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Math Help - How do I remove these discontinuities?

  1. #1
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    How do I remove these discontinuities?

    So I understand how to find where a discontinuity is, but I am not sure how to remove it. Any tips?

    Question 1: Identify and remove the discontinuities of f(x) = [sin(x+1) / tan(x+1)]
    a) discontinuity at x = -1, remove by making f(-1)= 1
    b) discontinuity at x = -1, can not removed
    c) discontinuity at x = 0, remove by making f(0) = 1
    d) discontinuity at x = 0, remove by making f(0) = 1

    Question 2: Identify and remove the discontinuities of f(x) = [(x^3 - 3x^2 + 2x - 6) / (x - 3)]
    a) discontinuity at x = 4, remove by making f(3) = 9
    b) discontinuity at x = 4, can not remove
    c) discontinuity at x = 3, remove by making f(3) = 11
    d) discontinuity at x = 4, remove by making f(4) = 11


    Any help is appreciated.
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  2. #2
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    Hello, iluvmathbutitshard!

    So I understand how to find where a discontinuity is,
    but I am not sure how to remove it. Any tips?

    (1) Identify and remove the discontinuities of: . f(x) \:=\:\dfrac{\sin(x+1)}{\tan(x+1)}

    (a) discontinuity at x = \text{-}1, remove by making f(\text{-}1)= 1
    (b) discontinuity at x = \text{-}1, cannot removed
    (c) discontinuity at x = 0, remove by making f(0) = 1
    (d) discontinuity at x = 0, remove by making f(0) = 1

    When x = \text{-}1, we have: . f(\text{-}1) \;=\;\dfrac{\sin0}{\tan0} \;=\;\dfrac{0}{0}

    The function is discontinuous at x = \text{-}1


    Note: . \dfrac{\sin(x+1)}{\tan(x+1)} \;=\;\dfrac{\sin(x+1)}{\frac{\sin(x+1)}{\cos(x+1)}  } \;=\; \cos(x +1)\;\text{ provided }x \ne \text{-}1

    The graph is the curve: y \:=\:\cos x with a "hole" at (-1, 1).


    We find that: . \displaystyle \lim_{x\to\text{-}1}\cos x \;=\;1

    If we let f(\text{-}1) \,=\,1, we can "plug the hole".


    The answer is (a).




    (2) Identify and remove the discontinuities of: . f(x) \:=\: \dfrac{x^3 - 3x^2 + 2x - 6}{x - 3}

    (a) discontinuity at x = 4, remove by making f(3) = 9
    (b) discontinuity at x = 4, cannot removed
    (c) discontinuity at x = 3, remove by making f(3) = 11
    (d) discontinuity at x = 4, remove by making f(4) = 11

    The function is: . f(x) \;=\;\dfrac{(x-3)(x^2+2)}{x-3}

    When x = 3, we have f(3) \,=\,\dfrac{0}{0}

    The function is discontinuous at x = 3.


    Note: . f(x) \:=\:\dfrac{(x-3)(x^2+2)}{x-3} \:=\:x^2+2 \;\text{ provided } x \ne 3.

    The graph is the curve: y \:=\:x^2+2 with a "hole" at (3, 11).


    We find that: . \displaystyle \lim_{x\to3}(x^2+2) \:=\:11

    If we let f(3) = 11, we can "plug the hole".


    The answer is (c).
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  3. #3
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    You should understand that discontinuities are only rarely "removable".

    For f(x) to be continuous at a point, x= a, three things must be true:
    1) f(a) must exist.
    2) \displaytype\lim_{x\to a} f(x) must exist.
    3) \displaytype\lim_{x\to a} f(x) must be equal to f(a).

    If (2) is true, if \displaytype\lim_{x\to a} f(x) exists, but (1) or (3) or both are not, we can make f continuous at x= a by defining \displaytype f(a)= \lim_{x\to a} f(x). If (2) is not true, there is no way to make f continous at x= a.

    What Soroban did was find the limits at each point and define the function to have that value.
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