Question

What values of a and b makes the following question continuous?

Answer 1

To be continuous, you need the limit to exist at, and be = to, f(x) at each point.

So for the "first chunk" of the domain, we have:

\(f(x)=\dfrac{x^2-4}{x-2}~~~for~x<2\)
So you need the limit at x=2 of that expression... can you tell me what that is? Now we'll call that L.

The "next chunk" of the domain gives us:
\(f(x)=ax^2-bx+1~~~for~2\le x<3\)

That's a polynomial so its limit is just = it's value at x=2, so we need L from above to equal that value, e.g, you need:
\(f(2)=ax^2-bx+1=L\)

Then finally, you have
\(f(x)=4x-a+b~~~for~x\ge 3\)

Again this is a polynomial, so you just need the value at 3 from the middle chunk to equal the value at 3 here, e.g., you need:

\(4(3)-a+b=a3^2-3b+1\)

So I'm thinking that gives you (after you find L) a system of 2 equations in 2 unknowns a & b to solve.
\(ax^2-bx+1=L\)
\(12-a+b=9a-3b+1\)

Answer 2

so for that second chunk would it be 0= 4a-2b+1

Answer 3

I'm not sure what you mean. Did you find L?

Answer 4

I found L to be 0/0 when using limit as x approaches 2 from the negative side

Answer 5

That's not a limit. YOu need to reduce, that rational function isn't in lowest form. Then find the limit.

Answer 6

I'm having trouble though... the solution is not working out for me, although the method seems sound.

Answer 7

ohhhh right oops :x thank you

Answer 8

...trying to figure out what's wrong....

Answer 9

so L is 4

Answer 10

yes

Answer 11

so then I would get 4a-2b=3 and then I'm completely lost lol

Answer 12

OK, I think I found my error.... Yes, that's right. Now that's one of your equations in 2 unknowns (that's what I should have put down at the bottom, where I re-capped the 2 equations).

Answer 13

But DO YOU UNDERSTAND WHY?? :) Do you understand how we came up with THAT equation?

Answer 14

The limits must be equal at the "crossover" points, e.g., when we cross from the first "chunk" to the middle "chunk" to the 3rd "chunk", the limits must be =

Answer 15

Yeah because youre using 2 for the first and second equation and 3 for second and third

Answer 16

well, yes... we have 2 unknowns, and 2 conditions:

limits = at x=2 is one condition
limits = at x=3 is the other condition

Each condition gives us one equation, hence we get 2 equations in 2 unknowns. :)

Answer 17

right :)

Answer 18

and then do i like solve the second condition

Answer 19

Well, you have to solve the SYSTEM. 2 equations, 2 unknowns.

Answer 20

would that be 11= 10a-4b

Answer 21

i mean for the second part with the 3

Answer 22

Yes, that looks right

Answer 23

What would I do after this?

Answer 24

In relation of the two equation when using 2 a= 3+2b/4 and when using 3 a= 11+4b/10 should I set them equal to eachother

Answer 25

Well, you can't set them equal unless you solve both for one of the variables, first. I think elimination method would work better here.

I'm so so sorry but I HAVE to go..... I do know that this method will work because I found a and b and can see from the graphs that they work.

Answer 26

its alrite thank you for your help :)