Definition :
Let f : A -> B and g : B -> C be two functions. Then a function g o f : A -> C defined by (g o f)(x) = g[f(x)], for all x ∈ A is called the composition of f and g.
Note : :
It should be noted that g o f exits if the range of f is a subset of g. Similarly, f o g exists if range of g is a subset of domain f.
Question 1 :
Let f : [2, 3, 4, 5] -> [3, 4, 5, 9] and g : [3, 4, 5, 9] -> [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 o f.
Answer :
Write the function f as a set of ordered pairs.
f = {(2, 3), (3, 4), (4, 5), (5, 5)}
Write the function g as a set of ordered pairs.
g = {(3, 7), (4, 7), (5, 11), (9, 11)}
g o f can be denoted as
(g o f) (2) = 7
(g o f) (3) = 7
(g o f) (4) = 11
(g o f) (5) = 11
Therefore, g o f = {(2, 7), (3, 7), (4, 11), (5, 11)}.
Question 2 :
Let f : {1, 3, 4} -> {1, 2, 5} and g : {1, 2, 5} -> {1, 3} be given by f = {(1, 2) (3, 5) (4, 1)} and g = {(1, 3) (2, 3) (5, 1)}. Find g o f.
Answer :
f = {(1, 2) (3, 5) (4, 1)}
g = {(1, 3) (2, 3) (5, 1)}
g o f can be denoted as
(g o f) (1) = 3
(g o f) (3) = 1
(g o f) (4) = 3
Therefore, g o f = {(1, 3), (3, 1), (4, 3)}.
Question 3 :
Let f = {(3, 1), (9, 3), (12, 4)} and g = {(1, 3), (3, 3), (4, 9), (5, 9)}. Show that g o f and f o g are defined. Also find f o g and g o f.
Answer :
f = {(3, 1), (9, 3), (12, 4)}
Domain of f = {3, 9, 12} and Range of f = {1, 3, 4}
g = {(1, 3), (3, 3), (4, 9), (5, 9)}
Domain of g = {1, 3, 4, 5} and Range of g = {3, 9}
If f o g is defined, then range of g must be a subset of domain of f. {3, 9} is a subset of {3, 9, 12}.
So, f o g is defined.
If g o f is defined, then range of f must be a subset of domain of g = {1, 3, 4} is a subset of {1, 3, 4, 5}.
So, g o f is defined.
g o f :
(g o f) (3) = 3
(g o f) (9) = 3
(g o f) (12) = 9
Therefore, g o f = {(3, 3), (9, 3), (12, 9)}.
f o g :
(f o g) (1) = 1
(f o g) (3) = 1
(f o g) (4) = 3
(f o g) (5) = 3
Therefore, f o g = {(1, 1), (3, 1), (4, 3), (5, 3)}.
Question 4 :
Let f = {1, -1), (4, -2), (9, -3), (16, 4)} and g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}. Show that g o f is defined while f o g is not defined. Also, find g o f.
Answer :
f = {1, -1), (4, -2), (9, -3), (16, 4)}
Domain of f = {1, 4, 9, 16} and Range of f = {-1, -2, -3, 4}
g = {(-1, -2), (-2, -4), (-3, -6), (4, 8)}.
Domain of g = {-1, -2, -3, 4} and Range of g = {-2, -4, -6, 8}
Since range of f is a subset of domain of g, g o f is defined.
Since range of g is not a subset of domain of f, f o g is not defined.
g o f :
(g o f) (1) = -2
(g o f) (4) = -4
(g o f) (9) = -6
(g o f) (16) = 8
Therefore, f o g = {(1, -2), (4, -4), (9, -6), (16, 8)}.
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 02, 24 01:19 AM
Dec 02, 24 01:14 AM
Nov 30, 24 07:08 AM