Find the values of a and b that make f continuous everywhere. F(x)= (x^2-4)/(x-2) x

Question

Find the values of  a and b that make f continuous everywhere. 
 F(x)=  (x^2-4)/(x-2) x< 2
           ax^2-bx+2    2 (uncluding 2) < x < 3
           4x-a+b     x>3 (including 3)

anonymous · Answer

do you know 
$$\lim_{x\to 2}\frac{x^2-4}{x-2}$$?

anonymous · Answer

yes!

anonymous · Answer

what is it?

anonymous · Answer

4

agent0smith · Answer

For the function to be continuous, all the endpoints must meet up.

When x=2, the (limit of the) first function equals 4. So ax^2-bx+2 must equal 4.

Then, when x=3, ax^2-bx+2 must equal 4x-a+b.

agent0smith · Answer

Plug all the x's in and you should be able to solve some simultaneous equations.

anonymous · Answer

so, for ax^2-bx+2.. I substitute 2 and 3 ..

anonymous · Answer

and then when i substitute 3, and get my result, the last equation must equal my result from the last one.

anonymous · Answer

yes

anonymous · Answer

$$4a+2b-2=4$$ is one equation

anonymous · Answer

$$9a-3b+2=12-a+b$$ is the other one

anonymous · Answer

but everything has variables, do they stay like that?

agent0smith · Answer

You have two equations and two variables, which you can solve simultaneously, just like if you had something like

x+y = 2
3x+2y=6

anonymous · Answer

ok. so I have being sitting here, trying to do the math, but I still don't understand where I'm getting at. I mean the second equation, I got 4a-2b+2=4 , do I pick a variable and solve for it?

agent0smith · Answer

Use simultaneous equations... remember those from algebra?

agent0smith · Answer

You have two equations which can be used to solve for a and b.