My Math Forum First Derivative Test

 Calculus Calculus Math Forum

 November 7th, 2013, 10:10 AM #1 Newbie   Joined: Oct 2013 Posts: 17 Thanks: 0 First Derivative Test SO im having difficulty with this question. The question is let g(x)=25x+((36)/(x+4)) I need to find the g'(x)________( I found it was =25-(36/(x+4)^2)), then it asked me to find g'(0) if x= ____ _____ ( i found that it equal -14/5 and -26/5 by simply makign the equation =0) We note tat g'(x) is not defined if x=___ ( i found that it was x= -4), but since this value is not in the domain of g it cannot be a critical value. Therefore the critical values of g are x= ____ ____ ( found that it was -2 and 2) I tried the bets i could, could someone double check the work that I have done and tell me where I am doing it wrong? Thanks
 November 7th, 2013, 12:04 PM #2 Member   Joined: Feb 2013 Posts: 80 Thanks: 8 Re: First Derivative Test If I told you that the derivative you stated, g'(x) = 25 - [36]/[(x+4)^2] could be put into a different form by finding a common denominator and inducting the 25 into the fraction as such, g'(x) = [25(x+4)^2]/[(x+4)^2] - [36]/[(x+4)^2] g'(x) = [25(x+4)^2 - 36]/[(x+4)^2] which then can be expanded in the numerator as follows, g'(x) = [25(x^2 + 8x + 16) - 36]/[(x+4)^2] g'(x) = [25x^2 + 200x + 400 - 36]/[(x+4)^2] g'(x) = [25x^2 + 200x + 364]/[(x+4)^2] which you can factor then in the numerator to, g'(x) = [(5x + 26)(5x + 14)]/[(x+4)^2]. Now simply repeat what you tried to do to find the x's where g'(0).

 Tags derivative, test

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Gerrit Calculus 13 October 14th, 2013 04:33 AM BrianMX34 Calculus 3 November 3rd, 2012 12:05 PM mathkid Calculus 5 October 6th, 2012 07:22 PM RealMadrid Calculus 1 July 19th, 2012 02:24 PM SyNtHeSiS Algebra 5 May 26th, 2010 04:28 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top