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Question 2 has to be wrong. The derivative of xlogx = logx + 1. Can't have a negative value for x. A stationary point exists at 1/e but not -1/eThanks for the quick replies ;D! I just re-did question 1 and 3 and i got it now w00t. Thanks!
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y'=ln x +1y = xlogx. Differentiating that gives y' = log x + 1. Equating that to 0 gives logx + 1 = 0, logx = -1. Using the rule:
y = logax
x = a ^ y
gives: x = e ^ -1 = 1 / e.
y'' = 1/x = pos. Therefore min point. So a min.point exists at 1/e.
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I read it again, I did it again, same result. They're saying, show the minimum point of y = xlogx = -1/e.
y' = logx + 1
= log(-1/e) + 1
= math error.
It isn't a real solution, therefore it can't be a min.
lol u beat me.......
the minimum occurs at x = 1/e
y = xlnx
= 1/e ln(1/e) at the minimum
= 1/e *-1
= -1/e
First of all, 4x-2 is NOT equal to 4x-2+1 / -1. What you've just done was integrate it already lol= 4x^-2
=4x^-2+1 / -1
=4x^-1 / -1
=-4/x
Yup. Don't i need to use logs to do it?