January 12th, 2013, 01:28 PM  #1 
Newbie Joined: Jan 2013 Posts: 2 Thanks: 0  inflection points analysis
Hello everybody, I have a hesitation about one problem. I have to analyse the following polynomial with respect to inflection points: x^4 + 2x^3  12x^2 + 14x  5 R > R According to the theorem I took the second derivative of this function: f``(x) = 12x^2 + 12x  24 and for x = 1 I have 0 Then I continue the derivation until I get nonzero. My question is do I plug in x = 1 in the third derivative f```(x) = 24x + 12 because if I do it, it's not a zero, hence x = 1 is an inflection point or I should continue to the 4th derivative when it is 24? And if x = 1 when I plug it into the 1st order polynomial it gives 0 too and it is a saddle point. Please tell me what you think. Thanks. 
January 12th, 2013, 02:05 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,887 Thanks: 1506  Re: inflection points analysis , inflection pts on the graph of f are at both zeros above since f'' changes sign at both locations. 
January 12th, 2013, 02:47 PM  #3  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: inflection points analysis Quote:
As skeeter says, the second derivative can be factored as so the derivative is 0 at x= 2 and at x= 1. Further, it is easy to see that if x< 2, both factors are negative so the product is negative. If x is between 2 and 1, x+ 2 is positive while x 1 is still negative and so the product is negative. The second derivative does change sign so x= 2 is an inflection point. if x> 1, both factors are positive so the second derivative is positive. The second derivative does change signso x= 1 is an inflection point. However, you can use your method: the third derivative is 24x+ 12. At x= 2 that is 48+ 12= 36. That means the second derivative is decreasing and, since it is 0 at x= 2, changes sign there. At x= 1, the third derivative is 24+ 12= 36. That means the second derivative is increasing and, since it is 0 at x= 1, changes sign there.  

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