Find the values of a and b that make f continuous everywhere
f(x) = (x^2 - 4)/(x-2) if x < 2
ax^2 - bx + 3 if 2 <= x < 3
2x - a + b if x >= 3
I have done questions similar to this before, but only using one variable. I'm not for certain how to go about one that involves two.
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Let `f(x)=(x^2-4)/(x-2)` for `x<2`
`f(x)=ax^2-bx+3` for `2 <= x <3`
`f(x)=2x-a+b` for `x >=3`
Find a,b so that f is continuos everywhere.
First note that we need only work at x=2 and x=3 as f is continuous on `(-oo,2),(2,3),(3,oo)` .(Polynomials and rationals are continuous everwhere on their domains)
(1) `lim_(x->2^-)(x^2-4)(x-2)=lim_(x->2^-)((x+2)(x-2))/(x-2)=lim_(x->2^-)x+2=4` . In order for f to be continuous we need `ax^2-bx+3=4` at `x=2` .
So `4a-2b+3=4` or `4a-2b=1` .
(2) `lim_(x->3^+)2x-a+b=6-a+b` . Thus we need `ax^2-bx+3=6-a+b` at `x=3` or `9a-3b+3=6-a+b` .
So `10a-4b=3` .
(3) Solve `4a-2b=1` and `10a-4b=3` simultaneously to get `a=1/2,b=1/2` which is the answer.
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