Mark M. answered • 01/06/15
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NO. For example, f(x) = x1/3 has an inflection point at (0,0), but f"(x) is undefined at (0,0).
Mekedi .
doesnt*
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01/06/15
Mark M.
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An inflection point of y = f(x) is a point on the graph where the concavity changes (i.e., the sign of the second derivative changes). This can only occur at a point on the graph of y = f(x) where f"(x) = 0 or is undefined. For example, f(x) = x1/3 is defined at (0,0), but f"(x) isn't defined there, For this function, the sign of f"(x) changes at (0, 0) (there is actually a vertical tangent at (0,0)). At a "cusp", the sign of the second derivative would not change at such a point, so a cusp would not be a point of inflection. By the way, if f"(x) = 0 at a point, P, that does not guarantee that there is an inflection point at P. We must also have a sign change of f"(x) at P. For example, for f(x) = x4
f"(x) = 0 when x = 0, but there isn't an inflection point when
x = 0, since the second derivative doesn't change sign.
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01/07/15
Mekedi .
01/06/15