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Thread: 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: .$\displaystyle f(x) \:=\:\dfrac{\sin(x+1)}{\tan(x+1)}$

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

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

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


    Note: .$\displaystyle \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: $\displaystyle y \:=\:\cos x$ with a "hole" at (-1, 1).


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

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


    The answer is (a).




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

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

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

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

    The function is discontinuous at $\displaystyle x = 3.$


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

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


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

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


    The answer is (c).
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  3. #3
    MHF Contributor

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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) $\displaystyle \displaytype\lim_{x\to a} f(x)$ must exist.
    3) $\displaystyle \displaytype\lim_{x\to a} f(x)$ must be equal to f(a).

    If (2) is true, if $\displaystyle \displaytype\lim_{x\to a} f(x)$ exists, but (1) or (3) or both are not, we can make f continuous at x= a by defining $\displaystyle \displaytype f(a)= \lim_{x\to a} f(x)$. 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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