Coco12]

Let's say we restricted the domain of the original so that we have an inverse. What would be the equations then?

 

[Mark44]

Have you sketched a graph of y = |-4(x + 3)| + 1? Note that this is exactly the same as y = |4(x + 3)| + 1 = 4|x + 3| + 1.

From the graph it should be fairly obvious how you need to divide the domain.