Exams Prep Master
Before understanding the inverse function formulae, here is a brief about what inverse functions are. When two functions are combined, basic or multiplied together, it might result in the formation of an identity function. In this case, one function is the inverse of the other. The inverse of any function f is denoted by f-1. This concept only exists for one-one and onto functions. It must be remembered that f-1 is not the reciprocal of the function f. There might be cases where the inverse of a function does not exist and in those scenarios, it is deemed as uninvertible. Here, we will be learning more about inverse function formula and discussing some sample questions.
| Table of Content |
Keyterms: Inverse Fuction, Functions, Domain, Graph, Symmetry, Line
Also read: Isosceles Triangle Theorems
What are Inverse Functions?
[Click Here for Sample Questions]
Inverse functions can be understood with a very basic mathematical idea. It says that if (x,y) is a point on the graph of a function f, then (y,x) is a point on f-1, provided f is invertible. It needs to be kept in mind that the domain of both functions needs to be considered. The domain of f should be the range of f-1 and vice versa should also be true. This can be further justified with an example.
If f: Aà B (where f is one-one and onto: invertible)
Then, f-1: B à A
Where: A is the domain of f and B is its range.
The formula of inverse functions also states that f and f-1 are only inverses of one another if their composition is x.
(f o f)(x) = (f-1 o f)(x) = x
Also read: Types of Functions
How to Calculate Inverse Function?
[Click Here for Sample Questions]
It is easy to find the inverse of a function with the simple replacement of y and x, then carrying further calculations. The concrete procedure has been stated below:
- X and y first need to be interchanged.
- Then we need to solve for y.
- Then we must replace y with f-1(x).
This ultimately gives us the final answer.
Also read: Calculus Formula
Identifying Inverse Functions from a Graph
[Click Here for Sample Questions]
Graphs can be used as an interesting method to check for the inverse of a function. If the graphs of two functions are given to us, it can be easily identified if they are inverses of one another or not. We need to check for symmetry about the line y = x. If the graphs are symmetric about this line, it means that they have existing inverses, or that they are invertible. This happens because of the idea that if (x,y) lies on the graph of function f, then (y,x) has to lie on its inverse f-1. This shows that f and f-1 are symmetric about this line as well.
Another interesting formula is,
If (fog)(x) = (gof)(x) = x
Then, f = g-1 and g = f-1
Also read: Circumference of a circle
Things to Remember
- This concept of inverse functions only exists for one-one and onto functions. It must be remembered that f-1 is not the reciprocal of the function f.
- Inverse functions can be understood with a very basic mathematical idea. It says that if (x,y) is a point on the graph of a function f, then (y,x) is a point on f-1, provided f is invertible.
- Graphs can be used as a method to check for the inverse of a function. We need to check for symmetry about the line y = x. If the graphs are symmetric about this line, it means that they have existing inverses, or that they are invertible.
- Another interesting formula is, if (fog)(x) = (gof)(x) = x, then, f = g-1 and g = f-1
Mathematics Related links:
Sample Questions
Ques: Verify whether f(x) = 3x + 2 and g(x) = (x-2)/3 are inverse of each other using the inverse function formulae. (3 marks)
Ans: We can solve this by verifying if ( f o g )(x) = ( g o f)(x) = x
Now, we know that ( f o g )(x) = f (g (x)
f ((x-2)/3)
= 3 ((x-2)/3) + 2
= x - 2 + 2
= x
We also know that (g o f)(x) = g (f(x))
= g (3x + 2)
= (3x + 2 - 2)/3
= 3x/3
= x
Hence, proved.
Ques: Find the inverse of the function f(x) = (2x+1)/(x+1). (4 marks)
Ans: The following steps need to be followed:
- X and y first need to be interchanged.
- Then we need to solve for y.
- Then we must replace y with f-1(x).
This ultimately gives us the final answer.
The given function is:
y = (2x+1)/(x+1)
Interchange x and y.
x = (2y+1)/(y+1)
Now we will solve this for y.
Multiplying both sides by (y + 1),
x (y + 1) = 2y + 1
xy + x = 2y + 1
xy - 2y = 1 - x
y (x - 2) = 1 - x
y = 1−xx−21−xx−2
Replace y with f-1(x).
f-1(x) = 1−x/ x−2
Ques: Find out the inverse function of f(x) = 2x + 3 ? (3 marks)
Ans: Given function is,
f(x) = y = 2x + 3
Inverse function equation is f-1(y) = x
So, x can be found out from the above expression.
2x = y – 3
X =y−3/2
So, f-1 = y−3/2
Ques: Prove that the function f : R → R, given by f(x) = 2x, is one-one and onto. (3 marks)
Ans: f is one-one, as f(x1 ) = f(x2 ) ⇒ 2x1 = 2x2 ⇒ x1 = x2 .
Also, given any real number y in R, there exists 2 y in R such that f(y/2 ) = 2 . (y/2 ) = y.
Hence, f is onto.
