If f(x)=e^{x}(x-2)^{2}, then | vedclass

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(A) Given function is $f(x)=e^{x}(x-2)^{2}$.To find the intervals of increase and decrease, we find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \f

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$f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in $(0, 2)$

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(A) Given function is $f(x)=e^{x}(x-2)^{2}$.To find the intervals of increase and decrease, we find the derivative $f^{\prime}(x)$:$f^{\prime}(x) = \frac{d}{dx}[e^{x}(x-2)^{2}]$Using the product rule:$f^{\prime}(x) = e^{x}(x-2)^{2} + e^{x} \cdot 2(x-2)$$f^{\prime}(x) = e^{x}(x-2)[(x-2) + 2]$$f^{\prime}(x) = x(x-2)e^{x}$Now, we determine the sign of $f^{\prime}(x)$:Since $e^{x} > 0$ for all $x \in \mathbb{R}$, the sign of $f^{\prime}(x)$ depends on $x(x-2)$.- For $x \in (-\infty, 0)$, $x  0$ (increasing).- For $x \in (0, 2)$, $x > 0$ and $(x-2) - For $x \in (2, \infty)$, $x > 0$ and $(x-2) > 0$, so $f^{\prime}(x) > 0$ (increasing).Thus, $f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in $(0, 2)$.

If $f(x)=e^{x}(x-2)^{2}$, then

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