SOLUTION: Find max/min using second-derivative test For: f (x) = xe^-0.2x

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Question 1035331: Find max/min using second-derivative test
For: f (x) = xe^-0.2x
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
f+%28x%29+=+xe%5E%28-0.2x%29 ==> df%28x%29%2Fdx+=+%281-0.2x%29e%5E%28-0.2x%29
Let df%28x%29%2Fdx+=+%281-0.2x%29e%5E%28-0.2x%29+=+0
==> 1-0.2x+=+0. Note that e%5E%28-0.2x%29 is always greater than 0.
==> x = 5. So there is a critical point at this x-value.
Now d%5E2f%28x%29%2Fdx%5E2+=+%28-0.4+%2B+0.04x%29e%5E%28-0.2x%29, and d%5E2f%285%29%2Fdx%5E2+=+%28-0.4+%2B+0.04%2A5%29e%5E-1+%3C+0,
hence there is a local max, which is also an absolute maximum at x = 5.
The maximum value is f+%285%29+=+5%2Fe.