how to find inverse of this

[allegansveritatem]

I need to find inverse function of:

Here is as near as I could get to isolating x:

I used the solve function on my calculator and the inverse I got I entered into a graphing window:

Now what I want to know is how do you get the red function out of the blue one? I mean, how algebraically?

 

[blamocur]

Re-write as [imath]y=\sqrt{x^2+2x}[/imath] and express x as a function of y. In fact, it looks like you started in the right direction in your first image, and f2(x) in your second image looks like the result of such transformation, especially if you replace x with y there.

 

[Dr.Peterson]

I need to find inverse function of:
View attachment 29621
Here is as near as I could get to isolating x:
View attachment 29622
I used the solve function on my calculator and the inverse I got I entered into a graphing window:
View attachment 29623
Now what I want to know is how do you get the red function out of the blue one? I mean, how algebraically?

You're trying to solve for x. The equation [imath]x^2 + 2x - y^2 = 0[/imath] is quadratic, so use the quadratic formula.

This will give [imath]x=f^{-1}(y)[/imath]. You can either replace your y with x now, or at the start of your work as many people teach, in order to get the calculator's version, [imath]y=f^{-1}(x)[/imath].

 

[Cubist]

I need to find inverse function of:
View attachment 29621
Here is as near as I could get to isolating x:
View attachment 29622
I used the solve function on my calculator and the inverse I got I entered into a graphing window:
View attachment 29623
Now what I want to know is how do you get the red function out of the blue one? I mean, how algebraically?

Like @Dr.Peterson suggests, you should spot the quadratic and then use an appropriate technique (for a quadratic). In this case I think completing the square is a good option. The RHS of your second line of work is [imath]x^2+2x[/imath]. Just complete the square on this expression which will leave a single "x" in the equation. 3 steps later you should have x=<something>

 

[allegansveritatem]

You're trying to solve for x. The equation [imath]x^2 + 2x - y^2 = 0[/imath] is quadratic, so use the quadratic formula.

This will give [imath]x=f^{-1}(y)[/imath]. You can either replace your y with x now, or at the start of your work as many people teach, in order to get the calculator's version, [imath]y=f^{-1}(x)[/imath].

I see it. I didn't think of using that method, I suppose, because I had just gone through 10 exercises of like nature--except that there was only one x--and so...the quadratic option did not figure into my noting of what to do to get a function's inverse. I will work this out tomorrow and post my results.Thanks

 

[allegansveritatem]

Like @Dr.Peterson suggests, you should spot the quadratic and then use an appropriate technique (for a quadratic). In this case I think completing the square is a good option. The RHS of your second line of work is [imath]x^2+2x[/imath]. Just complete the square on this expression which will leave a single "x" in the equation. 3 steps later you should have x=<something>

a good suggestion. That is how I will proceed I think. Thanks.

 

[allegansveritatem]

Re-write as [imath]y=\sqrt{x^2+2x}[/imath] and express x as a function of y. In fact, it looks like you started in the right direction in your first image, and f2(x) in your second image looks like the result of such transformation, especially if you replace x with y there.

I am not sure what you are saying. I will look at this again when I work on this tomorrow. Thanks

 

[blamocur]

Do you find it easier to solve if it is written as [imath]\sqrt{x^2+2x}=a[/imath] ? I.e., can you express [imath]x[/imath] as a function of [imath]a[/imath] ?

 

[allegansveritatem]

I went at it today and spent what must have been 45 mins trying to get [math]sqrt (x^2+1) -1[/math] but instead I kept getting this:

So I am still in the throes of my dilemma it seems, although maybe not quite as lost as before. At least now I see the outlines of what I am looking for. What is the problem here?

 

[allegansveritatem]

Do you find it easier to solve if it is written as [imath]\sqrt{x^2+2x}=a[/imath] ? I.e., can you express [imath]x[/imath] as a function of [imath]a[/imath] ?

I just posted the results of my work using the quadratic equation. I think somewhere in that method is the answer I want. I don't see that what you are proposing is going to go where I want to go, which is sqrt(x2+1)−1

 

[blamocur]

I just posted the results of my work using the quadratic equation. I think somewhere in that method is the answer I want. I don't see that what you are proposing is going to go where I want to go, which is sqrt(x2+1)−1

Lets try a different approach: swap [imath]x[/imath] and [imath]y[/imath] to rewrite the original function as [imath]x=\sqrt{y^2+2y}[/imath], then solve this equation for [imath]y[/imath], i.e. express [imath]y[/imath] in terms of [imath]x[/imath].

 

[JeffM]

Well, first you seem to have made an error.

[math]y^2 = x^2 + 2x + 1 - 1 = (x + 1)^2 - 1 \implies[/math]
[math]y = \sqrt{(x + 1)^2 - 1} \ne \sqrt{(x + 1)^2} - 1.[/math]
But that math, even done correctly, will not lead you where you want to go,

Basic idea

[math]f^{-1}(f(x)) = x.[/math]
[math]\text {But } y = f(x).[/math]
[math]\text {If } \exists \ g(y) \text { such that } g(y)= x, \text { then } g(f(x)) = g(y) = x.[/math]
Make sense? In this specific case, [imath]y = \sqrt{x^2 + 2x} \text { and } x > 0.[/imath]

Can we solve for x in terms of y?

[math]y = \sqrt{x^2 + 2x} \implies y^2 = x^2 + 2x \implies[/math]
[math]y^2 + 1= x^2 + 2x + 1 = (x + 1)^2 \implies [/math]
[math]x + 1 = \pm \sqrt{y^2 + 1} \implies x = \pm \sqrt{y^2 + 1} - 1.[/math]
[math]\text {But } x > 0 \implies x = \sqrt{y^2 + 1} - 1.[/math]
[math]\therefore g(y) = \sqrt{y^2 + 1} - 1 \implies g(x) = \sqrt{x^2 + 1} - 1.[/math]
Let's check.

