find the values of a, b and c for which the function f(x) is continuous in the real line f(x)

Question

f(x)= 3                 if       x<=0

         -x^2+3x+a if        0<x<1

         bx+c           if  1<=x<=2

         6                 if      x>=2

Answer 1

The function is continuous, so since f(x)=3 when x=0, -x^2+3x+a=3; when x=0, a=3. As x approaches 1 f(x) approaches 5, so bx+c=5 when x=1. Therefore b+c=5. Also when x=2, bx+c=6, so 2b+c=6. This makes b=1 and c=4. Thus, f(x)=3 when x<=0; f(x)=-x^2+3x+3 when 0<x<1; f(x)=x+4 when 1<=x<=2; f(x)=6 when x>=2. This makes f(x) continuous over the whole domain: a=3, b=1, c=4.