Inverse Image and Inverse Function

Table of Contents

  1. Inverse Image of an Element
  2. Inverse Function

Inverse Image of an Element

Consider a function ƒ: A➞B. The inverse image of an element y∈B with respect to ƒ is defined as the set of elements in A which have y as their image. It is denoted by ƒ-1(y). \[ƒ^{-1}(y)=\{x∈A: y=ƒ(x)\}\] Examples:

  1. Let ƒ: A➞B be defined by the arrow diagram,
Inverse image of an element

Then, the inverse of a under ƒ \[ƒ^{-1}(a)=\{1,2\}\] and, ƒ-1(b) is a null set. \[ƒ^{-1}(c)=3\]

2. Let a function ƒ: R➞R be defined by ƒ(x)=x2. Then, \[ƒ^{-1}(9)=\{-3,3\}\] ∵ 9 is the image of both -3 and 3. Also, \[ƒ^{-1}(-1)=Φ\] ∵ There is no real number whose square is -1.

Inverse Function

Let ƒ: A➞B be a bijective function. Since ƒ is onto, each element b∈B is an image of at least one element a∈A. But, ƒ is also one-one, so a is the only (or unique) element of A corresponding to the element b∈B. Inverse function, a function from B to A, associates each element b of B with a unique element a of A. In other words, a function of the type; \[ƒ^{-1}:B\rightarrow A\] This function is known as the inverse function of ƒ.

Thus, if ƒ: A➞B is one-one and onto, then ther exists a function ƒ-1: B➞A called the inverse function of ƒ and is defined as a function in which every element of B associates with a unique element of A.

Examples:

  1. Let ƒ: A➞B be one-one and onto function defined by the following diagram,
Inverse Function

2. Let a function ƒ: R➞R be defined by \[ƒ(x)=x^3\] This function is bijective. Hence, ƒ-1 exists and is defined by \[ƒ^{-1}(x)=\sqrt[3]{x}\] Q. Let a function ƒ: R➞R be defined by y=ƒ(x)=2x-3, x∈R. Find the inverse function.

Let x1, x2 ∈ R (domain). Then, \[ƒ(x_1)=2x_1-3\] \[ƒ(x_2)=2x_2-3\] Now, \[ƒ(x_1)=ƒ(x_2)\] \[\Rightarrow 2x_1-3=2x_2-3\] \[\Rightarrow x_1=x_2\] ∴ ƒ is one-one.

Again, let k ∈ R. Then, \[k=2x-3\] \[\Rightarrow x=\frac{k+3}{2}∈R\] ∴ ƒ is onto.

Since, ƒ is one-one and onto, it is a bijective function and ƒ-1 exists. We have, \[y=2x-3\] Interchanging the values of x and y, we get, \[x=2y-3\] \[y=\frac{x+3}{2}\] \[\therefore ƒ^{-1}(x)=\frac{x+3}{2}\]

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