Question 1117429
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Possible inflection points exist where *[tex \Large \frac{d^2}{dx^2}f(x)\ =\ 0]


Such a point is a point of inflection if the concavity of the function changes from one side of the critical point to the other.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ xe^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\ =\ xe^x\ =\ (x\,+\,1)e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2}{dx^2}f(x)\ = \frac{d}{dx}(x\,+\,1)e^x\ =\ (x\,+\,2)e^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\,+\,2)e^x\ =\ 0]


Since *[tex \LARGE e^x\ >\ 0\ \forall\ x\ \in\ \mathbb{R}], the only point where the second derivative is zero is at *[tex \LARGE x\ =\ -2], and therefore the only possible inflection point is at *[tex \LARGE \(-2,\, -2e^{-2}\)]


To test concavity at a point, evaluate the second derivative at the point in question.  If *[tex \LARGE \frac{d^2}{dx^2}f(x)\ >\ 0], the function is concave downward at that point.  If *[tex \LARGE \frac{d^2}{dx^2}f(x)\ >\ 0], the function is concave upward.


Since our critical point is at -2, we can choose -3 and -1 as points on either side to test


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2}{dx^2}f(-3)\ =\ (-3\,+\,2)e^{-3}\ <\ 0]


Since we have already determined that *[tex \LARGE e^x] is positive on the entire domain, the fact that the lead coefficient is negative makes the value of the second derivative at *[tex \LARGE x\ =\ -3] negative.  Consequently, to the left of the critical point, the original function is concave downward.


Similarly, and I will leave the details to you, evaluating the second derivative at *[tex \LARGE x\ =\ -1] results in a finding that the original function is concave upward to the right of the critical point.


Therefore, by the definition of inflection point, *[tex \LARGE \(-2,\, -2e^{-2}\)] is indeed a point of inflection, and since there was only one root to the equation *[tex \LARGE \frac{d^2}{dx^2}f(x)\ =\ 0], we are assured that this is the only point of inflection.


An examination of the graph of the second derivative of the original function superimposed on a graph of the original function should give you the warm fuzzy feeling that the answer derived above is correct.


*[illustration xe_xPtOfInflectionCrop.jpg]


A little closer look:


*[illustration xe_xPtOfInflectionZoomCrop.jpg]




John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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