Join Sarthaks eConnect Today - Largest Online Education Community!
0 votes
in Sets, Relations and Functions by (37.1k points)
closed by

Let f: R → R: f(x) = x2 and g: R → R: g(x) = (x + 1). Show that (g o f) ≠ (f o g).

1 Answer

+1 vote
by (20.3k points)
selected by
Best answer

It is given that

f(x) = x2 and g(x) = (x + 1)

We know that LHS

gof = gof (x) = g{f(x)} = g{ x2} = x2 + 1

We know that RHS

fog = fog (x) = f{g(x)} = f{x+1} = (x+1)2 = x2 + 2x + 1

Thus, gof ≠ fog

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.