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January 12th, 2013, 01:28 PM   #1
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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.
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January 12th, 2013, 02:05 PM   #2
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Re: inflection points analysis







,

inflection pts on the graph of f are at both zeros above since f'' changes sign at both locations.
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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.
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