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Math Help - Question regarding continuity

  1. #1
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    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

    Thx in advance!
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  2. #2
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    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
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  3. #3
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    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!
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  4. #4
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    Does someone have an approach in regards to how to solve this question?
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  5. #5
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    Quote Originally Posted by Solid8Snake View Post
    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?
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  6. #6
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    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?
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  7. #7
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    If f(x) = \left\{ {\begin{array}{rl}<br />
   {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.
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  8. #8
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    Thanks Plato I think its clearer now. Sorry if I was nagging too much.
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