SOLUTION: Find max/min when x belongs to the interval (0,infinite sign) For function: f (x) = (lnx)^2 - 3lnx

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Question 1035333: Find max/min when x belongs to the interval (0,infinite sign)
For function:
f (x) = (lnx)^2 - 3lnx
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
f+%28x%29+=+%28lnx%29%5E2+-+3lnx+ ==> df%28x%29%2Fdx+=+%282lnx-3%29%2Fx.
Now find the critical points of the function by letting %282lnx-3%29%2Fx+=+0
==> 2lnx+-+3=0, or x+=+e%5E%283%2F2%29 or x+=+e%5E1.5.
Now apply the first derivative test:
When x+%3E+e%5E1.5, df%28x%29%2Fdx+%3E+0 ==> f(x) is increasing over there.
When x+%3C+e%5E1.5, df%28x%29%2Fdx+%3C+0 ==> f(x) is decreasing over there.
Thus there is a local minimum at x+=+e%5E1.5, and since it is the only critical point in an open interval (0, infinity), it is also an absolute minimum.