Ques: Show that the function f: N à N, given by f(1) = f(2) = 1 and f(x) = x-1 for every x > 2, is onto but not one-one. (1 mark)
Ans: f is not one-one since it has the same value at two points, which are x = 1,2. However, f is onto, as given any y is any natural number, and not equal to 1, we can choose x as y+1 such that f(y+1) = y + 1 -1 = 1. Also, for 1, f(1) = 1.
Ques: Show that f: N à N, given by: (4 marks)
x + 1 (if x is odd)
x – 1 (if x is even)
is both one- one and onto.
Ans: Suppose f(x1 ) = f(x2 ). Note that if x1 is odd and x2 is even, then we will have x1 + 1 = x2 – 1, i.e., x2 – x1 = 2 which is impossible. Similarly, the possibility of x1 being even and x2 being odd can also be ruled out, using the similar argument. Therefore, both x1 and x2 must be either odd or even. Suppose both x1 and x2 are odd. Then f(x1 ) = f(x2 ) ⇒ x1 + 1 = x2 + 1 ⇒ x1 = x2 . Similarly, if both x1 and x2 are even, then also f(x1 ) = f(x2 ) ⇒ x1 – 1 = x2 – 1 ⇒ x1 = x2 . Thus, f is one-one. Also, any odd number 2r + 1 in the co-domain N is the image of 2r + 2 in the domain N and any even number 2r in the co-domain N is the image of 2r – 1 in the domain N. Thus, f is onto.
Ques: Show that-a is the inverse of a for the addition operation ‘+’ on R and 1/a is the inverse of a not equal to 0 for the multiplication operation ‘x’ on R. (3 marks)
Ans: As a + (– a) = a – a = 0 and (– a) + a = 0, – a is the inverse of a for addition. Similarly, for a ≠ 0, a × 1 a = 1 = 1 a × a implies that 1 a is the inverse of a for multiplication.
Ques: Show that – a is not the inverse of a ∈ N for the addition operation + on N and 1 a is not the inverse of a ∈ N for multiplication operation × on N, for a ≠ 1.(1 mark)
Ans: Since – a ∉ N, – a cannot be inverse of a for addition operation on N, although – a satisfies a + (– a) = 0 = (– a) + a. Similarly, for a ≠ 1 in N, 1 a ∉ N, which implies that other than 1 no element of N has inverse for multiplication operation on N.
Ques: Consider f : {1, 2, 3} → {a, b, c} and g : {a, b, c} → {apple, ball, cat} defined as f(1) = a, f(2) = b, f(3) = c, g(a) = apple, g(b) = ball and g(c) = cat. Show that f, g and gof are invertible. Find out f –1, g–1 and (gof)–1 and show that (gof) –1 = f –1o g–1. (3 marks)
Ans: Note that, f and g are bijective functions. Let f –1: {a, b, c} → (1, 2, 3} and g–1 : {apple, ball, cat} → {a, b, c} be defined as f –1{a} = 1, f –1{b} = 2, f –1{c} = 3, g –1{apple} = a, g –1{ball} = b and g –1{cat} = c. It is easy to verify that f –1o f = I {1, 2, 3}, f o f –1 = I {a, b, c} , g –1og = I {a, b, c} and g o g–1 = I D, where D = {apple, ball, cat}. Now, gof : {1, 2, 3} → {apple, ball, cat} is given by gof(1) = apple, gof(2) = ball, gof(3) = cat. We can define (gof) –1 : {apple, ball, cat} → {1, 2, 3} by (gof) –1 (apple) = 1,(gof) –1 (ball) = 2 and (gof)–1 (cat) = 3. It is easy to see that (g o f) –1 o (g o f) = I{1, 2, 3} and (gof) o (gof) –1 = ID. Thus, we have seen that f, g and gof are invertible. Now, f –1og–1 (apple)= f –1(g–1(apple)) = f –1(a) = 1 = (gof)–1 (apple) f –1og–1 (ball) = f –1(g–1(ball)) = f –1(b) = 2 = (gof) –1 (ball) and f –1og–1 (cat) = f –1(g–1(cat)) = f –1(c) = 3 = (gof)–1 (cat). Hence (gof) –1 = f –1og–1. The above result is true in general situation also
Ques: Let S = {1, 2, 3}. Determine whether the functions f: S → S defined as below have inverses. Find f –1, if it exists. (3 marks)
(a) f = {(1, 1), (2, 2), (3, 3)}
(b) f = {(1, 2), (2, 1), (3, 1)}
(c) f = {(1, 3), (3, 2), (2, 1)}
Ans: (a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse f –1 of f given by f –1 = {(1, 1), (2, 2), (3, 3)} = f.
(b) Since f (2) = f (3) = 1, f is not one-one so that f is not invertible.
(c) It is easy to see that f is one-one and onto so that f is invertible with f –1 = {(3, 1), (2, 3), (1, 2)}.
For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates
Check-Out:
Comments