{"id":908,"date":"2025-06-20T17:21:31","date_gmt":"2025-06-20T17:21:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=908"},"modified":"2025-08-28T13:22:16","modified_gmt":"2025-08-28T13:22:16","slug":"sequences-and-series-foundations-background-youll-need-1-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-series-foundations-background-youll-need-1-2\/","title":{"raw":"Power Series and Applications: Background You'll Need 1","rendered":"Power Series and Applications: Background You&#8217;ll Need 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve inequalities that include absolute values<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Absolute Value Inequalities<\/h2>\r\nAn <strong>absolute value inequality<\/strong> is an equation of the form\r\n<div style=\"text-align: center;\">[latex]|A| &lt; B,|A|\\le B,|A| &gt; B,\\text{or }|A|\\ge B[\/latex],<\/div>\r\nwhere [latex]A[\/latex], and sometimes [latex]B[\/latex], represents an algebraic expression dependent on a variable [latex]x[\/latex]<em>. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] <em>-<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>absolute value inequality<\/h3>\r\nFor an algebraic expression [latex]X[\/latex]<em>\u00a0<\/em>and [latex]k&gt;0[\/latex], an <strong>absolute value inequality<\/strong> is an inequality of the form:\r\n<div><\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c} |X| &lt; k \\text{ is equivalent to } -k &lt; X &lt; k \\\\ \\text{or} \\\\ |X| &gt; k \\text{ is equivalent to } X &lt; -k \\text{ or } X &gt; k \\\\ \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\r\nThese statements also apply to [latex]|X|\\le k[\/latex] and [latex]|X|\\ge k[\/latex].\r\n\r\n<\/section>There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Suppose we want to know all possible returns on an investment if we could earn some amount of money within [latex]$200[\/latex] of [latex]$600[\/latex].\r\n\r\n<hr \/>\r\n\r\nWe can solve algebraically for the set of [latex]x-[\/latex]values such that the distance between [latex]x[\/latex] and [latex]600 [\/latex] is less than [latex]200[\/latex]. We represent the distance between [latex]x[\/latex] and [latex]600 [\/latex] as [latex]|x - 600|[\/latex], and therefore,\r\n\r\n<center>[latex]|x - 600|\\le 200[\/latex]<\/center><center>or<\/center>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\r\nThis means our returns would be between [latex]$400[\/latex] and [latex]$800[\/latex].\r\n\r\n<\/section>To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.\r\n\r\n<section class=\"textbox example\">Describe all values [latex]x[\/latex] within a distance of [latex]4[\/latex] from the number [latex]5[\/latex].\r\n\r\n<hr \/>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225903\/CNX_Precalc_Figure_01_06_002.jpg\" alt=\"A number line with one tick mark in the center labeled: 5. The tick marks on either side of the center one are not marked. Arrows extend from the center tick mark to the outer tick marks, both are labeled 4.\" width=\"487\" height=\"81\" \/> Absolute value shows distance, so both directions from 5 are included[\/caption]\r\n\r\nWe want the distance between [latex]x[\/latex] and [latex]5[\/latex] to be less than or equal to [latex]4[\/latex]. We can draw a number line to represent the condition to be satisfied.The distance from [latex]x[\/latex] to [latex]5[\/latex] can be represented using an absolute value symbol, [latex]|x - 5|[\/latex]. Write the values of [latex]x[\/latex] that satisfy the condition as an absolute value inequality.\r\n<div style=\"text-align: center;\">[latex]|x - 5|\\le 4[\/latex]<\/div>\r\nWe need to write two inequalities as there are always two solutions to an absolute value equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}x - 5\\le 4\\hfill &amp; \\text{and}\\hfill &amp; x - 5\\ge -4\\hfill \\\\ x\\le 9\\hfill &amp; \\hfill &amp; x\\ge 1\\hfill \\end{array}[\/latex]<\/div>\r\nIf the solution set is [latex]x\\le 9[\/latex] and [latex]x\\ge 1[\/latex], then the solution set is an interval including all real numbers between and including [latex]1[\/latex] and [latex]9[\/latex].\r\n\r\nSo, [latex]|x - 5|\\le 4[\/latex] is equivalent to [latex]\\left[1,9\\right][\/latex] in interval notation.