Applications Of Derivatives Question 415
Question: OR When x is positive, the minimum value of $ x^{x} $ is
[EAMCET 1987]
Options:
A) $ {e^{-1}} $
B) $ {e^{-1/e}} $
C) $ {e^{1/e}} $
D) $ e^{e} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y=x^{x} $
therefore $ \log y=x.\log x,\ \ \ \ (x>0) $ Differentiating $ \frac{dy}{dx}=x^{x}(1+\log x) $ ;
$ \therefore \frac{dy}{dx}=0 $
therefore $ \log x=-1 $
therefore $ x={e^{-1}}=\frac{1}{e} $
$ \therefore $ Stationary point is $ x=\frac{1}{e} $
$ \frac{d^{2}y}{dx^{2}}=x^{x}{{(1+\log x)}^{2}}+x^{x}.\frac{1}{x} $ When $ x=\frac{1}{e},\ \ \frac{d^{2}y}{dx^{2}}={{( \frac{1}{e} )}^{(1/e)-1}}>0 $ Therefore y is minimum at $ x=\frac{1}{e} $ and minimum value $ ={{( \frac{1}{e} )}^{1/e}}={e^{-1/e}} $ .
BETA
