Question

What are the critical values of `f(x) = xlnx` ?

Answer 1

`f(x)` is defined for `x>0`
`f(x)` has a zero at `x=1`
`f(x)` has an absolute minimum of`-1/e` at `x=1/e`

Explanation:

`f(x) =xlnx`

Since `lnx` is defined for `x>0 -> f(x)` is defined for `x>0`

To find the zero: `f(x) = 0`

`xlnx = 0 -> lnx =0` Since `x>0`

`lnx = 0 -> x=1`

`f'(x) = x*1/x + lnx` (product rule and standard differential)

`f'(x) = 0 -> lnx = -1`

`x=e^-1`

`x=1/e ~= 0.3679`

`f''(x) = 1/x` which is `>0 forall x>0`

Hence: `f_min = f(e^-1) = -e^-1 ~= -0.3679`

These points can be seen on the graph of `f(x)` below:

graph{xlnx [-10, 10, -5, 5]}