Continuity of Composite Functions

Understanding the continuity of composite functions is an important topic in calculus, as it deals with the behavior of functions that are formed by combining two or more functions. A composite function is created when one function is applied to the result of another function.

Definition of Composite Function

Given two functions ( f: A \rightarrow B ) and ( g: B \rightarrow C ), the composite function ( g \circ f ) (read as "g composed with f") is defined by:

[ (g \circ f)(x) = g(f(x)) ]

for all ( x ) in the domain of ( f ) such that ( f(x) ) is in the domain of ( g ).

Continuity of Functions

Before diving into the continuity of composite functions, let's recall what it means for a function to be continuous. A function ( f ) is said to be continuous at a point ( c ) if the following three conditions are met:

1. ( f(c) ) is defined.
2. The limit of ( f(x) ) as ( x ) approaches ( c ) exists.
3. The limit of ( f(x) ) as ( x ) approaches ( c ) is equal to ( f(c) ).

In mathematical terms:

[ \lim_{x \to c} f(x) = f(c) ]

A function is continuous on an interval if it is continuous at every point in that interval.

Continuity of Composite Functions

The continuity of composite functions can be understood through the following theorem:

Theorem: If ( f ) is continuous at ( c ) and ( g ) is continuous at ( f(c) ), then the composite function ( g \circ f ) is continuous at ( c ).

Proof of Theorem

To prove that ( g \circ f ) is continuous at ( c ), we need to show that:

[ \lim_{x \to c} (g \circ f)(x) = (g \circ f)(c) ]

Given that ( f ) is continuous at ( c ), we have:

[ \lim_{x \to c} f(x) = f(c) ]

Since ( g ) is continuous at ( f(c) ), we can apply the limit to ( g ):

[ \lim_{x \to c} g(f(x)) = g(\lim_{x \to c} f(x)) = g(f(c)) = (g \circ f)(c) ]

Thus, ( g \circ f ) is continuous at ( c ).

Table of Differences and Important Points

Property	Function ( f )	Function ( g )	Composite Function ( g \circ f )
Definition	Maps from set ( A ) to set ( B )	Maps from set ( B ) to set ( C )	Maps from set ( A ) to set ( C ) by applying ( g ) to the result of ( f )
Continuity at a Point ( c )	Must satisfy ( \lim_{x \to c} f(x) = f(c) )	Must satisfy ( \lim_{y \to f(c)} g(y) = g(f(c)) )	Must satisfy ( \lim_{x \to c} (g \circ f)(x) = (g \circ f)(c) )
Continuity on an Interval	Continuous at every point in the interval	Continuous at every point in the interval	Continuous at every point in the interval if ( f ) and ( g ) are continuous

Examples

Example 1: Continuity of Composite Functions

Let ( f(x) = x^2 ) and ( g(x) = \sqrt{x} ). Determine if the composite function ( g \circ f ) is continuous everywhere.

Solution:

Function ( f ) is continuous everywhere because it is a polynomial. Function ( g ) is continuous for all ( x \geq 0 ) because it is a square root function.

The composite function ( (g \circ f)(x) = g(f(x)) = \sqrt{x^2} = |x| ).

The function ( |x| ) is continuous everywhere, hence ( g \circ f ) is continuous everywhere.

Example 2: Discontinuity of Composite Functions

Let ( f(x) = \begin{cases} x + 1 & \text{if } x < 0 \ x^2 & \text{if } x \geq 0 \end{cases} )

and ( g(x) = \frac{1}{x} ). Is the composite function ( g \circ f ) continuous at ( x = 0 )?

Solution:

Function ( f ) is continuous at ( x = 0 ) because both ( \lim_{x \to 0^-} f(x) = 1 ) and ( \lim_{x \to 0^+} f(x) = 0 ) are equal to ( f(0) = 0 ).

However, function ( g ) is not continuous at ( x = 0 ) because ( g(x) = \frac{1}{x} ) is not defined at ( x = 0 ).

Therefore, the composite function ( (g \circ f)(x) = g(f(x)) ) is not continuous at ( x = 0 ) because ( g ) is not continuous at ( f(0) = 0 ).

In conclusion, the continuity of composite functions depends on the continuity of the individual functions at the relevant points. It is important to check the continuity of each function separately before concluding the continuity of the composite function.