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# Composite Functions

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 Sub Topics A simple process to combine more than two functions with the help of addition, subtraction, multiplication and division known as composite function. Composite function is the result of one function in the form of other function. This is expressed by the small circle. Suppose that there are two functions ‘f’ and ‘g’, then the composite function ‘f’ to ‘g’ is denoted as (g o f) (x) or sometimes as g (f (x)). This cannot be denoted by a simple dot other wise the meaning would be the product in spite of composition. If g (x) is given in the problem then composite functions is denoted as f (g(x)) = f o g (x). Here it should be noticed that f (g(x)) is not equal to the g (f (x)).

## Limits of Composite Functions

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Limit of a composition is the composition of the limits provided the outside function is continuous at the limit of the inside function.

If lim g(x) = l and f is continuous at l; it follows that,

lim f [g(x)] = f [lim g(x)]

Here f and g are real functions and f is continuous.

Given below are the steps to be followed for finding the limit of a composite function.
1. If a composite function is of the form fog, then put the value of g in the function f.
2. Solve for finding the function f.
3. To get the result put the limit on the resultant function.
General form of limit on the composite function is lim f[g(x)] = f [lim g(x) ]

## Domain and Range of Composite Functions

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In a composite function y value of the inner function becomes the x value of the outer function. Since range of the inner function restricts the domain of the outer function.

In composite function it is not possible to simply look at the function and determine its domain and range. Domain of a composed function is either the same as the domain of the first function, or else lies inside it.

Range of a composed function is either the same as the range of the second function, or else lies inside it. It is determined by subbing its domain restrictions into the outer function.

If a composite function is of the form fog then the range of first function g(x) becomes the domain of second function f(x).

## Composition of Inverse Functions

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A function inverse can be thought of as the reversal of whatever our function does to its input. Composition of two inverse function results in identity function. Inverse of inverse of a function is that function itself
where (f o f$^{-1}$)(x) = (f$^{-1}$ o f)(x)

Steps to solve inverse function is summarized below.
• On both sides of the equation replace x with f$^{-1}$(x).
• Substitute f(f$^{-1}$ (x)) = x
• Solve f$^{-1}$(x) in terms of x.

## Derivative of Composite Function

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Chain rule to find the derivatives of composite functions.

Consider two functions y = f(u) and u = g(x), g(x) represents the sub function of f(u), then the rule is defined as
$\frac{dy}{dx}$ = $\frac{dy}{du}$ $\times$ $\frac{du}{dx}$.

This is applied to those functions which are composition of each other and the fractions as we get back to $\frac{dy}{dx}$.
First we have to find the derivative of the outer function and then, multiply this by derivative of inner function.

## Composition of Functions Examples

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### Solved Examples

Question 1: Find the derivative of Sin 10x
Solution:

Let y = sin u
u = 10x
$\frac{dy}{dx}$ = cos u and $\frac{du}{dx}$ = 10
Hence $\frac{dy}{dx}$ = $\frac{dy}{du}$ * $\frac{du}{dx}$
= cos u * 10
= 10 cos 10x

Question 2: The two functions f and g are defined on the set of real numbers such that: f(x)= x$^{2}$ + 5 and g(x) = $\sqrt{x}$. Find fog and gof and show that fog $\neq$ gof.
Solution:

(fog)(x) = f{g(x)}
(fog)(x) = f$(\sqrt{x})$
= $(\sqrt{x})^{2}$ + 5
= x + 5
(gof)(x) = g(f(x))
(gof)(x) = g(x$^{2}$ + 5)
= $\sqrt{x^{2} + 5}$
fog $\neq$ gof