Finding value of two constants that make a precewise function continuous

**s3a** · December 10th 2009, 02:18 PM

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!

**redsoxfan325** · December 10th 2009, 06:50 PM

Originally Posted by s3a

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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