# Thread: Finding value of two constants that make a precewise function continuous

1. ## 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!

2. 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!
$\displaystyle f(x)=\left\{\begin{array}{lr}a+bx-1:&x\leq1\\ax:&1<x<3\\bx^2-ax+3:&3\leq x\end{array}\right\}$

So you need to solve:

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

for $\displaystyle a$ and $\displaystyle b$. In other words, solve

$\displaystyle a+b-1=a$
$\displaystyle 3a=9b-3a+3$

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