Question regarding continuity

• Feb 17th 2010, 11:55 AM
Solid8Snake
Question 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

• Feb 17th 2010, 12:04 PM
running-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
• Feb 17th 2010, 03:46 PM
Solid8Snake
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!
• Feb 18th 2010, 08:36 AM
Solid8Snake
Does someone have an approach in regards to how to solve this question?
• Feb 18th 2010, 09:12 AM
Plato
Quote:

Originally Posted by Solid8Snake
Does someone have an approach in regards to how to solve this question?

This question has been answered very clearly.
You were given an example of a function which works both parts a) & b).
What don't you understand about that function?
• Feb 18th 2010, 09:32 AM
Solid8Snake
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?
• Feb 18th 2010, 09:46 AM
Plato
If $f(x) = \left\{ {\begin{array}{rl}
{1,} & {x \ne 0} \\ {-1,} & {x = 0} \\ \end{array} } \right.$
then $f\left( {f(x)} \right) = 1\quad ,\left( {\forall x} \right)$

Example: $f\left( {f(2)} \right) =f\left( 1 \right) =1$
$f\left( {f(0)} \right) =f\left( -1 \right) =1$
$f\left( {f(-2)} \right) =f\left( 1 \right) =1$.
• Feb 18th 2010, 09:49 AM
Solid8Snake
Thanks Plato I think its clearer now. Sorry if I was nagging too much.