component of composite function: given fcns (f o g)(x) and g(x), how to find f(x) ?

hellawowser

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so lets suppose i know the composite function : (f o g)(x) and one function g(x).
How do i calculate the other one? i mean i know that a composite function have several possible pairs of functions,but when i already have one of them stickied it is still many possibilities?

(f o g)(x) = f(g(x)) I've tried to fill in the functions on the equation but i just get stuck in the f cause this is not a variable like x or y i cant operate with it . I just wanna know if there is an algebric method to find the function or i just need to keep testing differents function that works?
 
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so lets suppose i know the composite function : (f o g)(x) and one function g(x).
How do i calculate the other one? i mean i know that a composite function have several possible pairs of functions,but when i already have one of them stickied it is still many possibilities?
I'm sorry, but I don't know what "having (something or other) stickied" means, nor what "it" is that has (or "is"?) "many possibilities"...?

(f o g)(x) = f(g(x)) I've tried to fill in the functions on the equation but i just get stuck in the f cause this is not a variable like x or y i cant operate with it . I just wanna know if there is an algebric method to find the function or i just need to keep testing differents function that works?
It might help if you provided an example of what you mean, showing what you've tried so far. Thank you! ;)
 
(f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3. That means f(2x - 3) = 2x^2 -4x + 1 right? How do i get to know the f equation?
 
I'm sorry, but I don't know what "having (something or other) stickied" means, nor what "it" is that has (or "is"?) "many possibilities"...?


It might help if you provided an example of what you mean, showing what you've tried so far. Thank you! ;)

I've meant that when you only know (f o g)(x) you have several pairs of equation for f(x) and g(x) , but when you know one of them, you still can have a lot of possibilities for the other one?

Like i said i've only tried to fill in the equation : f(2x - 3) = 2x^2 -4x + 1 but i couldnt operate with it.
Recently i've tried to analysis the domain and range for each but also couldnt get anything useful from it.
 
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(f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3.

That means f(2x - 3) = 2x^2 -4x + 1 right?
That's correct.


How do i get to know the f equation?
Ask it out for a date? ;)

One way is to experiment. Start by trying to obtain the second-degree term (2x^2).

(2x - 3) is the input to function f. What would function f need to do with the 2x part of the input, in order to output a 2x^2 term?
 
That's correct.


Ask it out for a date? ;)

One way is to experiment. Start by trying to obtain the second-degree term (2x^2).

(2x - 3) is the input to function f. What would function f need to do with the 2x part of the input, in order to output a 2x^2 term?

Multiply by x? But in this case wouldnt the x be (2x -3)?
 
Multiply by x … But in this case wouldnt the x be (2x -3)?
Correct, so let's see what happens, if function f multiplies the input by itself.

f(x) = x^2

f(g(x)) = (2x - 3)^2 = 4x^2 plus other stuff

So, multiplying by the input is not enough. The square of 2x is 4x^2, but we want 2x^2. What to do?

Answer: Take half the square.

1/2*(2x)^2 = 2x^2

So, let's start with that (i.e., make f the function that outputs half of the input's square):

f(x) = 1/2*x^2

Now we have:

f(g(x)) = f(2x - 3) = 2x^2 - 6x + 9/2

Next step: what can be done about that -6x term, knowing that function f has 2x-3, to work with? :cool:
 
Correct, so let's see what happens, if function f multiplies the input by itself.

f(x) = x^2

f(g(x)) = (2x - 3)^2 = 4x^2 plus other stuff

So, multiplying by the input is not enough. The square of 2x is 4x^2, but we want 2x^2. What to do?

Answer: Take half the square.

1/2*(2x)^2 = 2x^2

So, let's start with that (i.e., make f the function that outputs half of the input's square):

f(x) = 1/2*x^2

Now we have:

f(g(x)) = f(2x - 3) = 2x^2 - 6x + 9/2

Next step: what can be done about that -6x term, knowing that function f has 2x-3, to work with? :cool:


f(x) = 1/2x^2 + x - 1/2 ??
 
(f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3. That means f(2x - 3) = 2x^2 -4x + 1 right? How do i get to know the f equation?

Here the idea is to express the given function in terms of (2x-3). So

2x^2 -4x + 1 = 1/2 * (4x^2 -8x + 2) = 1/2 * [(4x^2 -12x + 9) + 4x - 7] = 1/2 * [(2x-3)^2 + 2(2x -3) - 1]

continue.....
 
I've meant that when you only know (f o g)(x) you have several pairs of equation for f(x) and g(x) , but when you know one of them, you still can have a lot of possibilities for the other one?

Like i said i've only tried to fill in the equation : f(2x - 3) = 2x^2 -4x + 1 but i couldnt operate with it.
Recently i've tried to analysis the domain and range for each but also couldnt get anything useful from it.

I observe that no one has suggested what to me is the best algebraic approach to this (no guessing).

You know that f(2x - 3) = 2x^2 - 4x + 1, and you want to find an expression for f(u). To do that, you can just substitute
u = 2x - 3

and solve to express x in terms of u. Put this in place of x on the right side, and simplify to obtain f(u).

Another way to describe this process, if you are comfortable with the notation of function composition, is that you want \(\displaystyle f \circ g = h\), where h(x) = 2x^2 - 4x + 1 and g(x) = 2x - 3, but you don't know f. You can just compose each side with \(\displaystyle g^{-1}\):

\(\displaystyle (f \circ g) \circ g^{-1} = h \circ g^{-1}\)

The left side simplifies to f, so the right side is the answer.

 
I observe that no one has suggested what to me is the best algebraic approach to this (no guessing).

You know that f(2x - 3) = 2x^2 - 4x + 1, and you want to find an expression for f(u). To do that, you can just substitute
u = 2x - 3

and solve to express x in terms of u. Put this in place of x on the right side, and simplify to obtain f(u).

Another way to describe this process, if you are comfortable with the notation of function composition, is that you want \(\displaystyle f \circ g = h\), where h(x) = 2x^2 - 4x + 1 and g(x) = 2x - 3, but you don't know f. You can just compose each side with \(\displaystyle g^{-1}\):

\(\displaystyle (f \circ g) \circ g^{-1} = h \circ g^{-1}\)

The left side simplifies to f, so the right side is the answer.


Awesome! Found the exact same result as the "guessing" and it is much easier. Thank you!
 
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