Applications Of Derivatives Question 425
Question: The curve $ y=xe^{x} $ has minimum value equal to
Options:
A) $ -\frac{1}{e} $
B) $ \frac{1}{e} $
C) $ -e $
D) e
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let $ y=xe^{x} $ . Differentiate both side w.r.t. -x………….
$ \Rightarrow \frac{dy}{dx}=e^{x}+xe^{x}=e^{x}(1+x) $ Put $ \frac{dy}{dx}=0 $
$ \Rightarrow e^{x}(1+x)=0\Rightarrow x=-1 $ Now, $ \frac{d^{2}y}{dx^{2}}=e^{x}+e^{x}(1+x)=e^{x}(x+2) $
$ {{( \frac{d^{2}y}{dx^{2}} )}_{(x=-1)}}=\frac{1}{e}+0>0 $
Hence, $ y=xe^{x} $ is minimum function and $ {y_{\min }}=-\frac{1}{e} $ .
BETA
