Let $f: \{2, 3, 4, 5\} \rightarrow \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \rightarrow \{7, 11, 15\}$ be functions defined as $f(2)=3, f(3)=4, f(4)=f(5)=5$ and $g(3)=g(4)=7$ and $g(5)=g(9)=11$. Find $g \circ f$.
The composition $g \circ f$ is defined as $(g \circ f)(x) = g(f(x))$.
We calculate the values for each element in the domain $\{2, 3, 4, 5\}$:
$(g \circ f)(2) = g(f(2)) = g(3) = 7$
$(g \circ f)(3) = g(f(3)) = g(4) = 7$
$(g \circ f)(4) = g(f(4)) = g(5) = 11$
$(g \circ f)(5) = g(f(5)) = g(5) = 11$
Thus,$g \circ f = \{(2, 7), (3, 7), (4, 11), (5, 11)\}$.
We calculate the values for each element in the domain $\{2, 3, 4, 5\}$:
$(g \circ f)(2) = g(f(2)) = g(3) = 7$
$(g \circ f)(3) = g(f(3)) = g(4) = 7$
$(g \circ f)(4) = g(f(4)) = g(5) = 11$
$(g \circ f)(5) = g(f(5)) = g(5) = 11$
Thus,$g \circ f = \{(2, 7), (3, 7), (4, 11), (5, 11)\}$.
Explore More
Similar Questions
Let $f: R \rightarrow R$ be defined as $f(x)=x-1$ and $g: R -\{1,-1\} \rightarrow R$ be defined as $g(x)=\frac{x^{2}}{x^{2}-1}$. Then the function $f \circ g$ is
If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then for every $x \in R, f(g(x))$ is equal to
Let $f, g: R \rightarrow R$ be defined as: $f(x)=|x-1|$ and $g(x)=\begin{cases} e^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$. Then the function $f(g(x))$ is
DifficultJEE MAIN 2024
View SolutionIf $f(x) = x^2$ and $g(x) = [x^2]$, where $[ \cdot ]$ represents the greatest integer function, then $(f \circ g)(\frac{3}{2}) + (g \circ f)(\frac{3}{2})$ is equal to ...
DifficultMHT CET 2026
View Solution$f: R \rightarrow R$ and $g:[0, \infty) \rightarrow R$ are defined by $f(x)=x^2$ and $g(x)=\sqrt{x}$. Which one of the following is not true?
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo