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 January 13th, 2016, 08:09 AM #1 Newbie   Joined: Sep 2015 From: mexico Posts: 10 Thanks: 0 Determine and classify the critical points of the function f(x) = x^4 - 3x^3 - 3sin(2x) + 2. Last edited by skipjack; January 14th, 2016 at 06:29 AM. January 14th, 2016, 01:02 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 The function has a minimum at x = 2.0251663 approximately. January 14th, 2016, 06:12 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Do you know what "critical points" are? Do you know how to differentiate this function? Is your difficulty with solving the resulting equation? That's not clear from what you write so it's not clear what help you need. January 14th, 2016, 06:26 AM #4 Newbie   Joined: Sep 2015 From: mexico Posts: 10 Thanks: 0 Skip jack thank you! Country Boy well I think my main problem here is the trig part, I mean I know how to derive, but to get my critical point and get to equal zero with that trig function in there Last edited by skipjack; January 14th, 2016 at 06:31 AM. January 14th, 2016, 06:34 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Instead of trying to obtain an exact solution, you will need to use some other method. I simply plotted the function using software and verified my answer by using W|A. January 14th, 2016, 10:41 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Okay, a "singular point" of a function is a point where the derivative does not exist or is 0. With $\displaystyle f(x) = x^4 - 3x^3 - 3\sin(2x) + 2$, $\displaystyle f'(x)= 4x^3- 9x^2- 6\cos(2x)$. That "exists" for all x, so we want to solve $\displaystyle 4x^3- 9x^2- 6\cos(2x)= 0$. There is no "formula" to solve that. As skipjack said, he plotted the function and, I presume, "zoomed in" on the point where the graph crossed the x-axis. You could also use a numerical method. The simplest would be to note that f(2)= -0.0781 while f(2.1)= 0.2955 (to four decimal places). Since this continuous function changes from negative to positive, it must be 0 somewhere between 2 and 2.1. The simplest thing to try is x= 2.05, half way between those two numbers. f(2.05)= 0.0869. Since that is positive, f must be 0 somewhere between 2 and 2.05. Again, try half way between, x= 2.025. f(2.025)= -0.000548 which is negative so f must be 0 somewhere between 2.025 and 2.05. Try (2.025+ 2.05)/2= 2.0375. Continue until you get two consecutive x values such that f(x) are closer than whatever accuracy you want. Last edited by skipjack; January 15th, 2016 at 09:18 AM. January 15th, 2016, 06:54 AM #7 Newbie   Joined: Sep 2015 From: mexico Posts: 10 Thanks: 0 Country Boy and skipjack thank you guys; I really do appreciate your help. Last edited by skipjack; January 15th, 2016 at 09:16 AM. Tags critical, points Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathe Calculus 2 November 11th, 2015 01:32 PM mathkid Calculus 1 November 11th, 2012 07:34 PM summerset353 Calculus 1 March 5th, 2010 02:50 AM SSmokinCamaro Calculus 2 April 3rd, 2009 08:04 PM stainsoftime Calculus 3 November 24th, 2008 05:24 AM

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