{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

composition - Composition of functions Given two functions...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Composition of functions Given two functions f and g, the composite function h = f o g is the function that consists of the action of g followed by the action of f and it is defined by = ( ) h x ( ) f ( ) g x , that is, ( f o g ) = ( ) x ( ) f ( ) g x . The domain of f o g is the subset of the domain of g consisting of those numbers x such that ( ) g x is in the domain of f, that is, D( f o g ) = { x | x D( g ) and ( ) g x D( f ) }. We need to restrict the domain of g to ensure that all output numbers from g are in the domain of f, so that f can then be applied to them. The following diagram illustrates the situation schematically. The blue shaded region represents R( g ) D( f ). Example 1 : Let the function f be given by = ( ) f x + x 3 and g be given by = ( ) g x x 2 . Find (a) (f o g) ( ) 5 (b) (f o g) ( ) u (c) (f o g) ( ) x (d) (g o f) ( ) x . Solution : (a) (f o g) = ( ) 5 ( ) f ( ) g 5 = ( ) f 25 = 28 (b) (f o g) = ( ) u ( ) f ( ) g u = ( ) f u 2 = + u 2 3 (c) (f o g) = ( ) x ( ) f ( ) g x = ( ) f x 2 = + x 2 3 , that is, (f o g) = ( ) x + x 2 3. Note : The action of the function f is to add 3 to any input number, while the action of g is to square any input number. The action of f o g is to perform both operations in the requisite order, namely to square any input number and then add 3 to the result. (d) (g o f) = ( ) x ( ) g ( ) f x = ( ) g + x 3 = ( ) + x 3 2 , that is, (g o f) = ( ) x ( ) + x 3 2 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Example 2 : Let the function f be given by = ( ) f x x and g be given by = ( ) g x - x 4. Find (f o g) ( ) x and (g o f) ( ) x . State the domain of each of the composite functions f o g and g o f. Solution : (f o g) = ( ) x ( ) f ( ) g x = ( ) f - x 4 = - x 4 , that is, (f o g) = ( ) x - x 4. _________ (g o f) = ( ) x ( ) g ( ) f x = ( ) g x = - x 4, that is, (g o f) = ( ) x - x 4. _________ In order to find the domain of f o g first note that the domain of g is the set of all real numbers ( ) , -∞ ∞ , and the domain of f is the set of real numbers that are greater than or equal to 0, that is, D(g) = ( ) , -∞ ∞ and D(f) = { x | x > 0 } = [ , 0 ). The domain of f o g is the set of all real numbers x such that = ( ) g x - x 4 is greater than or equal to 0. Now - x 4 > 0 exactly when x > 4. Hence D(f o g) = { x | x > 4 } = [ , 4 ). ______________ The domain of g o f is the same as the domain of f, namely, D( g o f ) = [ , 0 ). _________ Example 3 : Let the function f be given by = ( ) f x 2 x and g be given by = ( ) g x x + 1 x . Find (f o g) ( ) x and (g o f) ( ) x . State the domain of each of the composite functions f o g and g o f. Solution : (f o g) = ( ) x ( ) f ( ) g x = f x + 1 x = 2 x + 1 x = 2 . + 1 x x = 2 . = + 1 x 1 + 2 x 2 that is, (f o g) = ( ) x + 2 x 2. _________ (g o f) = ( ) x ( ) g ( ) f x = g 2 x = 2 x + 1 2 x = 2 + x 2 .
Image of page 2
The last equality follows by multiplying top and bottom of 2 x + 1 2 x by x , but there are other ways to achieve the same result. For example, = 2 x + 1 2 x 2 x + x 2 x = 2 x . = x + x 2 2 + x 2 . We should perhaps note that this simplification is not valid when = x 0. We have (g o f) = ( ) x 2 + x 2 .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern