{"id":203,"date":"2023-09-20T22:48:15","date_gmt":"2023-09-20T22:48:15","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/continuity-over-an-interval\/"},"modified":"2024-08-05T12:39:20","modified_gmt":"2024-08-05T12:39:20","slug":"continuity-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/continuity-learn-it-3\/","title":{"raw":"Continuity: Learn It 3","rendered":"Continuity: Learn It 3"},"content":{"raw":"<h2>Continuity Over an Interval<\/h2>\r\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, let's extend it to continuity over an interval. A function is continuous over an interval if you can trace it without lifting your pencil between any two points within that interval.<\/p>\r\n<ul>\r\n\t<li><strong>Open Interval<\/strong>: A function is continuous on an open interval [latex](a,b)[\/latex] if it is continuous at every point within that interval.<\/li>\r\n\t<li><strong>Closed Interval<\/strong>: A function is continuous on a closed interval [latex][a,b][\/latex] if it is continuous on [latex](a,b)[\/latex], continuous from the right at [latex]a[\/latex], and continuous from the left at [latex]b[\/latex].<\/li>\r\n<\/ul>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>continuity over an interval<\/h3>\r\n<ul>\r\n\t<li id=\"fs-id1170573352232\">A function is continuous over an open interval if it is continuous at every point in the interval.<\/li>\r\n\t<li>A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex][a,b][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex].<\/li>\r\n\t<li>Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to b^-}{\\lim}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to a^+}{\\lim}f(x)\\ne f(a)[\/latex], we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b][\/latex].<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Determine Continuity Over an Interval<\/strong><\/p>\r\n<ol>\r\n\t<li><strong>Check Continuity on Open Interval<\/strong>: Verify the function is continuous at all points within [latex](a, b)[\/latex].<\/li>\r\n\t<li><strong>Check Right Continuity at\u00a0 [latex]a[\/latex]<\/strong>: Ensure [latex]\\lim_{x \\to a^+} f(x) = f(a).[\/latex]<\/li>\r\n\t<li><strong>Check Left Continuity at [latex]b[\/latex]<\/strong>: Ensure [latex]\\lim_{x \\to b^-} f(x) = f(b)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\dfrac{x-1}{x^2+2x}[\/latex] is continuous.<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573426588\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573426588\"]<\/p>\r\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{x^2+2x}[\/latex] is a rational function, it is continuous at every point in its domain. <br \/>\r\n<br \/>\r\nThe domain of [latex]f(x)[\/latex] is the set [latex](\u2212\\infty ,-2) \\cup (-2,0) \\cup (0,+\\infty)[\/latex]. <br \/>\r\n<br \/>\r\nThus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\u2212\\infty ,-2), \\, (-2,0)[\/latex], and [latex](0,+\\infty)[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-x^2}[\/latex] is continuous.<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573439420\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573439420\"]<\/p>\r\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\lim}\\sqrt{4-x^2}=\\sqrt{4-a^2}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2)[\/latex]. <br \/>\r\n<br \/>\r\nWe also know that [latex]\\underset{x\\to -2^+}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists and [latex]\\underset{x\\to 2^-}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists. <br \/>\r\n<br \/>\r\nTherefore, [latex]f(x)[\/latex] is continuous over the interval [latex][-2,2][\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]204644[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<h2>Continuity Over an Interval<\/h2>\n<p id=\"fs-id1170573612125\">Now that we have explored the concept of continuity at a point, let&#8217;s extend it to continuity over an interval. A function is continuous over an interval if you can trace it without lifting your pencil between any two points within that interval.<\/p>\n<ul>\n<li><strong>Open Interval<\/strong>: A function is continuous on an open interval [latex](a,b)[\/latex] if it is continuous at every point within that interval.<\/li>\n<li><strong>Closed Interval<\/strong>: A function is continuous on a closed interval [latex][a,b][\/latex] if it is continuous on [latex](a,b)[\/latex], continuous from the right at [latex]a[\/latex], and continuous from the left at [latex]b[\/latex].