Finding the Inverse of Composite Functions

[Peter G.]

Hello guys

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

Find: (g ° f)^-1(x)

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

2 x e^2x - 1 = y
2 x e^2x = y + 1
e_2x = y + 1 / 2
(e^x)² = y + 1 / 2
ln (y+1/2) = x

Is that ok?

Thanks,
Peter G.

 

[Mark44]

Hello guys

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

Find: (g ° f)^-1(x)

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

2 x e^2x - 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 = 2e^2x - 1

2 x e^2x = y + 1
e_2x = y + 1 / 2

You need parentheses above.
It should be e^2x = (y + 1)/2

(e^x)² = y + 1 / 2
ln (y+1/2) = x

Is that ok?

No. You should have gotten x = (1/2) ln( (y + 1)/2)

Thanks,
Peter G.

 

[eumyang]

Hello guys

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

Find: (g ° f)^-1(x)

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

2 x e^2x - 1 = y

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

From here on, however, I would switch the x's and y's first, and then solve for y.

2 x e^2x = y + 1

Change to
$$2e^{2y} = x + 1$$

e^2x = y + 1 / 2

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

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

EDIT: Mark44 beat me to it. ;)

 

[Peter G.]

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:

(e^x)² = (y+1)/2
e^x = √(y+1)/2
ln √(y+1)/2 = x
y = ln √(x+1)/2

Which I suppose works too?

 

[Mark44]

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:

(e^x)² = (y+1)/2
e^x = √(y+1)/2

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

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]

Which I suppose works too?

 

[Peter G.]

Cool guys, thanks.

I guess I need to work on my notation though! :tongue2:

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.

 

[Mark44]

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

 

[Peter G.]

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.