# Continuity of Composite Functions

*F*(

*x*) =

*f*(

*g*(

*x*)) is discontinuous also at those values of

*x*where

*g*(

*x*) is discontinuous. For example,

*f*(

*x*) = 1/(1 â€“

*x*) is discontinuous at

*x*= 1. Now,

*f*(

*f*(

*x*)) =

is not only discontinuous at

*x*= 0 but also at*x*= 1. Now,*f*(

*f*(

*f*(

*x*))) =

seems to be continuous, but it is discontinuous at

*x*= 1 and*x*= 0 where*f*(*x*) and*f*(*f*(*x*)) respectively are discontinuous.

**Illustration**

If

*f*(*x*) = and*g*(*x*) = , then discuss the continuity of*f*(*x*),*g*(*x*), and fog(*x*).Solution

*f*(

*x*) =

*f*is not defined at

*x*= 1, therefore,

*f*is discontinuous at

*x*= 1.

*g*(

*x*) =

*g*(

*x*) is not defined at

*x*= 2, therefore,

*g*is discontinuous at

*x*= 2.

Now

*fog*(*x*) will be discontinuous at*x*= 2 (point of discontinuity of*g*(*x*)).*g*(*x*) = 1 (when*g*(*x*) = point of discontinuity of*f*(*x*)).

If

*g*(*x*) = 1 â‡’ 1/(*x*â€“ 2) = 1 â‡’*x*= 3, therefore,*fog*(*x*) is discontinuous at*x*= 2 and*x*= 3. Also*fog*(

*x*) =

and

*fog*(2) is not defined. =

=

Therefore,

*fog*(*x*) is discontinuous at*x*= 2 and it has removable discontinuity at*x*= 2 .For continuity at

*x*= 3, =

=

Therefore,

*fog*(*x*) is discontinuous at*x*= 3 and it has non-removable discontinuity at*x*= 3,# Properties of functions continuous in [a, b]

- If
*a*function*f*is continuous on a closed interval [*a*,*b*] then it is bounded. - A continuous function whose domain is some closed interval must have its range also in closed interval.
- If
*f*(*a*) and*f*(*b*) possess opposite signs then âˆƒ at least one solution of the equation*f*(*x*) = 0 in the open interval (*a*,*b*) provided*f*is continuous in [*a*,*b*]. - If
*f*is continuous on [*a*,*b*] then*f*^{â€“1}(from the range of*f*) is also continuous.