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F at g(a) versus f o g at a?

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  • #1
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Homework Statement



I have to understand this to understand my homework. What is the difference between these two function compositions?

Homework Equations



f at g(a)

f o g at a

The Attempt at a Solution



Let's make f(a) = a^2 and g(a) = 3a.
It seems that "f at g(a)" means f(g(a)) which equals (3a)^2.
And it also seems like "f o g at a" means f(g(a)) which equals (3a)^2.
 

Answers and Replies

  • #2
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Homework Statement



I have to understand this to understand my homework. What is the difference between these two function compositions?

Homework Equations



f at g(a)

f o g at a

The Attempt at a Solution



Let's make f(a) = a^2 and g(a) = 3a.
It seems that "f at g(a)" means f(g(a)) which equals (3a)^2.
And it also seems like "f o g at a" means f(g(a)) which equals (3a)^2.
They're both the same. It's just different notation for the same thing.
 
  • #3
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They're both the same. It's just different notation for the same thing.
Are you sure? Here is the reason why I asked this. In my notes, it says, "Corollary 3.1.7 (Composition of Continuous Functions): Suppose g is continuous at a and f is continuous at g(a). Then f ◦g is continuous at a".

Maybe this whole corollary somehow changes the meaning of the two function compositions?
 
  • #4
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Are you sure? Here is the reason why I asked this. In my notes, it says, "Corollary 3.1.7 (Composition of Continuous Functions): Suppose g is continuous at a and f is continuous at g(a). Then f ◦g is continuous at a".

Maybe this whole corollary somehow changes the meaning of the two function compositions?
f o g is a function in its own right. It's defined as: (f o g)(x) = f(g(x)).
 
  • #5
LCKurtz
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Are you sure? Here is the reason why I asked this. In my notes, it says, "Corollary 3.1.7 (Composition of Continuous Functions): Suppose g is continuous at a and f is continuous at g(a). Then f ◦g is continuous at a".

Maybe this whole corollary somehow changes the meaning of the two function compositions?
No, it doesn't change anything. The domain of ##h = f\circ g## is the domain of ##g##, so it makes sense to think about continuity of ##h## at points ##a## in the domain of ##g## where ##g## is continuous. I'm not really sure what is bothering you about this.
 
  • #6
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No, it doesn't change anything. The domain of ##h = f\circ g## is the domain of ##g##, so it makes sense to think about continuity of ##h## at points ##a## in the domain of ##g## where ##g## is continuous. I'm not really sure what is bothering you about this.
It bothers me because then the corollary is essentially saying that if g is continuous at x and f(g(x)) is continuous at x, then f(g(x)) is also continuous at x. There would be no reason to have the last part.
 
  • #7
Office_Shredder
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Let me give an example where g is not continuous to show what they mean. Suppose f(x) = x2 and g(x) = 1 for all x except g(0) = 4. Then f is continuous at 4 so f is continuous at g(0), but f(g(x)) is not continuous at 0 (because g is not continuous at 0).

Hopefully this clears things up
 
  • #8
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It bothers me because then the corollary is essentially saying that if g is continuous at x and f(g(x)) is continuous at x, then f(g(x)) is also continuous at x. There would be no reason to have the last part.
That's not what it's saying. It's saying that if g is continuous at a, and f is continuous at g(a), then f o g is continuous at a. That's different from what you wrote.
 
  • #9
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Let me give an example where g is not continuous to show what they mean. Suppose f(x) = x2 and g(x) = 1 for all x except g(0) = 4. Then f is continuous at 4 so f is continuous at g(0), but f(g(x)) is not continuous at 0 (because g is not continuous at 0).

Hopefully this clears things up
Would you agree that "f is continuous at g(a)" means the exact same thing as "f ◦g is continuous at a"? If so, then we can just change the former with the latter and then the corollary reads,
"Suppose g is continuous at a and f ◦g is continuous at a. Then f ◦g is continuous at a".

Normally I would question the notes, but my professor said that he made them over 20 years ago with another mathematician, and they have been using them ever since.
 
  • #10
Office_Shredder
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No, I don't agree it's the same thing. f is continuous at g(a) is talking about the continuity of the function f, at a value in the domain of f, which happens to be g(a). f◦g being continuous at a is talking about the continuity of f◦g at a value in its domain. They're referring to the continuity of two different functions at points in two different domains, they do not mean the same thing. Read my post again I give an example of where f IS continuous at g(0), but f◦g is NOT continuous at 0, so those two concepts must be different.

The value of f at g(a) is the same as the value of f◦g at a, but the value of f NEAR g(a) is not the same thing as the value of f◦g near a, and continuity is talking about the latter, not the former.
 
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  • #11
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That's not what it's saying. It's saying that if g is continuous at a, and f is continuous at g(a), then f o g is continuous at a. That's different from what you wrote.
I know; I changed "f is continuous at g(x)" and "f ◦g is continuous at a" to "f(g(x))" since I am told that they are all equivalent, and then the corollary would exactly mean, "If g is continuous at x and f(g(x)) is continuous at x, then f(g(x)) is also continuous at x.
 
  • #12
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The value of f at g(a) is the same as the value of f◦g at a, but the value of f NEAR g(a) is not the same thing as the value of f◦g near a, and continuity is talking about the latter, not the former.
Ohhhh, that's what was screwing me up. Now I see the point, thanks!
 
  • #13
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Thank-you everyone :)
 
  • #14
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I know; I changed "f is continuous at g(x)" and "f ◦g is continuous at a" to "f(g(x))" since I am told that they are all equivalent, and then the corollary would exactly mean, "If g is continuous at x and f(g(x)) is continuous at x, then f(g(x)) is also continuous at x.
No, they are not all equivalent, and your change has a different meaning from what you started with. The statement in your notes is exclusively about functions: namely, f, g, and f o g.

g(x) is not a function - it's the value of the function at a number x in the domain of f. Likewise, f(g(x)) is also not a function - it's the value of the function f o g at a number in its domain.
 

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