student34 said: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.
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".Mark44 said:They're both the same. It's just different notation for the same thing.
f o g is a function in its own right. It's defined as: (f o g)(x) = f(g(x)).student34 said: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?
student34 said: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?
LCKurtz said: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.
student34 said: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.
Office_Shredder said: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
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.Mark44 said: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.
Ohhhh, that's what was screwing me up. Now I see the point, thanks!Office_Shredder said: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.
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.student34 said: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.
F at g(a) is the composition of two functions, f and g, evaluated at the input value a. This means that the output of g(a) is used as the input for f. On the other hand, f o g at a represents the output of g(a) being used as the input for f, which is then evaluated at the input value a.
To calculate f at g(a), first evaluate g(a) to get a value, and then plug that value into the function f to get the final result. For f o g at a, evaluate g(a) to get a value, and then plug that value into the input of f to get the final result.
Yes, it is possible for f at g(a) and f o g at a to be equal, but it is not always the case. It depends on the specific functions f and g, as well as the input value a. In general, they will not be equal unless g(a) is a constant or identity function.
Composition of functions is used to combine two or more functions in order to create a new function. This can be useful in simplifying complex functions, expressing relationships between different functions, and solving certain mathematical problems.
The chain rule in calculus states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This directly relates to f at g(a) and f o g at a, as they represent the evaluation of the outer function and the inner function, respectively.