{"id":1370,"date":"2017-10-31T12:00:40","date_gmt":"2017-10-31T16:00:40","guid":{"rendered":"http:\/\/blog.richmond.edu\/math320\/?p=1370"},"modified":"2017-11-05T23:48:26","modified_gmt":"2017-11-06T04:48:26","slug":"hw-7-challenge-3","status":"publish","type":"post","link":"https:\/\/blog.richmond.edu\/math320\/2017\/10\/31\/hw-7-challenge-3\/","title":{"rendered":"HW 7: Challenge 3"},"content":{"rendered":"<p><strong>Problem:<\/strong> Let <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> be a differentiable function on an interval <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A\" class=\"latex\" \/>. Show that if\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28x%29+%5Cneq+0+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(x) &#92;neq 0 \" class=\"latex\" \/> on\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A \" class=\"latex\" \/>, then\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> is one-to-one. Provide an example to show that the converse of the statement need not be true.<\/p>\n<p><strong>Proof:\u00a0<\/strong> Suppose\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> is a differentiable function on some interval <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A+%5Cin+%5Cmathbb%7BR%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A &#92;in &#92;mathbb{R}\" class=\"latex\" \/>. Let <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+f%28y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = f(y)\" class=\"latex\" \/> for some arbitrary\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%2C+y+%5Cin+A&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x, y &#92;in A\" class=\"latex\" \/>. By Theorem 5.2.3, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f\" class=\"latex\" \/> is also continuous on <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A\" class=\"latex\" \/> and so by the Mean Value Theorem, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cexists+c+%5Cin+%28a%2Cb%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;exists c &#92;in (a,b)\" class=\"latex\" \/> such that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29+%3D+%5Cfrac%7Bf%28x%29-f%28y%29%7D%7Bx-y%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c) = &#92;frac{f(x)-f(y)}{x-y}\" class=\"latex\" \/>. By some algebra, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29-f%28y%29+%3D+f%27%28c%29%28x-y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)-f(y) = f&#039;(c)(x-y)\" class=\"latex\" \/>. Since\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+f%28y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = f(y)\" class=\"latex\" \/>,\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29%28x-y%29+%3D+0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c)(x-y) = 0\" class=\"latex\" \/>. So either\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29+%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c) =0\" class=\"latex\" \/> or <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%28x-y%29%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(x-y)=0\" class=\"latex\" \/>. However, by assumption, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28x%29+%5Cneq+0+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(x) &#92;neq 0 \" class=\"latex\" \/> so <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%28x-y%29%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(x-y)=0\" class=\"latex\" \/>, so\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%3Dy&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x=y\" class=\"latex\" \/> and thus\u00a0<img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> is one-to-one.<\/p>\n<p>&nbsp;<\/p>\n<p><strong><span style=\"color: #ff0000\">Corrected Proof: (Proof by Contradicition)\u00a0<\/span><\/strong><span style=\"color: #ff0000\">Suppose <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> is a differentiable function on some interval <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A+%5Cin+%5Cmathbb%7BR%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A &#92;in &#92;mathbb{R}\" class=\"latex\" \/> with the condition that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29+%5Cneq+0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c) &#92;neq 0\" class=\"latex\" \/>. Let <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+f%28y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = f(y)\" class=\"latex\" \/> for some arbitrary <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%2C+y+%5Cin+A&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x, y &#92;in A\" class=\"latex\" \/>. By Theorem 5.2.3, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f\" class=\"latex\" \/> is also continuous on <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=A&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"A\" class=\"latex\" \/> and so by the Mean Value Theorem, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%5Cexists+c+%5Cin+%28x%2Cy%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"&#92;exists c &#92;in (x,y)\" class=\"latex\" \/> such that (WLOG) for <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=y%3Ex&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"y&gt;x\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29+%3D+%5Cfrac%7Bf%28x%29-f%28y%29%7D%7Bx-y%7D&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c) = &#92;frac{f(x)-f(y)}{x-y}\" class=\"latex\" \/>. By some algebra, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29-f%28y%29+%3D+f%27%28c%29%28x-y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x)-f(y) = f&#039;(c)(x-y)\" class=\"latex\" \/>. Since <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+f%28y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = f(y)\" class=\"latex\" \/>, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29%28x-y%29+%3D+0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c)(x-y) = 0\" class=\"latex\" \/>. Since <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=y%3Ex&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"y&gt;x\" class=\"latex\" \/> by contruction of the mean value thereorem, <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c)=0\" class=\"latex\" \/> However, this is a contradiction to the assumption that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28c%29+%5Cneq+0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(c) &#92;neq 0\" class=\"latex\" \/>, thus <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%3Dy&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x=y\" class=\"latex\" \/> and thus <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f+&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f \" class=\"latex\" \/> is one-to-one.\u00a0<\/span><\/p>\n<p><strong>Example:\u00a0<\/strong>Let <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+x%5E5&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = x^5\" class=\"latex\" \/> on the interval <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%28-3%2C3%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(-3,3)\" class=\"latex\" \/>. We see that <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%27%28x%29+%3D+5x%5E4&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f&#039;(x) = 5x^4\" class=\"latex\" \/>. If <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=f%28x%29+%3D+f%28y%29&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"f(x) = f(y)\" class=\"latex\" \/>, then <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%5E5+%3D+y%5E5&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x^5 = y^5\" class=\"latex\" \/> and so <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%3Dy&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x=y\" class=\"latex\" \/> and we have that this is a 1-1 function. However at <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=x%3D0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"x=0\" class=\"latex\" \/> on <img decoding=\"async\" src=\"https:\/\/s0.wp.com\/latex.php?latex=%28-3%2C3%29%2C+f%27%28x%29+%3D+0&#038;bg=ffffff&#038;fg=000&#038;s=0&#038;c=20201002\" alt=\"(-3,3), f&#039;(x) = 0\" class=\"latex\" \/>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problem: Let be a differentiable function on an interval . Show that if\u00a0 on\u00a0, then\u00a0 is one-to-one. Provide an example to show that the converse of the statement need not be true. Proof:\u00a0 Suppose\u00a0 is a differentiable function on some interval . Let for some arbitrary\u00a0. By Theorem 5.2.3, is also continuous on and so [&hellip;]<\/p>\n","protected":false},"author":3525,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[58827,58826],"tags":[],"class_list":["post-1370","post","type-post","status-publish","format-standard","hentry","category-challenge-solutions","category-peer-grading"],"jetpack_publicize_connections":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"jetpack_shortlink":"https:\/\/wp.me\/p7L4E1-m6","jetpack-related-posts":[],"_links":{"self":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1370","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/users\/3525"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/comments?post=1370"}],"version-history":[{"count":0,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/posts\/1370\/revisions"}],"wp:attachment":[{"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/media?parent=1370"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/categories?post=1370"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.richmond.edu\/math320\/wp-json\/wp\/v2\/tags?post=1370"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}