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Math Help - Finding value of two constants that make a precewise function continuous

  1. #1
    s3a
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    Finding value of two constants that make a precewise function continuous

    I'm having trouble finding the value of "a" on question #3 on one of the attached "Calculus 1 Final Exam Review (=Practice Exam) (Questions).pdf" file.

    I found b = 1 but I don't know how to find a.

    My work is attached as the "mywork.pdf" file and the specific work for this problem is on page 10/11 of that pdf file.

    Any help would be greatly appreciated!
    Thanks in advance!
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by s3a View Post
    I'm having trouble finding the value of "a" on question #3 on one of the attached "Calculus 1 Final Exam Review (=Practice Exam) (Questions).pdf" file.

    I found b = 1 but I don't know how to find a.

    My work is attached as the "mywork.pdf" file and the specific work for this problem is on page 10/11 of that pdf file.

    Any help would be greatly appreciated!
    Thanks in advance!
    f(x)=\left\{\begin{array}{lr}a+bx-1:&x\leq1\\ax:&1<x<3\\bx^2-ax+3:&3\leq x\end{array}\right\}

    So you need to solve:

    a+bx-1=ax at the point x=1
    ax=bx^2-ax+3 at the point x=3

    for a and b. In other words, solve

    a+b-1=a
    3a=9b-3a+3

    for a and b. Once you know that b=1, plug it into the second equation to solve for a.
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