Hello! I have been busting my head over this question and can't for some reason solve it. Can anyone explain to me how it works or how to compute the answer.The question I am reffering to is number 5 both A and B

Thx in advance!

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- February 17th 2010, 11:55 AMSolid8SnakeQuestion regarding continuity
Hello! I have been busting my head over this question and can't for some reason solve it. Can anyone explain to me how it works or how to compute the answer.The question I am reffering to is number 5 both A and B

Thx in advance! - February 17th 2010, 12:04 PMrunning-gag
Hi

5a) Let f defined by

f(x) = 1 for x not equal to 0

f(0) = -1

f is clearly not continuous at 0

g(x) = f(x) f(x) = 1 for all x real, therefore is continuous at 0

h(x) = f(f(x)) = 1 as well - February 17th 2010, 03:46 PMSolid8Snake
I am still rather confused as to how to do this question. I'm not even sure why f(x)f(x)=1 and for b you wrote the same thing but i think it was asking for another example.

Help would be greatly appreciated! - February 18th 2010, 08:36 AMSolid8Snake
Does someone have an approach in regards to how to solve this question?

- February 18th 2010, 09:12 AMPlato
- February 18th 2010, 09:32 AMSolid8Snake
his answer is very good, however its just the part where it says f(fx) that i get slightly confused at. Like what exactly are we plugging in. Is it like based off the original rule

1 for x does not equal 0

f(x)= -1 for x equals 0

so when we do f(fx) is that essentialy defined as

h(x)= ????? for x=0

????? for x does not equal 0

so f(f(0) is f(-1)---> which is = 1

and f(f(4) is f(1)---> which is = 1

so h(x)=1

I'm I on the right track? - February 18th 2010, 09:46 AMPlato
If then

Example:

. - February 18th 2010, 09:49 AMSolid8Snake
Thanks Plato I think its clearer now. Sorry if I was nagging too much.