Continuity question..

**rodjav305** · January 31st 2010, 05:36 PM

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

**TWiX** · January 31st 2010, 05:48 PM

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.
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?

**VonNemo19** · January 31st 2010, 05:50 PM

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.
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.

**Plato** · January 31st 2010, 05:51 PM

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 $\lim _{x \to 1^ - } f(x) = 2$ then what are $a~\&~b$ so that $\lim _{x \to 1^ + } f(x) = 2?$

**rodjav305** · January 31st 2010, 05:53 PM

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

**General** · January 31st 2010, 05:54 PM

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.

**rodjav305** · January 31st 2010, 05:57 PM

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?

**VonNemo19** · January 31st 2010, 05:58 PM

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

**VonNemo19** · January 31st 2010, 06:02 PM

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?

$a-b=0\Rightarrow{a}=b$

Which further implies that

$10a-4a=7$

So, whats a?

**rodjav305** · January 31st 2010, 06:05 PM

Originally Posted by VonNemo19

$a-b=0\Rightarrow{a}=b$

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!!

**VonNemo19** · January 31st 2010, 06:08 PM

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!!

**rodjav305** · January 31st 2010, 06:16 PM

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

**VonNemo19** · January 31st 2010, 06:35 PM

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.

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

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

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

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