A) Decreasing for all x
B) Decreasing in \[(-\infty ,\,-1)\] and increasing in \[(-1,\infty )\]
C) Increasing for all x
D) Decreasing in \[(-1,\,\infty )\] and increasing in \[(-\infty ,\,-1)\]
Correct Answer: D
Solution :
\[f(x)=(x+2){{e}^{-x}}\] \[f'(x)={{e}^{-x}}-{{e}^{-x}}(x+2)\] \[f'(x)=-{{e}^{-x}}[x+1]\] For increasing, \[-{{e}^{-x}}(x+1)>0\] or \[{{e}^{-x}}(x+1)<0\] \[{{e}^{-x}}>0\] \[(x+1)<0\] \[x\in (-\infty ,\,\infty )\] and\[x\in (-\infty ,-1)\] \[\therefore x\in (-\infty ,-1)\] Hence, the function is increasing in \[(-\infty ,\,-1)\] For decreasing, \[-{{e}^{-x}}(x+1)<0\]or\[{{e}^{-x}}(x+1)>0\], \[x\in (-1,\,\infty )\] Hence the function is decreasing in \[(-1,\ \infty )\].You need to login to perform this action.
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