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It's a question from my Calculus text book (Single Variable Calculus: Early Transcendentals by James Stewart 6th Edition), and the question states:

I understand why x = 1 and x = 7 are inflection points, because it is at those x's that the f''(x) returns 0 and changes direction or sign to positive or negative, causing the f'(x) function to have increasing or decreasing slopes @ those points.

What I don't understand is why x = 4 is NOT an inflection point? f''(4) = 0, and turns from decreasing to increasing... so what makes x=4 a special excluded case?

I understand why x = 1 and x = 7 are inflection points, because it is at those x's that the f''(x) returns 0 and changes direction or sign to positive or negative, causing the f'(x) function to have increasing or decreasing slopes @ those points.

What I don't understand is why x = 4 is NOT an inflection point? f''(4) = 0, and turns from decreasing to increasing... so what makes x=4 a special excluded case?

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Start your 7-day free trialRemember that inflection point refers to function f(x), "decreasing to increasing" applies to f(x).

Another issue I just noticed is, the image I posted (from the PDF version of the book) and the image I'm reading in the book are a bit different. In the book, there is a very tiny bump right below the x=4 mark, so wouldn't that technically mean that it did in fact cross and change signs for a brief moment to cross and change back?

Math / Science

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http://en.wikipedia.org/wiki/Inflection_point#Categorization_of_points_of_inflection

f(x) = x^4 - x at (0,0)

f''(x) = 12x^2

so f'' is zero at (0,0), however the rate of change of gradient is never negative, the curvature never changes from concave to convex

f''(x)=0 does not have to be a max or min