{"id":1559,"date":"2024-05-28T18:15:53","date_gmt":"2024-05-28T18:15:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1559"},"modified":"2025-08-13T22:54:09","modified_gmt":"2025-08-13T22:54:09","slug":"combinations-and-compositions-of-functions-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/combinations-and-compositions-of-functions-learn-it-1\/","title":{"raw":"Combinations and Compositions of Functions: Learn It 1","rendered":"Combinations and Compositions of Functions: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Use algebraic operations to combine functions and create new expressions<\/li>\r\n \t<li>Build a new function by combining two or more functions together<\/li>\r\n \t<li>Calculate the output for composite functions for given values and determine the set of inputs that work for these functions<\/li>\r\n \t<li>Break down a composite function into the original functions that were combined to make it<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Combining Functions Using Algebraic Operations<\/h2>\r\nSuppose we need to add two columns of numbers that represent a husband and wife\u2019s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year\u2019s incomes and then collecting all the data in a new column.\r\n<p id=\"fs-id1165135533159\">If [latex]w\\left(y\\right)[\/latex] is the wife\u2019s income and [latex]h\\left(y\\right)[\/latex] is the husband\u2019s income in year [latex]y[\/latex], and we want [latex]T[\/latex] to represent the total income, then we can define a new function.<\/p>\r\n\r\n<div id=\"fs-id1165137641862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T\\left(y\\right)=h\\left(y\\right)+w\\left(y\\right)[\/latex]<\/div>\r\n<div>\r\n<p id=\"fs-id1165135347766\">If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write<\/p>\r\n\r\n<div id=\"fs-id1165137665765\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=h+w[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1165132957142\">Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>algebraic operations on functions<\/h3>\r\n<p id=\"fs-id1165137424116\">For two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] with real number outputs, we define new functions [latex]f+g,f-g,f\\cdot{g}[\/latex], and [latex]\\dfrac{f}{g}[\/latex] by the relations:<\/p>\r\n&nbsp;\r\n<div id=\"fs-id1165137543139\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align} \\text{Addition:} \\quad \\left(f+g\\right)\\left(x\\right) &amp;= f\\left(x\\right)+g\\left(x\\right) \\\\[2mm] \\text{Subtraction:} \\quad \\left(f-g\\right)\\left(x\\right) &amp;= f\\left(x\\right)-g\\left(x\\right) \\\\[2mm] \\text{Multiplication:} \\quad \\left(f\\cdot{g}\\right)\\left(x\\right) &amp;= f\\left(x\\right)\\cdot{g}\\left(x\\right) \\\\[2mm] \\text{Division:} \\quad \\left(\\dfrac{f}{g}\\right)\\left(x\\right) &amp;= \\dfrac{f\\left(x\\right)}{g\\left(x\\right)}, \\quad \\text{where } g(x) \\neq 0 \\end{align}[\/latex]<\/div>\r\n<\/section><section class=\"textbox example\">Find and simplify the functions<center>[latex](g-f)(x)[\/latex] and [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex],<\/center>given<center>[latex]f\\left(x\\right)=x - 1[\/latex] and [latex]g\\left(x\\right)={x}^{2}-1[\/latex].<\/center>Give the domain of your result. Are they the same function?[reveal-answer q=\"117163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"117163\"]\r\n<p id=\"fs-id1165137466263\">Begin by writing the general form, and then substitute the given functions.<\/p>\r\n\r\n<div id=\"fs-id1165135701567\" class=\"equation unnumbered\">\r\n<div id=\"fs-id1165135701567\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\left(g-f\\right)\\left(x\\right)&amp;=g\\left(x\\right)-f\\left(x\\right) \\\\[2mm] \\left(g-f\\right)\\left(x\\right)&amp;={x}^{2}-1-\\left(x - 1\\right)\\\\[2mm] \\text{ }&amp;={x}^{2}-x \\\\[2mm] \\text{ }&amp;=x\\left(x - 1\\right) \\\\[2mm]\\end{align}\\hspace{20mm}[\/latex] [latex]\\begin{align}\\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&amp;=\\frac{g\\left(x\\right)}{f\\left(x\\right)} \\\\[2mm] \\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&amp;=\\frac{{x}^{2}-1}{x - 1}\\\\[2mm] \\text{ }&amp;=\\frac{\\left(x+1\\right)\\left(x - 1\\right)}{x - 1}\\text{ where }x\\ne 1 \\\\[2mm] \\text{ }&amp;=x+1 \\end{align}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThe domain [latex](g-f)(x)[\/latex] is all real numbers and the domain of [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex] is [latex]x\\ne1[\/latex]. The functions are not the same.\r\n<p id=\"fs-id1165135351612\"><em>Note<\/em>: For [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], the condition [latex]x\\ne 1[\/latex] is necessary because when [latex]x=1[\/latex], [latex]f(x)=1-1=0[\/latex], which makes the quotient function undefined (dividing by zero).<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]19128[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]19129[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]19130[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Use algebraic operations to combine functions and create new expressions<\/li>\n<li>Build a new function by combining two or more functions together<\/li>\n<li>Calculate the output for composite functions for given values and determine the set of inputs that work for these functions<\/li>\n<li>Break down a composite function into the original functions that were combined to make it<\/li>\n<\/ul>\n<\/section>\n<h2>Combining Functions Using Algebraic Operations<\/h2>\n<p>Suppose we need to add two columns of numbers that represent a husband and wife\u2019s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year\u2019s incomes and then collecting all the data in a new column.