Let f(x)= casesx3+8; & x<0,\ x2-4; & x 0, cases and g(x)= cases(x-8)1/3; & x<0,\ (x+4)1/2; & x 0. cases Then the number of points, where the function g f is discontinuous, is .
Answer
3
Solution
[figure]
g f(x)=g(f(x)); the sign of f(x) decides which branch of g is used.
For x<0: f(x)=x3+8<0 x<-2
For x 0: f(x)=x2-4<0 0 x<2
g(f(x))= casesx; & x<-2\ x3+12; & -2 x<0\ (x2-12 )1/3; & 0 x<2\ x; & x 2 cases
(x<-2: (x3+8-8 )1/3=x; x 2: x2-4+4=x)
Each piece is continuous inside its own interval; check the joining points:
x=-2: LHL =-2, value = 4=2 discontinuous
x=0: LHL = 12, value =(-12)1/3 discontinuous
x=2: LHL =(-8)1/3=-2, value =2 discontinuous
Number of points of discontinuity =3
Answer: 3