i think you do f'(x) and use the product rule?

f'(x) = x (1/x) + lnx (1)

1+lnx = 0

lnx = -1

ok i dont kow where to go from there.

THANKS IN ADVANCE!

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- Feb 24th 2009, 04:20 PMLexiRaef(x) = xlnx. the minimum value attained by f is...
i think you do f'(x) and use the product rule?

f'(x) = x (1/x) + lnx (1)

1+lnx = 0

lnx = -1

ok i dont kow where to go from there.

THANKS IN ADVANCE! - Feb 24th 2009, 04:37 PMskeeter
$\displaystyle f'(x) = 1 + \ln{x} = 0$

$\displaystyle \ln{x} = -1$

$\displaystyle x = e^{-1} = \frac{1}{e}$

$\displaystyle f'(x) = 1 + \ln{x}$

$\displaystyle f''(x) = \frac{1}{x}$

$\displaystyle f''(x) > 0$ for all x in the domain of $\displaystyle f(x)$, so $\displaystyle f(x)$ is concave up everywhere, indicating $\displaystyle f\left(\frac{1}{e}\right)$ is a minimum.

value of the minimum is $\displaystyle f\left(\frac{1}{e}\right) = -\frac{1}{e}$