Let f(x) = (x - 3)^{2018}(x - 2)^{2019}, x in | vedclass

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(B) The given function is $f(x) = (x - 3)^{2018}(x - 2)^{2019}$.Applying the product rule, $f'(x) = 2018(x - 3)^{2017}(x - 2)^{2019} + 2019(x - 3)^{20

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$\frac{20186}{4037} - 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$

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(B) The given function is $f(x) = (x - 3)^{2018}(x - 2)^{2019}$.Applying the product rule, $f'(x) = 2018(x - 3)^{2017}(x - 2)^{2019} + 2019(x - 3)^{2018}(x - 2)^{2018}$.Factoring out common terms, $f'(x) = (x - 3)^{2017}(x - 2)^{2018} \{2018(x - 2) + 2019(x - 3)\}$.Simplifying the expression inside the braces, $2018x - 4036 + 2019x - 6057 = 4037x - 10093$.Thus, $f'(x) = (x - 3)^{2017}(x - 2)^{2018} \cdot 4037 \left( x - \frac{10093}{4037} \right)$.The critical points are $x = 3, 2, \frac{10093}{4037}$.Analyzing the sign of $f'(x)$ around $x = \alpha = \frac{10093}{4037}$, the derivative changes from positive to negative, indicating a local maximum.At $x = \alpha = \frac{10093}{4037}$, we have $x - 3 = \frac{10093 - 12111}{4037} = -\frac{2018}{4037}$ and $x - 2 = \frac{10093 - 8074}{4037} = \frac{2019}{4037}$.Thus, $f(\alpha) = \left( -\frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.Therefore, $2\alpha + 3f(\alpha) = 2 \left( \frac{10093}{4037} \right) + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019} = \frac{20186}{4037} + 3 \left( \frac{2018}{4037} \right)^{2018} \left( \frac{2019}{4037} \right)^{2019}$.Note: The provided option $B$ has a minus sign, but based on the calculation, the result is positive.

Let $f(x) = (x - 3)^{2018}(x - 2)^{2019}, x \in R$. If $f(\alpha)$ is a relative maximum of $f$ at $x = \alpha$, then $2\alpha + 3f(\alpha) =$

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