<\/li>\n<\/ul>\n<section class=\"textbox keyTakeaway\">\n<h3>continuity over an interval<\/h3>\n<ul>\n<li id=\"fs-id1170573352232\">A function is continuous over an open interval if it is continuous at every point in the interval.<\/li>\n<li>A function [latex]f(x)[\/latex] is continuous over a closed interval of the form [latex][a,b][\/latex] if it is continuous at every point in [latex](a,b)[\/latex] and is continuous from the right at [latex]a[\/latex] and is continuous from the left at [latex]b[\/latex].<\/li>\n<li>Analogously, a function [latex]f(x)[\/latex] is continuous over an interval of the form [latex](a,b][\/latex] if it is continuous over [latex](a,b)[\/latex] and is continuous from the left at [latex]b[\/latex].<\/li>\n<\/ul>\n<\/section>\n<p id=\"fs-id1170573533852\">Requiring that [latex]\\underset{x\\to a^+}{\\lim}f(x)=f(a)[\/latex] and [latex]\\underset{x\\to b^-}{\\lim}f(x)=f(b)[\/latex] ensures that we can trace the graph of the function from the point [latex](a,f(a))[\/latex] to the point [latex](b,f(b))[\/latex] without lifting the pencil. If, for example, [latex]\\underset{x\\to a^+}{\\lim}f(x)\\ne f(a)[\/latex], we would need to lift our pencil to jump from [latex]f(a)[\/latex] to the graph of the rest of the function over [latex](a,b][\/latex].<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Determine Continuity Over an Interval<\/strong><\/p>\n<ol>\n<li><strong>Check Continuity on Open Interval<\/strong>: Verify the function is continuous at all points within [latex](a, b)[\/latex].<\/li>\n<li><strong>Check Right Continuity at\u00a0 [latex]a[\/latex]<\/strong>: Ensure [latex]\\lim_{x \\to a^+} f(x) = f(a).[\/latex]<\/li>\n<li><strong>Check Left Continuity at [latex]b[\/latex]<\/strong>: Ensure [latex]\\lim_{x \\to b^-} f(x) = f(b)[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170570990867\">State the interval(s) over which the function [latex]f(x)=\\dfrac{x-1}{x^2+2x}[\/latex] is continuous.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573426588\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573426588\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573426588\">Since [latex]f(x)=\\frac{x-1}{x^2+2x}[\/latex] is a rational function, it is continuous at every point in its domain. <\/p>\n<p>The domain of [latex]f(x)[\/latex] is the set [latex](\u2212\\infty ,-2) \\cup (-2,0) \\cup (0,+\\infty)[\/latex]. <\/p>\n<p>Thus, [latex]f(x)[\/latex] is continuous over each of the intervals [latex](\u2212\\infty ,-2), \\, (-2,0)[\/latex], and [latex](0,+\\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573406820\">State the interval(s) over which the function [latex]f(x)=\\sqrt{4-x^2}[\/latex] is continuous.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573439420\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573439420\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573439420\">From the limit laws, we know that [latex]\\underset{x\\to a}{\\lim}\\sqrt{4-x^2}=\\sqrt{4-a^2}[\/latex] for all values of [latex]a[\/latex] in [latex](-2,2)[\/latex]. <\/p>\n<p>We also know that [latex]\\underset{x\\to -2^+}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists and [latex]\\underset{x\\to 2^-}{\\lim}\\sqrt{4-x^2}=0[\/latex] exists. <\/p>\n<p>Therefore, [latex]f(x)[\/latex] is continuous over the interval [latex][-2,2][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm204644\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=204644&theme=lumen&iframe_resize_id=ohm204644&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"2.4 Continuity\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"2.4 Continuity","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/203"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":12,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/203\/revisions"}],"predecessor-version":[{"id":4497,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/203\/revisions\/4497"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/203\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=203"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=203"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=203"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=203"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}