Detailed solutions to Composition of Functions Questions are presented.
Solution to Question 1: The composition (f o g) (x) of f(x) = x + 1 , g(x) = 3x is given by (f o g) (x) = f(g(x)) = g(x) + 1 = 3x + 1 Since the domain of both functions is the set of all real numbers, the composition (f o g) (x) also have the set of all real numbers as its domain.
Solution to Question 2: The composition (f o g) (x) of f(x) = x 2 + 1 , g(x) = sqrt(2 x) is given by (f o g) (x) = f(g(x)) = g(x) 2 + 1 = [ sqrt(2 x) ] 2 + 1 = 2 x + 1 To find the domain of the composition of the two functions, we proceed as follows: x must be in the domain of g. 2 x > = 0 which is equivalent to x > = 0 g(x) must be in the domain of f(x). The domain of f is the set of all real numbers. Th condition x > = 0 makes g(x) real and is therefore in the domain of f. Hence the domain of the composition is the set of all values defined by the interval [0 , +infinity)
Solution to Question 3: The composition (f o g) (x) of f(x) = sqrt(-x + 1) , g(x) = x 2 - 8 is given by (f o g) (x) = f(g(x)) = sqrt(-g(x) + 1) = sqrt(-(x 2 - 8) + 1) = [ sqrt(9 - x 2) ] The domain of g is the set of real numbers. Hence to find the domain of the composition, x must satisfy the condition 9 - x 2 > = 0 The solution set of the above inequality is also the domain given by the interval [-3 , 3]
Solution to Question 4: Given f(x) = 2x + 1 and g(x) = x 2 , (f o g) (-2) is calculated as follows (f o g) (-2) = f(g(-2)) = f(4) = 9
Solution to Question 5: Given f(x) = 1 / (x + 1) and g(x) = 1 / (x - 1) , (f o g) (0) is calculated as follows (f o g) (0) = f(g(0)) = f(-1) = undefined since x = -1 makes the denominator of f equal to zero.
SEARCH THIS SITE
Copyright © 2003 - 2011 - All rights reserved - A Dendane