f (x) = x^x has a stationary point at

Given equation is f (x) = x^x

Let y= x^x………(i)

Taking logarithm on both sides we get

log y=log (x^x)

⇒ log y=x log x

Now applying first derivative, we get

Now applying the product rule of differentiation we get

Now applying the deravative we get

Substituting the value of y from equation (i), we get

Now we will find the critical point by equating equation (i) to 0, we get

x^x (1+log x)=0

⇒ 1+log x =0 as x^x cannot be equal to 0

⇒ log x=-1

But -1=log e^-1

⇒ log x=log e^-1

Equating the terms we get

x= e^-1

Hence f(x) has a stationary point at

So the correct option is option B.