tag:blogger.com,1999:blog-7806043650549721218.post8485052736613609324..comments2020-07-09T01:58:15.844-05:00Comments on EPSILON-DELTA: A case I hadn't given much thought to until recentlyRebecka Petersonhttp://www.blogger.com/profile/12227797437296056645noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-7806043650549721218.post-27936992402100602472014-10-27T12:19:41.310-05:002014-10-27T12:19:41.310-05:00I found your blog as I was searching for the answe...I found your blog as I was searching for the answer to a similar problem. I was using a removable discontinuity graph in the intro to my extrema lesson. I used y=x^2 with open circle at (0, 0) and the point located at (0, 1). I was trying to emphasize that (0, 0) is not an absolute min, but then the question of it being a relative max came up. According to the definition, I agree that it is a relative max. I wanted to find confirmation...I'm glad we agree.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-7806043650549721218.post-76584445898672997752013-05-08T16:59:59.807-05:002013-05-08T16:59:59.807-05:00Exactly. Definitions, definitions! Exactly. Definitions, definitions! Rebecka Petersonhttps://www.blogger.com/profile/12227797437296056645noreply@blogger.comtag:blogger.com,1999:blog-7806043650549721218.post-19507460012285053152013-05-06T06:57:29.216-05:002013-05-06T06:57:29.216-05:00Yeah, you're right, first derivative test is c...Yeah, you're right, first derivative test is conditioned on continuity. Hmm.. I guess we're in the wild on this one, where anything goes! Maybe we can come up with some other pathological examples.<br /><br />I guess the first derivative test isn't the be-all and end-all when looking for extrema. The most important thing is f(b)<=f(x) in some neighborhood of b. When in doubt, return to the definition!Mr. Chasehttps://www.blogger.com/profile/17707024524634807456noreply@blogger.comtag:blogger.com,1999:blog-7806043650549721218.post-44822282707219663962013-04-29T20:30:26.097-05:002013-04-29T20:30:26.097-05:00I agree-we have a relative min at x=b, as it does ...I agree-we have a relative min at x=b, as it does the follow the definition.<br /><br />But, I don't think we can use the first derivative test here since f needs to be continuous at x=b in order to apply the test. So, in that sense, I wouldn't say it's similar to finding relative extrema that occur at corners or cusps. Or, what do you think?Rebecka Petersonhttps://www.blogger.com/profile/12227797437296056645noreply@blogger.comtag:blogger.com,1999:blog-7806043650549721218.post-61087099101822634402013-04-29T13:45:51.566-05:002013-04-29T13:45:51.566-05:00Yes, I think. We only require that f(b)<=f(x) o...Yes, I think. We only require that f(b)<=f(x) on some open interval containing b, right?<br /><br />And this doesn't throw off our first derivative test, either. We check places where the first derivative is zero or undefined (like it is here). If the derivative is undefined, we have to do a bit more careful analysis and check the actual value of the function. <br /><br />I think this is similar to if you were finding extrema on y=abs(x) or y=x^(2/3). Or maybe piecewise-defined functions like your picture or like y=x (x>=0), y=-x+1 (x<0).<br /><br />What's your take?<br />Mr. Chasehttps://www.blogger.com/profile/17707024524634807456noreply@blogger.com