Let S = {1, 2, 3}. Determine whether the func | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) function $f: S \rightarrow S$ is invertible if and only if it is a bijection (both one-one and onto). Given $f = \{(1, 1), (2, 2), (3, 3)\}$,we ob

Vedclass · Accepted Answer

$f$ is invertible and $f^{-1} = \{(1, 1), (2, 2), (3, 3)\}$.

Vedclass · Answer

(B) function $f: S \rightarrow S$ is invertible if and only if it is a bijection (both one-one and onto). Given $f = \{(1, 1), (2, 2), (3, 3)\}$,we observe that each element in the domain $S$ is mapped to a unique element in the codomain $S$,and every element in the codomain has a pre-image. Thus,$f$ is one-one and onto. Since $f$ is a bijection,it is invertible. The inverse function $f^{-1}$ is obtained by interchanging the elements of the ordered pairs $(x, y)$ to $(y, x)$. Therefore,$f^{-1} = \{(1, 1), (2, 2), (3, 3)\} = f$.

Let $S = \{1, 2, 3\}$. Determine whether the function $f: S \rightarrow S$ defined as below has an inverse. Find $f^{-1}$,if it exists: $f = \{(1, 1), (2, 2), (3, 3)\}$.

Similar Questions

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to

Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x},$ where $b$ is a constant such that $0 < b < 1$. Then

If $f: R \rightarrow R$ is a mapping defined by $f(x)=x^{3}+5$,then $f^{-1}(x)$ is equal to

If $f:[1, \infty) \rightarrow[5, \infty)$ is given by $f(x)=3x+\frac{2}{x}$,then $f^{-1}(x)=$

If the function $f(x) = x^3 + e^{x/2}$ and $g(x) = f^{-1}(x)$,then the value of $g^{\prime}(1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module