Finding the domain of a composite function

AI Thread Summary
The discussion focuses on finding the domain of the composite function f(g(x)), specifically analyzing the expression f(g(x)) = f(4/(3x - 2)). Participants highlight that the domain is restricted by values that make g(x) undefined, particularly noting that x cannot equal 2/3. The importance of simplifying the function correctly to identify these restrictions is emphasized, as multiplying by terms that equal zero is not permissible. The conversation reveals a misunderstanding of the simplification process and the necessity of considering both g(x) and f's domain when determining the overall domain of the composite function. Ultimately, understanding these principles is crucial for accurately solving composite function problems.
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Homework Statement
Pls see below
Relevant Equations
Pls see below
For this problem,
1677528081908.png

The solution is,
1677528105661.png

However, I tried solving this problem by using the definition of composite function

##f(g(x)) = f(\frac{4}{3x -2}) = \frac{5}{\frac{4}{3x - 2} - 1} = \frac{5}{\frac{6 - 3x}{3x - 2}} = \frac {15x - 10}{6 - 3x}## which only gives a domain ##x ≠ 2##. Would some please know how to find the solution from the composite function?

Many thanks!
 
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Consider: <br /> \frac{5}{\frac{4}{3x-2}-1} = \frac{5(3x-2)}{\left(\frac{4}{3x-2}-1\right)(3x-2)}. This is fine if x \neq 2/3, but if x = 2/3 then multiplying numerator and denominator by 3x - 2 is multiplying numerator and denominator by zero, which is not allowed.
 
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pasmith said:
Consider: <br /> \frac{5}{\frac{4}{3x-2}-1} = \frac{5(3x-2)}{\left(\frac{4}{3x-2}-1\right)(3x-2)}. This is fine if x \neq 2/3, but if x = 2/3 then multiplying numerator and denominator by 3x - 2 is multiplying numerator and denominator by zero, which is not allowed.
Thank you for your reply @pasmith!

Sorry, why did you not multiply ##\frac{4}{3x - 2}## individually by the ##\frac{3x-2}{3x-2}##?

Many thanks!
 
Callumnc1 said:
Sorry, why did you not multiply ##\frac{4}{3x - 2}## individually by the ##\frac{3x-2}{3x-2}##?
Here's from your work:
Callumnc1 said:
However, I tried solving this problem by using the definition of composite function ##f(g(x)) = f(\frac{4}{3x -2})##
What do you get if x = 2/3 when you evaluate f(g(2/3))?
 
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Callumnc1 said:
pasmith said:
Consider: <br /> \frac{5}{\frac{4}{3x-2}-1} = \frac{5(3x-2)}{\left(\frac{4}{3x-2}-1\right)(3x-2)}. This is fine if x \neq 2/3, but if x = 2/3 then multiplying numerator and denominator by 3x - 2 is multiplying numerator and denominator by zero, which is not allowed.

Thank you for your reply @pasmith!

Sorry, why did you not multiply ##\frac{4}{3x - 2}## individually by the ##\frac{3x-2}{3x-2}##?

Many thanks!
Sure. If you distribute the ##(3x-2)## in the denominator and then simplify by "cancelling" and collecting terms, you will indeed get ## \dfrac {15x - 10}{6 - 3x}## which is ## \dfrac {5(3x - 2)}{3(2 - x)}## .

However, the point made by @pasmith is that the step he shows, which is required to get your final result;
that step is valid, only if ##x\ne \dfrac 2 3 ##.

@Mark44 shows in a different way, why ##x\ne \dfrac 2 3 ##.

If you study the given solution (the one you posted) seriously, you should notice that the first sentence there says much the same thing as Mark says.
 
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Mark44 said:
Here's from your work:

What do you get if x = 2/3 when you evaluate f(g(2/3))?
Thank you for your reply @Mark44!

I think I get undefined. However, I thought we were meant to simplify the composite function to that rational function in post #1 before substituting in values?

Many thanks!
 
SammyS said:
Sure. If you distribute the ##(3x-2)## in the denominator and then simplify by "cancelling" and collecting terms, you will indeed get ## \dfrac {15x - 10}{6 - 3x}## which is ## \dfrac {5(3x - 2)}{3(2 - x)}## .

However, the point made by @pasmith is that the step he shows, which is required to get your final result;
that step is valid, only if ##x\ne \dfrac 2 3 ##.

@Mark44 shows in a different way, why ##x\ne \dfrac 2 3 ##.

If you study the given solution (the one you posted) seriously, you should notice that the first sentence there says much the same thing as Mark says.
Thank you for your reply @SammyS!

I think I understand @pasmith method now! I guess it interesting that he did not simplify the fraction at first to see what values it was zero for, and then to find the other x-value you have to reduce it more to get the rational function I got. I guess this is interesting. Is there a way to remember this technique? I have not seen it in my calculus textbook.

With other composite function problems (without fractions) I have done, I just had to find the x-values that would not work for the final answer without having to do that in intermediate steps.

Many thanks!
 
Callumnc1 said:
I think I get undefined. However, I thought we were meant to simplify the composite function to that rational function in post #1 before substituting in values?
No, not quite. You're asked to find the domain of f(g(x)), which means you have to find values of x for which g(x) is undefined, and values of g(x) for which f is undefined. These values will determine the domain of f(g(x)).
 
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Callumnc1 said:
With other composite function problems (without fractions) I have done, I just had to find the x-values that would not work for the final answer without having to do that in intermediate steps.
Well, for those problems, it looks like you were simply fortunate.
 
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Mark44 said:
No, not quite. You're asked to find the domain of f(g(x)), which means you have to find values of x for which g(x) is undefined, and values of g(x) for which f is undefined. These values will determine the domain of f(g(x)).
Thank you for your help @Mark44!
 
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SammyS said:
Well, for those problems, it looks like you were simply fortunate.
Thank you for your help @SammyS! Oh ok, that's quite interesting because I solved quite a few of them using that method.

Many thanks!
 
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