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The functions $f(x)=x e^{-x}, \forall(x \in R)$ attains a maximum value at $x$ is equal to
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Verified Answer
The correct answer is:
$1$
We have, $f(x)=x e^{-x}$
$$
f^{\prime}(x)=-x e^{-x}+e^{-x}
$$
For maximum or minimum, put $f^{\prime}(x)=0$
$$
\begin{aligned}
& \Rightarrow \quad-x e^{-x}+e^{-x}=0 \Rightarrow x=1 \\
& f^{\prime \prime}(x)=x e^{-x}-e^{-x}-e^{-x}=(x-2) e^{-x} \\
& f^{\prime \prime}(1)=(1-2) e^{-1}=-\mathrm{ve}
\end{aligned}
$$
$f(x)$ is maximum at $x=1$.
$$
f^{\prime}(x)=-x e^{-x}+e^{-x}
$$
For maximum or minimum, put $f^{\prime}(x)=0$
$$
\begin{aligned}
& \Rightarrow \quad-x e^{-x}+e^{-x}=0 \Rightarrow x=1 \\
& f^{\prime \prime}(x)=x e^{-x}-e^{-x}-e^{-x}=(x-2) e^{-x} \\
& f^{\prime \prime}(1)=(1-2) e^{-1}=-\mathrm{ve}
\end{aligned}
$$
$f(x)$ is maximum at $x=1$.
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