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Finding removable discontinuity

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Let f(x)= {x^2-7x+10, for x^2 ≠ 25
    { A, for x = 5
    { B, for x = -5
    Is there a value of A that makes f continuous at x= 5?
    Is there a value of B that makes f continuous at x= -5?
     
  2. jcsd
  3. Oct 7, 2012 #2

    Zondrina

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    You should draw out the graph of f, it will be easier to see.

    So f(x) = x2-7x+10 when x≠±5. So f(x) is a parabola at every point except when x=5 or x=-5.

    At those two points it attains a value A for x=5 and B for x=-5 respectively. Are there values of A and B you can choose to make the graph smooth? As in have no jumps or breaks in it?

    Big hint : What is f(5) and f(-5) ? That should tell you something about the roots of f and the values you need.
     
  4. Oct 7, 2012 #3
    Thank you!
    There is a Vertical asymptote at x=-5 so does that mean there are no values to make it continuos?
     
  5. Oct 7, 2012 #4
    i made a mistake when writing the question, it is f(x) ={ x^2-7x+10 / x^2 - 25
     
  6. Oct 7, 2012 #5

    Zondrina

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    Ah even better then.

    Can you factor : x^2-7x+10 ?
    Can you factor : x^2-25?

    Now simplify your f(x) after factoring, what do you get and what do you notice?
     
  7. Oct 7, 2012 #6
    Yes, you can factor so there is a value for a when x = 5 you get 3/10. But there is not a value for when x=-5 because after plugging in -5 you get undefined therefore there are no values for B.
    Thank you so much!
     
  8. Oct 7, 2012 #7

    Zondrina

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    No problem bud :)
     
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