Applications Of Derivatives Question 271
Question: The function $ x^{x} $ is increasing, when
[MP PET 2003]
Options:
A) $ x>\frac{1}{e} $
B) $ x<\frac{1}{e} $
C) $ x<0 $
D) For all real x
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y=x^{x} $
therefore $ \frac{dy}{dx}=x^{x}(1+\log x) $ For $ \frac{dy}{dx}>0 $ ; $ x^{x}(1+\log x)>0 $
therefore $ 1+\log x>0\Rightarrow {\log_{e}}x>{\log_{e}}\frac{1}{e} $
For this to be positive, x should be greater than $ \frac{1}{e} $ .
BETA
