Consider $f: N \rightarrow N$,$g: N \rightarrow N$,and $h: N \rightarrow R$ defined as $f(x) = 2x$,$g(y) = 3y + 4$,and $h(z) = \sin z$,$\forall x, y, z \in N$. Show that $h \circ (g \circ f) = (h \circ g) \circ f$.
(N/A) We have:
$h \circ (g \circ f)(x) = h(g(f(x)))$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Also,$((h \circ g) \circ f)(x) = (h \circ g)(f(x))$
$= (h \circ g)(2x)$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Since both expressions yield the same result,it is shown that $h \circ (g \circ f) = (h \circ g) \circ f$.
This result holds true in general for the composition of functions.
$h \circ (g \circ f)(x) = h(g(f(x)))$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Also,$((h \circ g) \circ f)(x) = (h \circ g)(f(x))$
$= (h \circ g)(2x)$
$= h(g(2x))$
$= h(3(2x) + 4)$
$= h(6x + 4)$
$= \sin(6x + 4), \forall x \in N$
Since both expressions yield the same result,it is shown that $h \circ (g \circ f) = (h \circ g) \circ f$.
This result holds true in general for the composition of functions.
Explore More
Similar Questions
Let $D = \mathbb{R} - \{0, 1\}$ and $f: D \rightarrow D$,$g: D \rightarrow D$,and $h: D \rightarrow D$ be three functions defined by $f(x) = \frac{1}{x}$,$g(x) = 1 - x$,and $h(x) = \frac{1}{1 - x}$. If $j: D \rightarrow D$ is such that $(g \circ j \circ f)(x) = f(x)$ for all $x \in D$,then which one of the following is $j(x)$?
Let $f$ and $g$ be functions defined by $f(x) = \frac{x}{x + 1}$ and $g(x) = \frac{x}{1 - x}$. Then $(fog)(x)$ is
Easy
View SolutionLet $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that $f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$ and $g(n) = n - (-1)^n$. Then $fog$ is
DifficultJEE MAIN 2019
View SolutionLet $f: \{2, 3, 4, 5\} \rightarrow \{3, 4, 5, 9\}$ 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 $g \circ f$.
Easy
View SolutionLet $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ and the composed mapping $g \circ f: S \rightarrow U$ be defined. If $g \circ f$ is an injective mapping, then:
MediumWBJEE 2022
View SolutionVedclass 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