Pre-Calc Combination of Functions Question

[tboo]

If Im given two functions: f(x) & g(x), how can I go about finding the domain of f o g without finding the function of f o g itself?

 

[habib89]

isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

 

[tboo]

Originally posted by: habib89
isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

Yes I have a function for both f & g. f(x) is 5-x divided by x g(x) is 1 divided by x-5

The domain for f(x) is x cannot equal zero & the domain for g(x) is x cant equal 5.

I need to find the domain of f o g without working out the function itself.

 

[Legendary]

intersection of domain of f and domain of g

 

[tboo]

Originally posted by: Legendary
intersection of domain of f and domain of g

Can you elaborate?

 

[Gibson486]

man...you are so close too...you pratically already have the answer....keep working man, it will come to you.

 

[artikk]

Originally posted by: tboo

Originally posted by: Legendary
intersection of domain of f and domain of g

Can you elaborate?

The intersection would be the values that are located in both domains of the functions.
wiki link
Therefore the shared domain would contain a set of all Real numbers except 5
corrected edit

 

[Hacp]

I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

 

[tboo]

Originally posted by: artikk

Originally posted by: tboo

Originally posted by: Legendary
intersection of domain of f and domain of g

Can you elaborate?

The intersection would be the values that are located in both domains of the functions.
wiki link
Therefore the shared domain would contain a set of all Real numbers except 0 and 5

I thought that rule applied to the algebra of functions(f-g, f*g..etc) & not the composition of functions?

 

[Gibson486]

Originally posted by: Hacp
I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

well....i wanted him to figure it out, but artikk had to go off on a tangent and ruin the party.

You are supposed to realize that if a function has a restricted domain, the same restrictions will apply even if you plug into another function.

 

[Hacp]

Originally posted by: Gibson486

Originally posted by: Hacp
I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

well....i wanted him to figure it out, but artikk had to go off on a tangent and ruin the party.

You are supposed to realize that if a function has a restricted domain, the same restrictions will apply even if you plug into another function.

So if f(x) were 1/x and g(x) were (x+20), and you wanted f(g(x)) x can not be 0?

 

[Gibson486]

Originally posted by: Hacp

Originally posted by: Gibson486

Originally posted by: Hacp
I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

well....i wanted him to figure it out, but artikk had to go off on a tangent and ruin the party.

You are supposed to realize that if a function has a restricted domain, the same restrictions will apply even if you plug into another function.

So if f(x) were 1/x and g(x) were (x+20), and you wanted f(g(x)) x can not be 0?

You are missing the point.....


x can't be 0 in f(x). Instead of just plugging in g(x) into the equation, you can just equate the undefined values to g(x). In your case, it's so simple that the it's a duh, but when it gets more complex, then it is no longer a duh.

 

[Hacp]

Originally posted by: Gibson486

Originally posted by: Hacp

Originally posted by: Gibson486

Originally posted by: Hacp
I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

well....i wanted him to figure it out, but artikk had to go off on a tangent and ruin the party.

You are supposed to realize that if a function has a restricted domain, the same restrictions will apply even if you plug into another function.

So if f(x) were 1/x and g(x) were (x+20), and you wanted f(g(x)) x can not be 0?

You are missing the point.....


x can't be 0 in f(x). Instead of just plugging in g(x) into the equation, you can just equate the undefined values to g(x). In your case, it's so simple that the it's a duh, but when it gets more complex, then it is no longer a duh.

So x can't be 0 in that example right? I'm just confused, precalc was 5 years ago. I hate math btw .

 

[Hacp]

Now this is 10x more confusing. I just checked it on the web, and it confirmed what I had originally thought, that only 5 excluded from the domain in the OP's original function.

http://regentsprep.org/Regents.../f/domaincomposite.htm

 

[tboo]

Shouldnt the domain of f o g just be 5 just like Hacp said?

I think I figured this out. Its actually very easy. Too bad my prof cant explain it that way.