<\/p>\n<p id=\"fs-id1165135533159\">If [latex]w\\left(y\\right)[\/latex] is the wife\u2019s income and [latex]h\\left(y\\right)[\/latex] is the husband\u2019s income in year [latex]y[\/latex], and we want [latex]T[\/latex] to represent the total income, then we can define a new function.<\/p>\n<div id=\"fs-id1165137641862\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T\\left(y\\right)=h\\left(y\\right)+w\\left(y\\right)[\/latex]<\/div>\n<div>\n<p id=\"fs-id1165135347766\">If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write<\/p>\n<div id=\"fs-id1165137665765\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]T=h+w[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1165132957142\">Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>algebraic operations on functions<\/h3>\n<p id=\"fs-id1165137424116\">For two functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] with real number outputs, we define new functions [latex]f+g,f-g,f\\cdot{g}[\/latex], and [latex]\\dfrac{f}{g}[\/latex] by the relations:<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1165137543139\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align} \\text{Addition:} \\quad \\left(f+g\\right)\\left(x\\right) &= f\\left(x\\right)+g\\left(x\\right) \\\\[2mm] \\text{Subtraction:} \\quad \\left(f-g\\right)\\left(x\\right) &= f\\left(x\\right)-g\\left(x\\right) \\\\[2mm] \\text{Multiplication:} \\quad \\left(f\\cdot{g}\\right)\\left(x\\right) &= f\\left(x\\right)\\cdot{g}\\left(x\\right) \\\\[2mm] \\text{Division:} \\quad \\left(\\dfrac{f}{g}\\right)\\left(x\\right) &= \\dfrac{f\\left(x\\right)}{g\\left(x\\right)}, \\quad \\text{where } g(x) \\neq 0 \\end{align}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\">Find and simplify the functions<\/p>\n<div style=\"text-align: center;\">[latex](g-f)(x)[\/latex] and [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex],<\/div>\n<p>given<\/p>\n<div style=\"text-align: center;\">[latex]f\\left(x\\right)=x - 1[\/latex] and [latex]g\\left(x\\right)={x}^{2}-1[\/latex].<\/div>\n<p>Give the domain of your result. Are they the same function?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q117163\">Show Solution<\/button><\/p>\n<div id=\"q117163\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137466263\">Begin by writing the general form, and then substitute the given functions.<\/p>\n<div id=\"fs-id1165135701567\" class=\"equation unnumbered\">\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}\\left(g-f\\right)\\left(x\\right)&=g\\left(x\\right)-f\\left(x\\right) \\\\[2mm] \\left(g-f\\right)\\left(x\\right)&={x}^{2}-1-\\left(x - 1\\right)\\\\[2mm] \\text{ }&={x}^{2}-x \\\\[2mm] \\text{ }&=x\\left(x - 1\\right) \\\\[2mm]\\end{align}\\hspace{20mm}[\/latex] [latex]\\begin{align}\\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&=\\frac{g\\left(x\\right)}{f\\left(x\\right)} \\\\[2mm] \\text{ }\\left(\\frac{g}{f}\\right)\\left(x\\right)&=\\frac{{x}^{2}-1}{x - 1}\\\\[2mm] \\text{ }&=\\frac{\\left(x+1\\right)\\left(x - 1\\right)}{x - 1}\\text{ where }x\\ne 1 \\\\[2mm] \\text{ }&=x+1 \\end{align}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>The domain [latex](g-f)(x)[\/latex] is all real numbers and the domain of [latex]\\left(\\dfrac{g}{f}\\right)(x)[\/latex] is [latex]x\\ne1[\/latex]. The functions are not the same.<\/p>\n<p id=\"fs-id1165135351612\"><em>Note<\/em>: For [latex]\\left(\\dfrac{g}{f}\\right)\\left(x\\right)[\/latex], the condition [latex]x\\ne 1[\/latex] is necessary because when [latex]x=1[\/latex], [latex]f(x)=1-1=0[\/latex], which makes the quotient function undefined (dividing by zero).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm19128\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=19128&theme=lumen&iframe_resize_id=ohm19128&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm19129\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=19129&theme=lumen&iframe_resize_id=ohm19129&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm19130\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=19130&theme=lumen&iframe_resize_id=ohm19130&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":5,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":142,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1559"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1559\/revisions"}],"predecessor-version":[{"id":6674,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1559\/revisions\/6674"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/142"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1559\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1559"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1559"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1559"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1559"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}