Find the values of a and b that make f continuous everywhere

1. Sep 12, 2010

1. The problem statement, all variables and given/known data
Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)

2. Relevant equations

3. The attempt at a solution

lim x->2 (x2-4)/(x-2) = 0

lim x->2 4a - 2b +5
lim x->3 9a - 3b +5

lim x->3 12 - a + b

4a - 2b + 5 = 0
4a - 2b = -5

9a - 3b +5 = 12 - a + b
10a - 4b = 7

4a - 2b = -5
a = 2.5b/4

10(2.5b/4) - 4b = 7
6.25b - 4b = 7
2.25b = 7
b = 28/9

I'm obviously missing something because I've been screwing around with this problem for well over an hour already and I haven't been able to solve it. I need to get this done by tomorrow morning so any help is appreciated.

Attached Files:

• 1mLJ9.gif
File size:
1.6 KB
Views:
16
2. Sep 12, 2010

Dick

You've got the right method, but there's a problem with the first step. lim x->2 (x^2-4)/(x-2) isn't zero.

3. Sep 12, 2010

I'm really not sure what to do if x->2 for the first equation is undefined...

4. Sep 12, 2010

Dick

Try and factor the numerator and cancel the factor which is going to zero.

5. Sep 12, 2010

(x+2)(x-2)/(x-2)
(x+2)
x->2 = 4

That would mean that...

4a - 2b + 5 = 4
4a - 2b = -1
4a = 1/2b
a = 1/8b

But plugging that into 10a - 4b = 7 gives me...

10(1/8b) - 4b = 7
(5/4)b - 4b = 7
(-11/4)b = 7
b = -28/11

...Okay, that isn't right...I did this problem 3 different times and came up with a new answer each time.

6. Sep 12, 2010

Dick

You are goofing up your algebra. 4a-2b=(-1) doesn't lead to 4a=(1/2)b. What are you doing?

7. Sep 12, 2010