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
f(x) =ax^2 +bx +3,     2<x<3
f(x) =2x -a+b,     x >= 3

myininaya · Answer

this looks like a problem for my paper lol

anonymous · Answer

what school do you go to?

myininaya · Answer

scanning 
i don't go anymore

myininaya · Answer

see we have two equations
8a+2b=-1
10a+2b=3

myininaya · Answer

if we solve the first one for b, we get
2b=-1-8a
b=(-1-8a)/2
then we can plug this into the other equation where the other b is
10a+2(-1-8a)/2=3
10a+(-1-8a)=3
10a-1-8a=3
2a-1=3
2a=4
a=2
so b=(-1-8*2)/2=(-1-16)/2=-17/2

myininaya · Answer

cool? :)

anonymous · Answer

yeah i think so. Thanks a lot

myininaya · Answer

np i like the continuous problems they are fun

anonymous · Answer

I think you made a slight mistake in the first equation. It should be $4a+2b+3=4$. Right?!

anonymous · Answer

So, we have the two equations $4a+2b=1$ and $10a+2b=3$. Subtract equation (1) from equation (2), you get $6a=2 \implies a=\frac{1}{3}$. Substitute in either equation, you will find that $b=-\frac{1}{6}$.