Question

What are the values and types of the critical points, if any, of `f(x)=xlnx`?

Answer 1

Local minimum: `(1/e, -1/e)`

Explanation:

Take the first derivative, noting that the domain of the original function is `(0, oo).`

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

`f'(x)=1+lnx`

The domain of the first derivative is also `(0, oo),` so there won't be any critical points where the first derivative does not exist.

Set to zero and solve for `x.`

`1+lnx=0`

`lnx=-1`

`e^lnx=e^-1`

Recalling that `e^lnx=x:`

`x=1/e`

This is not a very complex function, so we can take the second derivative and apply the Second Derivative Test, which tells us that if `x=a` is a critical value and `f''(a)>0,` then it is a minimum, and if `f''(a)<0,` then it is a maximum.

`f''(x)=1/x`

`f''(1/e)=1/(1/e)=e>0`

Thus, we have a local minimum at `x=1/e.` Find the `y-`coordinate:

`f(1/e)=1/eln(1/e)=1/e(-lne)=-1/e`

Local minimum: `(1/e, -1/e)`