document.write( "Question 1035333: Find max/min when x belongs to the interval (0,infinite sign)
\n" ); document.write( "For function:\r
\n" ); document.write( "\n" ); document.write( "f (x) = (lnx)^2 - 3lnx
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Algebra.Com's Answer #650008 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
\"f+%28x%29+=+%28lnx%29%5E2+-+3lnx+\" ==> \"df%28x%29%2Fdx+=+%282lnx-3%29%2Fx\".\r
\n" ); document.write( "\n" ); document.write( "Now find the critical points of the function by letting \"%282lnx-3%29%2Fx+=+0\"\r
\n" ); document.write( "\n" ); document.write( "==> \"2lnx+-+3=0\", or \"x+=+e%5E%283%2F2%29\" or \"x+=+e%5E1.5\".\r
\n" ); document.write( "\n" ); document.write( "Now apply the first derivative test:\r
\n" ); document.write( "\n" ); document.write( "When \"x+%3E+e%5E1.5\", \"df%28x%29%2Fdx+%3E+0\" ==> f(x) is increasing over there.
\n" ); document.write( "When \"x+%3C+e%5E1.5\", \"df%28x%29%2Fdx+%3C+0\" ==> f(x) is decreasing over there.\r
\n" ); document.write( "\n" ); document.write( "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.
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