A) \[x=-1\]is minimum
B) \[x=0\]is minimum
C) \[x=-1\]is maximum
D) \[x=0\]is maximum
Correct Answer: A
Solution :
Given equation (curve) \[y=x{{e}^{x}}\] \[\therefore \frac{dy}{dx}=x{{e}^{x}}+{{e}^{x}}={{e}^{x}}(1+x)\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(x+2)\,{{e}^{x}}\] For maximum or minimum value of \[f(x)\], Þ \[\frac{dy}{dx}=0\Rightarrow x=-1\]. \[\therefore {{\left\{ {f}''(x) \right\}}_{x=-1}}=+ve\] Hence \[f(x)\]is minimum at \[x=-1\].You need to login to perform this action.
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