Question

what is the minimum value of f(x) =xlnx

Answer 1

\(f'(x) = \log(x)+\frac{1}{x}x=\log(x)+1\)
now set that equal to zero

\(\log(x)+1=0\implies \log(x)=-1\implies e^{-1}=x\)

Answer 2

thanks so would -1/e be corrct

Answer 3

\(e^{-1}=\frac{1}{e}\)

Answer 4

my choices a
re -e
-1
-1/e
0
no minimum value

Answer 5

yes the minimum value is at x=1\e
you need to plug that into the original question to obtain the value.

Answer 6

so the answer is \(f(\frac{1}{e})\)

Answer 7

thanks