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Homework Help: Finding a value to make piecewise continuous

  1. Feb 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Find c such that it makes f(x) continuous.

    2. Relevant equations

    2x+c&x < -5\\
    3x^2&-5 \leq x < 0\\
    cx^2&0 \leq x\\

    3. The attempt at a solution

    I know that
    [tex] \lim_{x\to 5^-}3x^2 [/tex] = 2x+c
    [tex] \lim_{x\to 0^+}3x^2 [/tex] = cx^2

    Which makes the 2 points where it disconnects at (-5, 75) and (0,0)
    Given that I make 2x+c = 75, where x=-5. I get C= 85, and since cx^2=0 is any real, does that mean the answer is c=85?
    Last edited: Feb 2, 2009
  2. jcsd
  3. Feb 2, 2009 #2


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    Welcome to PF!

    Hi Brian! Welcome to PF! :smile:

    Yes, c = 85.

    (though you have a strange way of using lim …

    you might as well say lim 3x2 = 3*(-5)2, and so on. :wink:)
  4. Feb 2, 2009 #3
    Re: Welcome to PF!

    My teacher has a definition where

    [tex]\lim_{x\to a}f(x) [/tex] = f(c)

    It was my understanding that f(x) = 3x^2, a = -5, and f(c) = 2x+c. So once I knew the answer to the limit, i knew that f(c) = 75, thus 2x+c must also be 75. Maybe I am misunderstanding the definition. Can anyone clarify?
  5. Feb 2, 2009 #4


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    :confused: that doesn't make any sense …

    what is f(c) supposed to mean? :confused:

    (f(c) = 2c + c or 3c2 or cc2)

    does he mean [tex]\lim_{x\to a}f(x) [/tex] = f(a)?​
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