\r\n\r\n<\/section><section class=\"textbox example\">Solve the following:<center>[latex]|x - 1|\\le 3[\/latex]<\/center>[reveal-answer q=\"4865\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"4865\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}|x - 1|\\le 3\\hfill \\\\ \\hfill \\\\ -3\\le x - 1\\le 3\\hfill \\\\ \\hfill \\\\ -2\\le x\\le 4\\hfill \\\\ \\hfill \\\\ \\left[-2,4\\right]\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div>[\/hidden-answer]<\/div>\r\n<\/section><section class=\"textbox example\">Given the equation:<center>[latex]y=-\\frac{1}{2}|4x - 5|+3[\/latex],<\/center>determine the [latex]x[\/latex]-values for which the [latex]y[\/latex]-values are negative.[reveal-answer q=\"624558\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624558\"]We are trying to determine where [latex]y&lt;0[\/latex] which is when [latex]-\\frac{1}{2}|4x - 5|+3&lt;0[\/latex]. We begin by isolating the absolute value.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}-\\frac{1}{2}|4x - 5|&lt; -3\\hfill &amp; \\text{Multiply both sides by -2, and reverse the inequality}.\\hfill \\\\ |4x - 5|&gt; 6\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\nNext, we solve [latex]|4x - 5|=6[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}4x - 5=6\\hfill &amp; \\hfill &amp; 4x - 5=-6\\hfill \\\\ 4x=11\\hfill &amp; \\text{or}\\hfill &amp; 4x=-1\\hfill \\\\ x=\\frac{11}{4}\\hfill &amp; \\hfill &amp; x=-\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/div>\r\nNow, we can examine the graph to observe where the [latex]y[\/latex]<em>-<\/em>values are negative. We observe where the branches are below the [latex]x[\/latex]<em>-<\/em>axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=-\\frac{1}{4}[\/latex] and [latex]x=\\frac{11}{4}[\/latex] and that the graph opens downward.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225906\/CNX_CAT_Figure_02_07_006.jpg\" alt=\"A coordinate plan with the x-axis ranging from -5 to 5 and the y-axis ranging from -4 to 4. The function y = -1\/2|4x \u2013 5| + 3 is graphed. An open circle appears at the point -0.25 and an arrow\" width=\"487\" height=\"363\" \/> Graph of a V-shaped function indicating where \ud835\udc66&lt;0 on the x-axis.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18993[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve inequalities that include absolute values<\/li>\n<\/ul>\n<\/section>\n<h2>Absolute Value Inequalities<\/h2>\n<p>An <strong>absolute value inequality<\/strong> is an equation of the form<\/p>\n<div style=\"text-align: center;\">[latex]|A| < B,|A|\\le B,|A| > B,\\text{or }|A|\\ge B[\/latex],<\/div>\n<p>where [latex]A[\/latex], and sometimes [latex]B[\/latex], represents an algebraic expression dependent on a variable [latex]x[\/latex]<em>. <\/em>Solving the inequality means finding the set of all [latex]x[\/latex] <em>&#8211;<\/em>values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>absolute value inequality<\/h3>\n<p>For an algebraic expression [latex]X[\/latex]<em>\u00a0<\/em>and [latex]k>0[\/latex], an <strong>absolute value inequality<\/strong> is an inequality of the form:<\/p>\n<div><\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c} |X| < k \\text{ is equivalent to } -k < X < k \\\\ \\text{or} \\\\ |X| > k \\text{ is equivalent to } X < -k \\text{ or } X > k \\\\ \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>These statements also apply to [latex]|X|\\le k[\/latex] and [latex]|X|\\ge k[\/latex].<\/p>\n<\/section>\n<p>There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Suppose we want to know all possible returns on an investment if we could earn some amount of money within [latex]$200[\/latex] of [latex]$600[\/latex].<\/p>\n<hr \/>\n<p>We can solve algebraically for the set of [latex]x-[\/latex]values such that the distance between [latex]x[\/latex] and [latex]600[\/latex] is less than [latex]200[\/latex]. We represent the distance between [latex]x[\/latex] and [latex]600[\/latex] as [latex]|x - 600|[\/latex], and therefore,<\/p>\n<div style=\"text-align: center;\">[latex]|x - 600|\\le 200[\/latex]<\/div>\n<div style=\"text-align: center;\">or<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{c}-200\\le x - 600\\le 200\\\\ -200+600\\le x - 600+600\\le 200+600\\\\ 400\\le x\\le 800\\end{array}[\/latex]<\/div>\n<p>This means our returns would be between [latex]$400[\/latex] and [latex]$800[\/latex].<\/p>\n<\/section>\n<p>To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.<\/p>\n<section class=\"textbox example\">Describe all values [latex]x[\/latex] within a distance of [latex]4[\/latex] from the number [latex]5[\/latex].