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November 7th, 2013, 10:10 AM   #1
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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
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November 7th, 2013, 12:04 PM   #2
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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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