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

[badatmaths]

Homework Statement

Find the values of a and b that make f continuous everywhere. (Enter your answers as fractions.)

Homework Equations

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.

 

[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.

 

[badatmaths]

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.

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

 

[Dick]

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

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

 

[badatmaths]

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

(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.

 

[Dick]

(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.

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

 

[badatmaths]

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

I'm going to blame it on the embarrassingly long amount of time I've been looking at this problem. I just solved, finally. Thanks a lot! I really appreciate it!

 

[Dick]

I'm going to blame it on the embarrassingly long amount of time I've been looking at this problem. I just solved, finally. Thanks a lot! I really appreciate it!

Staring at a problem for too long does make this stuff happen. Glad you got it.