 

[TruePaige]

Originally posted by: Hacp

Originally posted by: Gibson486

Originally posted by: Hacp

Originally posted by: Gibson486

Originally posted by: Hacp
I'm confused as well. If he prevents you from plugging g(x) into f(x), how the fuck are you going to get the answer ever?

well....i wanted him to figure it out, but artikk had to go off on a tangent and ruin the party.

You are supposed to realize that if a function has a restricted domain, the same restrictions will apply even if you plug into another function.

So if f(x) were 1/x and g(x) were (x+20), and you wanted f(g(x)) x can not be 0?

You are missing the point.....


x can't be 0 in f(x). Instead of just plugging in g(x) into the equation, you can just equate the undefined values to g(x). In your case, it's so simple that the it's a duh, but when it gets more complex, then it is no longer a duh.

So x can't be 0 in that example right? I'm just confused, precalc was 5 years ago. I hate math btw .

See, x can't be zero, ever, because you cannot have 0 as the denominator. It also cannot be 5 because of the 5-x would = 0.

otherwise X can be whatever you want that is a real number and it works. These problems don't have purely numerical answers.


BTW, I hate math courses.

 

[LOUISSSSS]

Originally posted by: tboo

Originally posted by: habib89
isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

Yes I have a function for both f & g. f(x) is 5-x divided by x g(x) is 1 divided by x-5

The domain for f(x) is x cannot equal zero & the domain for g(x) is x cant equal 5.

I need to find the domain of f o g without working out the function itself.

f o g = [5-(1/x-5)]/(1/x-5)

 

[LOUISSSSS]

cake

 

[chuckywang]

Originally posted by: tboo

Originally posted by: habib89
isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

Yes I have a function for both f & g. f(x) is 5-x divided by x g(x) is 1 divided by x-5

The domain for f(x) is x cannot equal zero & the domain for g(x) is x cant equal 5.

I need to find the domain of f o g without working out the function itself.

Think about f(g(x)).

As before, x can't equal to 5, but also the input of f cannot be 0. So when is g(x) = 0? Never.

Therefore the domain is everything but 5.

 

[chuckywang]

Originally posted by: artikk

Originally posted by: tboo

Originally posted by: Legendary
intersection of domain of f and domain of g

Can you elaborate?

The intersection would be the values that are located in both domains of the functions.
wiki link
Therefore the shared domain would contain a set of all Real numbers except 0 and 5

No. x=0 is a perfectly valid input to the composition f(g(x)).

 

[Kirby]

math sucks giant cocks

 

[Acanthus]

Originally posted by: nkgreen
math sucks giant cocks

If a giant cock is perfectly cylindrical and has a diameter of 5' 3" and its length is 5/3x^3 the diameter. How many pictures of your cousin would you have to jerk it to in order to fill a 5 gallon bucket?

 

[tboo]

Originally posted by: chuckywang

Originally posted by: tboo

Originally posted by: habib89
isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

Yes I have a function for both f & g. f(x) is 5-x divided by x g(x) is 1 divided by x-5

The domain for f(x) is x cannot equal zero & the domain for g(x) is x cant equal 5.

I need to find the domain of f o g without working out the function itself.

Think about f(g(x)).

As before, x can't equal to 5, but also the input of f cannot be 0. So when is g(x) = 0? Never.

Therefore the domain is everything but 5.

Thats what I came up with.

Want to teach my Pre-calc class?

 

[Turin39789]

Originally posted by: tboo

Originally posted by: chuckywang

Originally posted by: tboo

Originally posted by: habib89
isn't the domain of f & g just the overlapping domain of both f and g? you should be given a function for both f and g. pre-calc was too long ago

Yes I have a function for both f & g. f(x) is 5-x divided by x g(x) is 1 divided by x-5

The domain for f(x) is x cannot equal zero & the domain for g(x) is x cant equal 5.

I need to find the domain of f o g without working out the function itself.

Think about f(g(x)).

As before, x can't equal to 5, but also the input of f cannot be 0. So when is g(x) = 0? Never.

Therefore the domain is everything but 5.

Thats what I came up with.

Want to teach my Pre-calc class?

Yup, this is right.

You figure out what would fuck up each equation. since you are plugging G into F, the domain is everything except what will fuck up G and everything that will cause G to have the values that will fuck up F.