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Math Help - Continuity question..

  1. #1
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    Continuity question..

    Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)



    a=?
    b=?

    Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
    Thank you in advance
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  2. #2
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    Quote Originally Posted by rodjav305 View Post
    Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)



    a=?
    b=?

    Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
    Thank you in advance
    Continuous everywhere means its continuous at x=1 and x=3.
    What is the definition of the continuity of a function at a point?
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rodjav305 View Post
    Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)



    a=?
    b=?

    Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
    Thank you in advance
    In order to be continuous we must have

    \lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}(ax^2+bx+2)

    and

    \lim_{x\to3}(3x-a+b)=\lim_{x\to3}(ax^2+bx+2)

    Substitute and solve.
    Last edited by VonNemo19; January 31st 2010 at 06:55 PM. Reason: Little innaccuracy.
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  4. #4
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    Quote Originally Posted by rodjav305 View Post
    Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)



    a=?
    b=?

    Can someone show me the steps on how to do it? I think i messed up in substituting (-2) in both equations.
    The answer to the question in red is no.
    That is not the object of this site. We want you to learn something.
    So here is some help.
    If \lim _{x \to 1^ -  } f(x) = 2 then what are a~\&~b so that \lim _{x \to 1^ +  } f(x) = 2?
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  5. #5
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    i keep on getting (7/5) for a though, when I substitute my end result is -22a+10b=-14. I know it's wrong but i do not know where i went wrong
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  6. #6
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    Quote Originally Posted by rodjav305 View Post
    i keep on getting (7/5) for a though, when I substitute my end result is -22a+10b=-14. I know it's wrong but i do not know where i went wrong
    We should see your work to tell you what are your mistakes.
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  7. #7
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    at x=1, I get 1a-1b+2=2 ,or 1a-1b=0
    at x=3 I get 9a-3b+2=9-a+b, or 10a-4b=7

    From there on I substituted (-2) into equation 1 and 2. Can someone confirm me if substituting (-2) is correct or no?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Evaluating the limits in my above post, and then solving the system of the equations that remain is all you need to do.
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rodjav305 View Post
    at x=1, I get 1a-1b+2=2 ,or 1a-1b=0
    at x=3 I get 9a-3b+2=9-a+b, or 10a-4b=7

    From there on I substituted (-2) into equation 1 and 2. Can someone confirm me if substituting (-2) is correct or no?
    a-b=0\Rightarrow{a}=b

    Which further implies that

    10a-4a=7

    So, whats a?
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  10. #10
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    Quote Originally Posted by VonNemo19 View Post
    a-b=0\Rightarrow{a}=b

    Which further implies that

    10a-4a=7

    So, whats a?

    Wow thank you very much VonNemo19, sorry for having to put up with my trouble. I thought the problem was more complicated and no i notice i should have stopped. Thank you!!
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rodjav305 View Post
    Wow thank you very much VonNemo19, sorry for having to put up with my trouble. I thought the problem was more complicated and no i notice i should have stopped. Thank you!!
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  12. #12
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    Sorry to bother but for x=1 it is 1a-1b=0
    and i cannot get a fraction out of it.
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  13. #13
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by rodjav305 View Post
    Sorry to bother but for x=1 it is 1a-1b=0
    and i cannot get a fraction out of it.

    \lim_{x\to1^-}\frac{x^2-1}{x-1}=\lim_{x\to1^+}(ax^2+bx+2)

    Which implies

    1. 0=a+b or a=-b


    \lim_{x\to3^+}(3x-a+b)=\lim_{x\to3^-}(ax^2+bx+2)

    implies

    2. 9-a+b=9a+3b+2

    From 1. we see that a=-b. So we substitute for a in 2.

    9-(-b)+b=9(-b)+3b+2

    We now solve for b

    9+2b=-6b+2

    8b=-7

    b=\frac{-7}{8}.

    And from 1.

    a=-\left(\frac{-7}{8}\right).

    Therefore

    f(x)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},&\mbox{ if }<br />
x<1\\\\{\frac{7}{8}x^2+\frac{7}{8}x+2},&\mbox{ if }1\leq{x}<3\\\\{3x-\frac{7}{4}},&\mbox{ if }x\geq3\end{array}\right.
    Last edited by VonNemo19; January 31st 2010 at 07:49 PM.
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