The function xx decreases on the interval
Answer & explanation
Correct answer: option 3
The correct answer is Option 3: $(0, \frac{1}{e})$
⇒ $f'(x) = x^x[1+\log x]$
1+ log$x$ <0
⇒log$x$ < −1
⇒x<1/e
Hence, decreasing on (0, 1/e)
The function xx decreases on the interval
Correct answer: option 3
The correct answer is Option 3: $(0, \frac{1}{e})$
⇒ $f'(x) = x^x[1+\log x]$
1+ log$x$ <0
⇒log$x$ < −1
⇒x<1/e
Hence, decreasing on (0, 1/e)