As, polynomial function is continuous everywhere in its domain. So, f[g(x)] will be continuous everywhere at x<0, 0<x<1 and x>1. We will check the behaviour of fog(x) only at boundary points which is x=0 and x=1.
lim
x=0,
x6=0−→0−(x3+2)=2 Clearly, L+HL≠RHL at x=0 So, f∘g(x) is discontinuous at x=0.
lim
x→1+
(3x−2)2x→1−(lim6=1 Also f(1)=1 fog (x) is continuous at x=1 Derivative test at x=1, LHD=
lim
h→0
f(1)−f(1−h)
h
=
lim
h→0
1−(1−h)6
h
=
lim
h→0
6(1−h)5=6 RHD=
lim
h→0
f(1+h)−f(1)
h
=
lim
h→0
[3(1+h)−2]2−1
h
=
lim
h→0
2[3(1+h)−2]⋅3=6 ∴ fog (x) is continuous and differentiable at x=1.