Composition of a function and its inverse are two mathematical concepts with practical applicability. The objective of these two concepts is to increase the understanding of functions and all the terms related to it. Students will better understand the definitions of domain and range after going through these two concepts.
The objective of the composition of functions and inverse of a function is to develop an application based thinking of how the functions work. Both of these concepts have a real-life application. Students are advised to regularly give time and effort to mathematics and increase their score in it.
Let us try and understand both of these mathematical concepts in detail. Composition of functions and inverse functions are easy concepts to understand and apply. Students are advised to try as many examples as possible to solidify their learning and understanding of both the concepts. Given below is the detailed explanation of both the concepts:
Composition of function is defined when the result of a function is obtained by applying another function. The independent variable is another function. Let us try to understand the composition of functions with the help of an example.
Let there be two functions, f(x) and g(x).
f(x) = 2x + 1 g(x)=x2
Let us find the value of g(x) with the result obtained from f(x).
Calculate f(x) at 1.
Let us calculate g(x) at 3.
To streamline the above process and understand it better, we create a new function. This is how we represent the composition of functions.
Where o is the composition operator and is used to define the composition of functions. Let us try and solve the above problem with this representation at x=1.
=4x2+ 4x + 1
The notation (fog) is read as f of g or f is composed of g.
Inverse functions, as the name suggests, is to describe an inverse relationship between two functions. The two functions are opposite of each other. Let us try and understand this concept using a common example.
Let us take the case of temperature scales. Two scales, degree Celsius scale and the Fahrenheit scale, are used to measuring temperature.
C(x)=5/9(x-32) converts fahrenheit to degree celsius.
F(x)=9/5(x) + 32 converts degree celsius into degree fahrenheit.
We wish to convert 77oF into degree celsius.
Now to understand the inverse function, let us convert this degree celsius into degree fahrenheit.
We get the same Fahrenheit value we began with. Inverse functions are opposite of each other. We can obtain any of the functions.
C(x) = 5/9(x-32)
9/5(C(x)) = x-32
9/5(x) + 32 = F(x)
This is how the inverse functions work.
1. f(x)=x2 - x + 3 g(x)= 2x - 1.Solve (a) (fog)(x) (b) (gof)(x).
We have been given two functions f(x) and g(x). These are composition function problems. Composition functions are defined as functions whose result is obtained from the results of another function.
This is a composition function. We read it as f of g. We have to substitute g into the function f and obtain the result.
This is a composition function. We read it as g of f. We have to substitute f into the function g and obtain the result.
2. Find the inverse of f(x)=x4,xR.
The function f(x)=x4is a one-one and onto function. It is defined in the range R. So the inverse of the function exists. To find the inverse of the function,
We can write this function as,
Now we solve this equation for x. We get,
We write this as a new function g,
This function is the inverse of the above function as the domain is the same as the range of the function. So we can write this as the function,
It can also be written as (f-1(x)), which is the inverse of the function.
3. Explain the Composition of inverse trigonometric functions.
According to the Composition of inverse functions we know that,
f(f-1(x))=x for all values of x for which f inverse is defined.
Applying this property to trigonometric functions, we obtain
These are also inverse composite functions and are only restricted to true values of x in its domain. It’s not always that only the same composition of function exists. We can learn about the composition of inverse trigonometric functions in detail using some examples.
It is true because (sin-1(√2/2) = π/4
It is true because tan-1(√3= π/3 and Cos π/3=½
So it is possible to have a composition of a trigonometric function with a different trigonometric function.