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Math Help - Continuous

  1. #1
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    Continuous

    Hello i got problem with problem nr 41 . Ik how i shall think but that c making it difficoult for me. To se if its continuous is limit right side=limit left side=f(a).
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  2. #2
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    Re: Continuous

    Quote Originally Posted by Petrus View Post
    Hello i got problem with problem nr 41 . Ik how i shall think but that c making it difficoult for me. To se if its continuous is limit right side=limit left side=f(a).
    Very good! Not enough people know the basic definitions.
    For x< 2, f(x)= cx^2+ 2x. What is \lim_{x\to 2^-} f(x)?
    For x> 2, f(x)= x^3- cx. What is \lim_{x\to 2^+} f(x)?
    What is f(2)?
    Thanks from Petrus
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  3. #3
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    Re: Continuous

    Hello Hallsofivy!
    The problem is idk how to calculate limit when i got c, well im kinda trying but does not work, im stuck there.
    well this is what i think from left side we get 2(2+c)(negative way)
    Right side 2(4-c) possitive way and f(2)=2(4-c)
    Last edited by Petrus; February 8th 2013 at 12:38 PM.
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  4. #4
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    Re: Continuous

    Quote Originally Posted by Petrus View Post
    Hello Hallsofivy!
    The problem is idk how to calculate limit when i got c, well im kinda trying but does not work, im stuck there.
    well this is what i think from left side we get 2(2+c)(negative way)
    Right side 2(4-c) possitive way and f(2)=2(4-c)
    You want \lim _{x \to 2 ^ +  } cx^2  + 2x = \lim _{x \to 2^ -  } x^3  - cx.

    \begin{gathered}  \lim _{x \to 2^ -  } cx^2  + 2x = 4c + 4 \hfill \\   \hfill \\  \lim _{x \to 2^ +  } x^3  - cx = 8 - 2c \hfill \\ \end{gathered}
    Last edited by Plato; February 8th 2013 at 02:23 PM.
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    Re: Continuous

    If i make it to a equation then i get c=1 so?
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  6. #6
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    Re: Continuous

    Quote Originally Posted by Plato View Post
    You want \lim _{x \to 2 ^ +  } cx^2  + 2x = \lim _{x \to 2^ -  } x^3  - cx.

    \begin{gathered}  \lim _{x \to 2^ -  } cx^2  + 2x = 4c + 4 \hfill \\   \hfill \\  \lim _{x \to 2^ +  } x^3  - cx = 8 - 2c \hfill \\ \end{gathered}
    Why didn't you substitute x = 2 into the LHS of your equation too?
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  7. #7
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    Re: Continuous

    Quote Originally Posted by Petrus View Post
    If i make it to a equation then i get c=1 so?
    Surly you get c=\frac{2}{3}~?
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  8. #8
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    Re: Continuous

    Sorry My bad, accident did write a wrong number on equation:/ ty for all help,explain i got! Have a nice day
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