Applications Of Derivatives Question 358
Question: The point for the curve $ y=xe^{x} $
[MNR 1990]
Options:
A) $ x=-1 $ is minimum
B) $ x=0 $ is minimum
C) $ x=-1 $ is maximum
D) $ x=0 $ is maximum
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation (curve) $ y=xe^{x} $
$ \therefore \frac{dy}{dx}=xe^{x}+e^{x}=e^{x}(1+x) $ and $ \frac{d^{2}y}{dx^{2}}=(x+2),e^{x} $
For maximum or minimum value of $ f(x) $ ,
therefore $ \frac{dy}{dx}=0\Rightarrow x=-1 $ .
$ \therefore {{{ {f}’’(x) }}_{x=-1}}=+ve $
Hence $ f(x) $ is minimum at $ x=-1 $ .
BETA
