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Math Help - Continuity

  1. #1
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    Continuity

    The function Abs[x - 2]/(x^2 - 5x + 6) is not defined for x=2. Is there a value that could be given for f(2) that would make the function continuous at 2?
    Give reasons for your answer.
    Last edited by Monster32432421; May 1st 2010 at 10:51 PM.
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  2. #2
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    You dropped an x on the denominator I think. Try factoring.
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  3. #3
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    Quote Originally Posted by maddas View Post
    You dropped an x on the denominator I think. Try factoring.
    yea my bad. You factorise it and simplify to get 1/(x-3) but then what would that mean?
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  4. #4
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    Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....
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  5. #5
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    Quote Originally Posted by maddas View Post
    Are you sure that's how it reduces? Make sure to consider both x>2 and x<2....
    |x-2|/(x-3)(x-2) = 1/(x-3)
    or cant i cancel them cause its abs?
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  6. #6
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    You tell me. Are both sides of that equation actually equal for all x? What about x=1?
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  7. #7
    MHF Contributor chisigma's Avatar
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    Because is...

    f(x)= \frac{|x-2|}{x^{2}- 5 x + 6} = \left\{\begin{array}{ll} \frac{1}{x-3} ,\,\, x >2\\{}\\ \frac{1}{3-x} ,\,\, x<2\end{array}\right. (1)

    ... is...

    \lim_{x \rightarrow 2 +} f(x)= -1

    \lim_{x \rightarrow 2 -} f(x)= 1 (2)

    ... so that f(x) isn't continous for x=2 no matter which is the value You assign to it in that point...

    Kind regards

    \chi \sigma
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  8. #8
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    ^
    Thank you


    Quote Originally Posted by maddas View Post
    You tell me. Are both sides of that equation actually equal for all x? What about x=1?
    time waster...
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  9. #9
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    Then they're not equal. Anyway, chisigma posted a solution now.

    edit: yea, cause its such a waste of time to know that 1/2 is not -1/2.
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  10. #10
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    Maddas' point when he wrote "You tell me. Are both sides of that equation actually equal for all x?" was that you cannot cancel "x- 2" because at x= 2, that would be dividing by 0.

    [tex]\frac{(x- 3)(x- 2)}{x- 2}= x- 3[/math for all x except x= 2. That's why the original function, \frac{|x- 2|}{x^2- 5x+ 6} is not defined at x= 2.

    Fortunately, the definition of " \lim_{x\to a} f(x)= L" says that "Given \epsilon> 0, there exist \delta> 0 such that if 0< |x- a|< \delta, then |f(x)- L|< \epsilon". That "0< |x- a|" (which many people forget to write) means that the value of f(x) at x= a has no effect whatsoever on the lim as x goes to a.

    Although [tex]\frac{(x- 3)(x- 2)}{x- 2}[tex] is not defined at x= 2, it has a limit there and you can make the function continuous by defining f(2) to be that limit.
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