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Math Help - Continuous

  1. #1
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    Continuous

    Hello! This problem is almost same as last problem i posted but still i am stuck. Its problem nr 42Continuous-image.jpg i need help with
    My progress:
    (x^2-4)/(x-2)=(x+2) so at when lim goes to x->2 negative we know it shall be 4 so now we use number two that is in range 2<x<3 what we know is that when x=2 the answer shall be 4 because if its continuous there need to be lim same on negative and possitive. So we got the equation a2^2-2b=1 now im pretty stuck and dont know what to do.
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  2. #2
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    Re: Continuous

    Quote Originally Posted by Petrus View Post
    Hello! This problem is almost same as last problem i posted but still i am stuck. Its problem nr 42Click image for larger version. 

Name:	image.jpg 
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ID:	26919 i need help with

    Note that f(x) = \left\{ {\begin{array}{rr}   {x + 2,} & {x \leqslant 2}  \\   {ax^2  - bx + 3,} & {2 < x < 3}  \\   {2x - a + b,} & {3 \leqslant x}  \\ \end{array} } \right.

    \begin{gathered}  \lim _{x \to 2^ -  } f(x) = 4 \hfill \\  \lim _{x \to 2^ +  } f(x) = 4a - 4b + 3 \hfill \\  \lim _{x \to 3^ -  } f(x) = 9a - 3b + 3 \hfill \\  \lim _{x \to 3^ +  } f(x) = 6x - a + b \hfill \\ \end{gathered}
    Thanks from Petrus
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  3. #3
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    Re: Continuous

    Just solved it ty! Did just forget about that
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