The minimum value of the function `f(x)=xlogx` is

Options

A. `-(1)/(e)`
B. `-e`
C. `(1)/(e)`
D. e

Answer

A

Explanation

We have,
`f(x)= x log x`
`f'(x)=1+log x`
For maxima or minima, put `f'(x)=0`
`:.1+log x = 0 `
`rArr log x = - 1`
`rArr x=e^(-1)`
Now, `f''(x)=(1)/(x)`
`rArr f''(e^(-1))=(1)/(e^(-1))=e gt 0`
`:.f(x)` is decreasing.
Hence , minimum value of f(x) at `x=e^(-1)` is
`f(e^(-1))=e^(-1) log e^(-1)=(1)/(e)(- log e) = - (1)/(e)`