document.write( "Question 1033076: Find the point of inflexion on the
\n" ); document.write( "curve y=xln(x)-x^2.\r
\n" ); document.write( "\n" ); document.write( "I found the derivative: y'= ln(x+1) -2x
\n" ); document.write( "But I can't find the point of inflexion because I can't solve for y'=0
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Algebra.Com's Answer #647698 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
\"y=xlnx-x%5E2\" ==> y' = lnx + 1 - 2x ==> y\" = \"1%2Fx-2\"\r
\n" ); document.write( "\n" ); document.write( "The critical values of the 2nd derivative are x = 0 and x = 1/2, but the function \"y=xlnx-x%5E2\" is undefined at x = 0, so there is no inflection point there.\r
\n" ); document.write( "\n" ); document.write( "Now in (\"-infinity\", 0), y\" < 0.
\n" ); document.write( "In (0, 1/2), y\" > 0, and
\n" ); document.write( "at (1/2, \"infinity\"), y\" < 0.
\n" ); document.write( "(At x = 1/2, y\" = 0.)\r
\n" ); document.write( "\n" ); document.write( "Hence there is only one inflection point at (1/2, \"-ln2%2F2-1%2F4\"). (Even though there is a change in the direction of concavity from the interval (\"-infinity\", 0) to (0, 1/2).)\r
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