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If $f(x)=e^{x}(x-2)^{2},$ then
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Verified Answer
The correct answer is:
$f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing $\operatorname{lin}(0,2)$
Given function is, $f(x)=e^{x}(x-2)^{2}$
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=e^{x}(x-2)^{2}+2(x-2) e^{x} \\
&=e^{x}(x-2)(x-2+2)=x(x-2) e^{x}
\end{aligned}
$$
Now, sign scheme of $f^{\prime}(x)$ is
So, $t$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in (0,2)
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=e^{x}(x-2)^{2}+2(x-2) e^{x} \\
&=e^{x}(x-2)(x-2+2)=x(x-2) e^{x}
\end{aligned}
$$
Now, sign scheme of $f^{\prime}(x)$ is
So, $t$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in (0,2)
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