SOLUTION: Find the coordinates of the inflection points for f(x)=xe^x?
Calculus
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Question 1117429: Find the coordinates of the inflection points for f(x)=xe^x?
Calculus
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Possible inflection points exist where \ =\ 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.
\ =\ xe^x)
e^x)
\ = \frac{d}{dx}(x\,+\,1)e^x\ =\ (x\,+\,2)e^x)
e^x\ =\ 0)
Since 
, the only point where the second derivative is zero is at 
, and therefore the only possible inflection point is at )
To test concavity at a point, evaluate the second derivative at the point in question. If \ >\ 0)
, the function is concave downward at that point. If , 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
\ =\ (-3\,+\,2)e^{-3}\ <\ 0)
Since we have already determined that 
is positive on the entire domain, the fact that the lead coefficient is negative makes the value of the second derivative at 
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 
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, is indeed a point of inflection, and since there was only one root to the equation \ =\ 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.
A little closer look:
John
My calculator said it, I believe it, that settles it
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