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?
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good example at wikipedia

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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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).
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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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.
what you have to do is think about curvature, if the second derivative changes sign then the curve changes between concave and convex
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?
f''(x)=0 and f'''(x) < 0 or f'''(x) > 0
Yes, this is why the derivative can be understand as calculating the slope of a function
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