November 7th, 2013, 11:10 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, 01:04 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). 

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