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!

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.