Question

Let f(x) = xln(x). The minimum value attained by f is?

Answer 1

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

Explanation:

Take the first derivative, determine for what values of `x` it equals zero.

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

`f'(x)=1+lnx`

`1+lnx=0`

`lnx=-1`

`e^lnx=e^-1`

`x=1/e`

Now, calculate `f(1/e).`

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

Furthermore, calculate the second derivative.

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

Now, the Second Derivative Test tells us if `x=a` is a critical point, we can take `f''(a).` If `f''(a)>0, x=a` is a minimum, if `f''(a)<0, x=a` is a maximum. Here, `a=1/e`

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

Then, the minimum value is at the coordinates `(1/e, -1/e)`, the value itself is `f(1/e)=-1/e`