The function f(x) =$x^x$ has a critical point at :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x=\frac{1}{e}$
$f(x) =x^x$
$f'(x)=x^x(1+\log x)=0$
$⇒\log x=-1$
$x=\frac{1}{e}$ → critical point
The function f(x) =$x^x$ has a critical point at :
Correct answer: option 2
The correct answer is Option (2) → $x=\frac{1}{e}$
$f(x) =x^x$
$f'(x)=x^x(1+\log x)=0$
$⇒\log x=-1$
$x=\frac{1}{e}$ → critical point