Finding the Inverse of Composite Functions

I hope I don't become annoying :PThanks again,Peter G.In summary, the conversation was about finding the inverse of the composition of two functions, given the functions f(x) = e2x and g(x) = (2x-1). The conversation included a discussion on notation and equations, and ultimately, the final answer was y = (1/2) ln[(x + 1)/2] as the inverse. The use of LaTeX was also mentioned as a helpful tool for presenting equations in a clear and organized manner.
  • #1
Peter G.
442
0
Hello guys :smile:

Given that: f(x) = e2x and g(x) = (2x-1)

Find: (g ° f)-1(x)

So, what I did first was to put f into g:

2 x e2x - 1 = y
2 x e2x = y + 1
e2x = y + 1 / 2
(ex)2 = y + 1 / 2
ln (y+1/2) = x

Is that ok?

Thanks,
Peter G.
 
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  • #2
Peter G. said:
Hello guys :smile:

Given that: f(x) = e2x and g(x) = (2x-1)

Find: (g ° f)-1(x)

So, what I did first was to put f into g:

2 x e2x - 1 = y
It's bad practice to use x for multiplication and as a variable.
What you have above is g(f(x)), but would be better written as g(f(x)) = y = 2e2x - 1
Peter G. said:
2 x e2x = y + 1
e2x = y + 1 / 2
You need parentheses above.
It should be e2x = (y + 1)/2
Peter G. said:
(ex)2 = y + 1 / 2
ln (y+1/2) = x

Is that ok?
No. You should have gotten x = (1/2) ln( (y + 1)/2)
Peter G. said:
Thanks,
Peter G.
 
  • #3
Peter G. said:
Hello guys :smile:

Given that: f(x) = e2x and g(x) = (2x-1)

Find: (g ° f)-1(x)

So, what I did first was to put f into g:

2 x e2x - 1 = y

Can you not use "x" for multiplication? Looks like the variable "x." So you have
[tex](g \circ f)(x) = 2e^{2x} - 1[/tex]
. Looks good.

From here on, however, I would switch the x's and y's first, and then solve for y.
2 x e2x = y + 1
Change to
[tex]2e^{2y} = x + 1[/tex]

e2x = y + 1 / 2
That looks like
[tex]y + \frac{1}{2}[/tex]
which is wrong. You mean
[tex]\frac{y + 1}{2}[/tex]
of course. So the next step should be
[tex]e^{2y} = \frac{x + 1}{2}[/tex]

(ex)2 = y + 1 / 2
Don't do this. Convert to the logarithm right away:
[tex]2y = \ln \left(\frac{x + 1}{2} \right)[/tex]
Then divide by 2:
[tex]y = \frac{1}{2}\ln \left(\frac{x + 1}{2} \right)[/tex]

EDIT: Mark44 beat me to it. ;)
 
  • #4
Argh... sorry Mark44, towards the end I made a typo.

Well, first and foremost, yes, the x's are confusing, I apologize.

I understood what you guys did, but, what I did originally was:

(ex)2 = (y+1)/2
ex = √(y+1)/2
ln √(y+1)/2 = x
y = ln √(x+1)/2

Which I suppose works too?
 
  • #5
Peter G. said:
Argh... sorry Mark44, towards the end I made a typo.

Well, first and foremost, yes, the x's are confusing, I apologize.

I understood what you guys did, but, what I did originally was:

(ex)2 = (y+1)/2
ex = √(y+1)/2
You have some parentheses, but you need more to indicate that (y + 1)/2 is inside the radical.
ex = √((y+1)/2)

Peter G. said:
ln √(y+1)/2 = x
y = ln √(x+1)/2
ln √[(y+1)/2] = x
y = ln √[(x+1)/2]

or
y = ln [(x+1)/2]1/2 = (1/2) ln[(x + 1)/2]


Peter G. said:
Which I suppose works too?
 
  • #6
Cool guys, thanks.

I guess I need to work on my notation though! :-p

And by the way, in order to make things simpler next time for me and the ones helping in order to use that kind of format eumyang used I have to use the latex reference function?

Thanks once again
Peter G.
 
  • #7
No, you don't have to use LaTeX as long as what you write is clear. I.e., writing (x + 1)/2 instead of x + 1/2, if the first is what you mean.

If you plan to post here often, though, it's a good idea to learn some LaTeX. Here's a link to get you started: https://www.physicsforums.com/showthread.php?t=386951
 
Last edited by a moderator:
  • #8
Thanks for the Link Mark44.

I understand I don't need the latex, but, since I really like the concept of a forum, even though I'd love to help some people but I find few doubts I am comfortable explaining due to my level of expertise...

So yeah, I plan on using this website quite often so the latex function would be a nice addition, especially when I am reviewing my own work.
 

FAQ: Finding the Inverse of Composite Functions

What are composite functions?

Composite functions are functions that are created by combining two or more functions together. The output of one function is used as the input for the other function.

What is an inverse function?

An inverse function is a function that "undoes" the actions of another function. It switches the input and output values of the original function.

How do you find the composite function of two functions?

To find the composite function of two functions, you need to substitute the output of the first function into the input of the second function. The resulting function will be the composite function.

How do you find the inverse of a function?

To find the inverse of a function, you need to switch the input and output values of the original function. This means that the output of the original function becomes the input for the inverse function, and vice versa.

What is the relationship between composite and inverse functions?

The relationship between composite and inverse functions is that they "cancel" each other out. When a composite function is applied to the output of another function, and then the inverse function is applied to the resulting output, the original input value is obtained.

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