<\/p>\n<hr \/>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225903\/CNX_Precalc_Figure_01_06_002.jpg\" alt=\"A number line with one tick mark in the center labeled: 5. The tick marks on either side of the center one are not marked. Arrows extend from the center tick mark to the outer tick marks, both are labeled 4.\" width=\"487\" height=\"81\" \/><figcaption class=\"wp-caption-text\">Absolute value shows distance, so both directions from 5 are included<\/figcaption><\/figure>\n<p>We want the distance between [latex]x[\/latex] and [latex]5[\/latex] to be less than or equal to [latex]4[\/latex]. We can draw a number line to represent the condition to be satisfied.The distance from [latex]x[\/latex] to [latex]5[\/latex] can be represented using an absolute value symbol, [latex]|x - 5|[\/latex]. Write the values of [latex]x[\/latex] that satisfy the condition as an absolute value inequality.<\/p>\n<div style=\"text-align: center;\">[latex]|x - 5|\\le 4[\/latex]<\/div>\n<p>We need to write two inequalities as there are always two solutions to an absolute value equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}x - 5\\le 4\\hfill & \\text{and}\\hfill & x - 5\\ge -4\\hfill \\\\ x\\le 9\\hfill & \\hfill & x\\ge 1\\hfill \\end{array}[\/latex]<\/div>\n<p>If the solution set is [latex]x\\le 9[\/latex] and [latex]x\\ge 1[\/latex], then the solution set is an interval including all real numbers between and including [latex]1[\/latex] and [latex]9[\/latex].<\/p>\n<p>So, [latex]|x - 5|\\le 4[\/latex] is equivalent to [latex]\\left[1,9\\right][\/latex] in interval notation.<\/p>\n<\/section>\n<section class=\"textbox example\">Solve the following:<\/p>\n<div style=\"text-align: center;\">[latex]|x - 1|\\le 3[\/latex]<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q4865\">Show Solution<\/button><\/p>\n<div id=\"q4865\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}|x - 1|\\le 3\\hfill \\\\ \\hfill \\\\ -3\\le x - 1\\le 3\\hfill \\\\ \\hfill \\\\ -2\\le x\\le 4\\hfill \\\\ \\hfill \\\\ \\left[-2,4\\right]\\hfill \\end{array}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Given the equation:<\/p>\n<div style=\"text-align: center;\">[latex]y=-\\frac{1}{2}|4x - 5|+3[\/latex],<\/div>\n<p>determine the [latex]x[\/latex]-values for which the [latex]y[\/latex]-values are negative.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q624558\">Show Solution<\/button><\/p>\n<div id=\"q624558\" class=\"hidden-answer\" style=\"display: none\">We are trying to determine where [latex]y<0[\/latex] which is when [latex]-\\frac{1}{2}|4x - 5|+3<0[\/latex]. We begin by isolating the absolute value.\n\n\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}-\\frac{1}{2}|4x - 5|< -3\\hfill & \\text{Multiply both sides by -2, and reverse the inequality}.\\hfill \\\\ |4x - 5|> 6\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p>Next, we solve [latex]|4x - 5|=6[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}4x - 5=6\\hfill & \\hfill & 4x - 5=-6\\hfill \\\\ 4x=11\\hfill & \\text{or}\\hfill & 4x=-1\\hfill \\\\ x=\\frac{11}{4}\\hfill & \\hfill & x=-\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, we can examine the graph to observe where the [latex]y[\/latex]<em>&#8211;<\/em>values are negative. We observe where the branches are below the [latex]x[\/latex]<em>&#8211;<\/em>axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at [latex]x=-\\frac{1}{4}[\/latex] and [latex]x=\\frac{11}{4}[\/latex] and that the graph opens downward.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/24225906\/CNX_CAT_Figure_02_07_006.jpg\" alt=\"A coordinate plan with the x-axis ranging from -5 to 5 and the y-axis ranging from -4 to 4. The function y = -1\/2|4x \u2013 5| + 3 is graphed. An open circle appears at the point -0.25 and an arrow\" width=\"487\" height=\"363\" \/><figcaption class=\"wp-caption-text\">Graph of a V-shaped function indicating where \ud835\udc66&lt;0 on the x-axis.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18993\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18993&theme=lumen&iframe_resize_id=ohm18993&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/908"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/908\/revisions"}],"predecessor-version":[{"id":2052,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/908\/revisions\/2052"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/908\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=908"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=908"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=908"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=908"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}