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2011 Calculus AB free response #4c

Finding the points of inflection for a strangely defined function. Created by Sal Khan.

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• Wouldn't the point (-3, 0) also be a point of inflection, because the rate of change of the slope at that point changes from positive to negative? Just visually looking at the graph leads me to believe this must be the case.
• (-3, 0) is a point of inflection for f(x), but not for g(x). In this problem, the graph shown is the derivative of g(x), so to find inflection of g(x) from this graph, we have to see where the slope of this graph goes from positive to negative (or vice versa). I think you're thinking this graph represents g(x), when it is actually g'(x).
• My textbook says that there's more than one definition of inflection point and that the one they use in the book requires that there exists a tangent line and the tangent line intersects the graph. According to my book's definition, would the answer be that g has no inflection points?.
• Yes. This is because this is an odd function, that, like Sal said, is not differentiable at f(0) because the slope jumps. So the book's definition doesn't work for this strange problem.
• I'm a little confused on this. Since f '(x) does not exist at this point, then does this not imply that g ' ' (x) also does not exist? So, therefore, the function has no inflection point?
• Generally... an inflection point is understood to be a point where the concavity/acceleration changes from positive to negative or from negative to positive.

It does not matter whether or not a function's output value exists.
In this case, we just want a point where f'(x) changes between positive and negative.
• In order for a given point to qualify as an Inflection Point of a function f, must the second derivative of f (with respect to x) be Continuous at said point? Or is it simply enough for the function to be convex on one side and concave on the other (and with a limit that exists at that point)?
(1 vote)
• A point by definition is an inflection point if the concavity changes at that point(from concave up to concave down or vice versa). f does not have to be continuous or differentiable at that point. Nor does the limit have to exist there.
(1 vote)
• I've been led to believe that any justification on the AP Calculus AB Exam requires us to use the given information. If that's the case, this would be my justification:

"x = 0 is an extremum point of f(x) and thus an extremum point of f(x) + 2 = g'(x).

x = 0 is, therefore, an inflection point of g(x)."

Would this be okay?
(1 vote)