1. ## 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.

2. Originally Posted by rodjav305
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.
Continuous everywhere means its continuous at x=1 and x=3.
What is the definition of the continuity of a function at a point?

3. Originally Posted by rodjav305
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.
In order to be continuous we must have

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

and

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

Substitute and solve.

4. Originally Posted by rodjav305
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 $\displaystyle \lim _{x \to 1^ - } f(x) = 2$ then what are $\displaystyle a~\&~b$ so that $\displaystyle \lim _{x \to 1^ + } f(x) = 2?$

5. 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

6. Originally Posted by rodjav305
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.

7. 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?

8. Evaluating the limits in my above post, and then solving the system of the equations that remain is all you need to do.

9. Originally Posted by rodjav305
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?
$\displaystyle a-b=0\Rightarrow{a}=b$

Which further implies that

$\displaystyle 10a-4a=7$

So, whats a?

10. Originally Posted by VonNemo19
$\displaystyle a-b=0\Rightarrow{a}=b$

Which further implies that

$\displaystyle 10a-4a=7$

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!!

11. Originally Posted by rodjav305
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!!

12. Sorry to bother but for x=1 it is 1a-1b=0
and i cannot get a fraction out of it.

13. Originally Posted by rodjav305
Sorry to bother but for x=1 it is 1a-1b=0
and i cannot get a fraction out of it.

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

Which implies

1. $\displaystyle 0=a+b$ or $\displaystyle a=-b$

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

implies

2. $\displaystyle 9-a+b=9a+3b+2$

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

$\displaystyle 9-(-b)+b=9(-b)+3b+2$

We now solve for b

$\displaystyle 9+2b=-6b+2$

$\displaystyle 8b=-7$

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

And from 1.

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

Therefore

$\displaystyle f(x)=\left\{\begin{array}{cc}\frac{x^2-1}{x-1},&\mbox{ if } x<1\\\\{\frac{7}{8}x^2+\frac{7}{8}x+2},&\mbox{ if }1\leq{x}<3\\\\{3x-\frac{7}{4}},&\mbox{ if }x\geq3\end{array}\right.$