[math]g(x) = f^{-1}(x) \implies f(g(x)) = x.[/math]
[math]f(f^{-1}(x)) = f(\sqrt{x^2 + 1} - 1) = \sqrt{(\sqrt{x^2 + 1} - 1)^2 + 2(\sqrt{x^2 + 1} - 1)} =[/math]
[math]\sqrt{(\sqrt{x^2 + 1})^2 - 2\sqrt{x^2 + 1} + 1 + 2\sqrt{x^2 + 1} - 2} =[/math]
[math]\sqrt{x^2 + 1 + 1 - 2} = \sqrt{x^2} = x.[/math]

 

[allegansveritatem]

Lets try a different approach: swap [imath]x[/imath] and [imath]y[/imath] to rewrite the original function as [imath]x=\sqrt{y^2+2y}[/imath], then solve this equation for [imath]y[/imath], i.e. express [imath]y[/imath] in terms of [imath]x[/imath].

yes, I have done this also and did get an inverse out of it--once the constraint on x was applied, but even with that constraint this expression was supplying one too many lines....but my quest is not really to find an expression that will produce an inverse. What I want to know is how, when I asked it for the inverse of y=sqrt(x^2+2x) my calculator came up with y=sqrt(x2+1)-1 I have been trying to get the one from the other and can't do it.

 

[allegansveritatem]

Well, first you seem to have made an error.

[math]y^2 = x^2 + 2x + 1 - 1 = (x + 1)^2 - 1 \implies[/math]
[math]y = \sqrt{(x + 1)^2 - 1} \ne \sqrt{(x + 1)^2} - 1.[/math]
But that math, even done correctly, will not lead you where you want to go,

Basic idea

[math]f^{-1}(f(x)) = x.[/math]
[math]\text {But } y = f(x).[/math]
[math]\text {If } \exists \ g(y) \text { such that } g(y)= x, \text { then } g(f(x)) = g(y) = x.[/math]
Make sense? In this specific case, [imath]y = \sqrt{x^2 + 2x} \text { and } x > 0.[/imath]

Can we solve for x in terms of y?

[math]y = \sqrt{x^2 + 2x} \implies y^2 = x^2 + 2x \implies[/math]
[math]y^2 + 1= x^2 + 2x + 1 = (x + 1)^2 \implies [/math]
[math]x + 1 = \pm \sqrt{y^2 + 1} \implies x = \pm \sqrt{y^2 + 1} - 1.[/math]
[math]\text {But } x > 0 \implies x = \sqrt{y^2 + 1} - 1.[/math]
[math]\therefore g(y) = \sqrt{y^2 + 1} - 1 \implies g(x) = \sqrt{x^2 + 1} - 1.[/math]
Let's check.

[math]g(x) = f^{-1}(x) \implies f(g(x)) = x.[/math]
[math]f(f^{-1}(x)) = f(\sqrt{x^2 + 1} - 1) = \sqrt{(\sqrt{x^2 + 1} - 1)^2 + 2(\sqrt{x^2 + 1} - 1)} =[/math]
[math]\sqrt{(\sqrt{x^2 + 1})^2 - 2\sqrt{x^2 + 1} + 1 + 2\sqrt{x^2 + 1} - 2} =[/math]
[math]\sqrt{x^2 + 1 + 1 - 2} = \sqrt{x^2} = x.[/math]

well, I know that those two expressions don't = one another but I recall I elided certain steps in producing that result, but I won't go into that here. It was a stunt that misfired. I have copied your post and I will study it over today. It looks promising. Thanks.

 

[blamocur]

yes, I have done this also and did get an inverse out of it--once the constraint on x was applied, but even with that constraint this expression was supplying one too many lines....but my quest is not really to find an expression that will produce an inverse. What I want to know is how, when I asked it for the inverse of y=sqrt(x^2+2x) my calculator came up with y=sqrt(x2+1)-1 I have been trying to get the one from the other and can't do it.

"...calculator came up with y=sqrt(x2+1)-1" -- because this is what you get if you solve [imath]x=\sqrt{y^2+2y}[/imath] for [imath]y[/imath].

 

[allegansveritatem]

"...calculator came up with y=sqrt(x2+1)-1" -- because this is what you get if you solve [imath]x=\sqrt{y^2+2y}[/imath] for [imath]y[/imath].

Exactly. I was unable to see how they got that because I couldn't come up with it. Now however I see, thanks to the contributors to this thread, how easy it should have been.

 

[allegansveritatem]

I went at it again today, armed with a copy of relevant portions of JeffM's post and saw the light!

I should have seen this long before now but I was pig-headedly following a different tack. Thanks to Jeff and to all the others who took the time to enlighten me.

 

[Deleted member 4993]

I went at it again today, armed with a copy of relevant portions of JeffM's post and saw the light!
View attachment 29656
I should have seen this long before now but I was pig-headedly following a different tack. Thanks to Jeff and to all the others who took the time to enlighten me.

Pig-headed !! Insulting pigs in open forum? Next you will say "frog-headed" !! Miss Piggy and Kermit will sue you for slander ....

 

[allegansveritatem]

Pig-headed !! Insulting pigs in open forum? Next you will say "frog-headed" !! Miss Piggy and Kermit will sue you for slander ....

True...it's unfair to the animals...but I didn't want to say anything that might sound political....and yet, for all that, that's what I seem to have done anyway!

 

[Steven G]

I learned a long time ago that you can't win on this forum.