SOLUTION: Find the number which when raised by itself, it will become minimum.

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Question 390517: Find the number which when raised by itself, it will become minimum.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let y+=+x%5Ex. We have to restrict the domain of y to (0, infinity), otherwise we encounter a densely infinite number of problems, haha.(E.g., %28-1%2F4%29%5E%28-1%2F4%29+=+1%2F%28%28-1%2F4%29%5E%281%2F4%29%29, or %28-pi%29%5E%28-pi%29+=+1%2F%28%28-pi%29%5Epi%29). Note also that y > 0 for all x in the domain.
That said, we can take ln of both sides of y+=+x%5Ex:
ln y = x lnx.
Differentiating both sides:
%28dy%2Fdx%29%2Fy+=+lnx+%2B1 <-----(A)
Set dy%2Fdx+=+0, so that the equation (A) becomes lnx = -1, or x+=+e%5E%28-1%29+=+1%2Fe. Since there is only 1 critical value this is either an absolute min or absolute max. We find out using the 2nd derivative test.
Differentiate (A) again:
%28y%28d%5E2y%2Fdx%5E2%29+-+%28dy%2Fdx%29%5E2%29%2Fy%5E2+=+1%2Fx.
When x = 1/e, the equation becomes %28d%5E2y%2Fdx%5E2%29%2Fy+=+e, or d%5E2y%2Fdx%5E2+=+ye+%3E+0. Thus the graph of the function y is concave up at x=1/e, and so there is an absolute minimum there for y.