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Prove 2x^2+5 is continuous at x=3

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the de finition of continuity show that f(x) = 2x2 + 5 is continuous at x = 3

    2. Relevant equations



    3. The attempt at a solution
    For all ε>0 there exists δ>0 such that |x-3|<δ implies that |2x2 + 5 -23| = |2(x2-9)| = |2(x+3)(x-3)| < |2(x+3)δ|

    Could anybody give me advice as to where I should take it from here to show that |2x2 + 5 -23|<ε
     
  2. jcsd
  3. Oct 25, 2012 #2

    tiny-tim

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    hi gottfried! :smile:

    if eg you insist that δ < 1, then |x+3| is between 5 and 7 :wink:
     
  4. Oct 25, 2012 #3

    Zondrina

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    |2(x+3)(x-3)| = 2|x+3||x-3| < 2δ|x+3|

    Now is |x+3| = |x - 3 + 6| ? Also remember to use the triangle inequality.
     
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