continuos everywhere

Question

find the value of a and b that make f continuous everywhere

f( x ) =  x^2 - 4 / (x - 2) ; x < 2

            ax^2- bx + 3 ; 2< x < 3

           2x - a + b  ;  > 3

Answer 1

For continuity at x=2: (x^2-4)/(x-2)=ax^2-bx+3 when x=2; so:

(x-2)(x+2)/(x-2)=x+2 in the limit as x→2, so the left-hand expression evaluates to 4.

4a-2b+3=4, 4a-2b=1 and 8a-4b=2. Call this equation A.

At x=3: 9a-3b+3=6-a+b for continuity. So 10a-4b=3. Call this equation B.

B-A; 2a=1, a=1/2 and 2b=4a-1=1, so b=1/2.

The graph will appear continuous with a=b=1/2, but the function is undefined at x=2. If the second condition had been 2≤x<3 it would have been continuous everywhere because the anomaly of (x^2-4)/(x-2) would have been avoided.