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 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 non-zero. 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: 3,002 Thanks: 1587 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
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Re: inflection points analysis

Quote:
 Originally Posted by m.nikolov 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 non-zero. 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.
An "inflection point" is a point where the curvature changes sign. Since the curvature is determined by the second derivative, we are looking for points where the secod derivative changes sign. That means that the second derivative must be 0 at any inflection point, although that is not a "sufficient condition".
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. Tags analysis, inflection, 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 cheyb93 Calculus 4 November 8th, 2012 11:30 AM soulrain Calculus 5 June 22nd, 2012 05:13 PM mathhelp123 Calculus 2 December 5th, 2009 07:34 AM stainsoftime Calculus 3 December 15th, 2008 12:19 AM lovetolearn Algebra 0 December 31st, 1969 04:00 PM

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