Composition of Functions Composition of Functions Suppose we have a function f: (P) ?Q and other function g: (Q) ?R, it is composed by putting the output values of g when it has an argument value of f (P) instead of P, suppose R is a function g of Q and Q is a function f of P, then we can say that R is the function of P. And the Composition of Functions is said to be associative. If we have three function P, Q, R so we can write in the function form as: P.(Q.R) = (P.Q).R, the values in the parenthesis which indicate the composition is solved first for the parenthesis function. Now we see types of function: Some types of function are given below: 1. One â€“ to â€“ one function; 2. onto function; Now we will see some of the introduction about the function:

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One – to – one function: - A function is said to be one – to – one function if every element of domain has one unique pre image in range. For example: suppose we have domain value (2, 6, 9, 5) and the range value is (1, 3, 2, 5, 6, 8, 9); Here all the domain values (2, 6, 9, 5) are present in the range value (1, 3, 2, 5, 6, 8, 9), so the given function is one – to – one function. So these given values follow above definition so the function is one – to – one. Now we will see onto function: If two or more than two values of domain have same element in the range then the function is said to be onto function. Suppose we have domain values (P, Q, R, S) and the range values (2, 8, 9); Here all the domain values have more than two pre image in the range values. So the given function is onto function. We can also plot the graph of functions with the help of function values: We have some of the steps which are given below: Step1: First we have some function value. Step2: Then we know that the values present in the domain are plotted towards the x – axis. Step3: And range value are plotted towards the y – axis.

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Step4: Al last me join all the points in a graph. By using above steps we can plot the graph of function. Example 2: The two functions f, g : R â†’ R are defined by f(x) = x2 + 1, g(x) = x - 1. Find fog and gof and show that fog ≠ gof. Solution: (fog) (x) = f(g(x)) = f(x - 1) = (x - 1)2 + 1 = x2 - 2x + 2 (gof) (x) = g(f(x)) = g(x2 + 1) = (x2 + 1) - 1 = x2 Thus (fog) (x) = x2 - 2x + 2 (gof) (x) = x2 fog ≠ gof Composition of Functions Practice ProblemsBack to Top Problem 1: Let f, g: R → R be defined by f(x) = $\frac{(x – 1)}{2}$, and g(x) =2x-1. Show that (fog) ≠ (gof). Problem 2: Let f: R → R be a function defined by f(x) = 2x + 1. Find f -1.

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Composition of Functions

Published on Jun 25, 2012

Suppose we have a function f: (P) ?Q and other function g: (Q) ?R, it is composed by putting the output values of g when it has an argument...

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