inflection points on f(x) using f''(x) graph

It's a question from my Calculus text book (Single Variable Calculus: Early Transcendentals by James Stewart 6th Edition), and the question states:
 inflection point question
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?
ZenotureAsked:
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deightonprogCommented:
good example at wikipedia

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
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parnassoCommented:
Because the third derivative at 4 iszero, f'''(4)  = 0; so 4 is not an inflection point, neither x = 2 nor x = 6,

Remember that inflection point refers to function f(x),  "decreasing to increasing" applies to f(x).
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ZenotureAuthor Commented:
Ok but my issue is that when I look at this graph for the first time, I think f''(x) = 0 @ x = 1, 4, and 7 so I would put those down as my points, but 4 would be wrong. Is there a faster way to just look at it and know, rather than taking another derivative of the given, or determining where the given graph crosses axes and changes sign?

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?
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ozoCommented:
If it goes slightly below 0 at x=4, then there would be two inflection points there, one when it crosses below 0 and another when it crosses back above.
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deightonprogCommented:
what you have to do is think about curvature, if the second derivative changes sign then the curve changes between concave and convex
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ZenotureAuthor Commented:
So if I understood that right, the determining factor of inflection points ( and "decreasing" or "increasing" of f(x) ) is when f''(x) < 0 or f''(x) > 0?
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ozoCommented:
f''(x)=0 and f'''(x) < 0 or f'''(x) > 0
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parnassoCommented:
Yes, this is why the derivative can be understand as calculating the slope of a function
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