The complete set of values of x for which f(x | vedclass

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(B) Given $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$. For $f(x)$ to be defined,we must have $x - 2 > 0$,which implies $x > 2$. To find the interval where

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$(2, 3)$

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(B) Given $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$. For $f(x)$ to be defined,we must have $x - 2 > 0$,which implies $x > 2$. To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$: $f'(x) = \frac{2}{x - 2} - 2x + 4$. Simplifying the expression: $f'(x) = \frac{2 - 2x(x - 2) + 4(x - 2)}{x - 2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x - 2} = \frac{-2x^2 + 8x - 6}{x - 2}$. Factoring the numerator: $f'(x) = \frac{-2(x^2 - 4x + 3)}{x - 2} = \frac{-2(x - 3)(x - 1)}{x - 2}$. For $f(x)$ to be increasing,$f'(x) > 0$: $\frac{-2(x - 3)(x - 1)}{x - 2} > 0 \Rightarrow \frac{(x - 3)(x - 1)}{x - 2} Using the wavy curve method for the intervals $(-\infty, 1) \cup (2, 3)$,we have $f'(x) > 0$. Since the domain of $f(x)$ is $x > 2$,the intersection of $(-\infty, 1) \cup (2, 3)$ and $(2, \infty)$ is $(2, 3)$. Thus,the function is increasing on $(2, 3)$.

The complete set of values of $x$ for which $f(x) = 2 \log_e(x - 2) - x^2 + 4x + 1$ is increasing is:

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