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(B) To find the interval where the function $f(x) = 2 \ln(x-2) - x^2 + 4x + 1$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx} [2 \

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$(2, 3)$

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(B) To find the interval where the function $f(x) = 2 \ln(x-2) - x^2 + 4x + 1$ is increasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx} [2 \ln(x-2) - x^2 + 4x + 1] = \frac{2}{x-2} - 2x + 4$.Setting $f'(x) > 0$ for the function to be increasing:$\frac{2}{x-2} - 2(x-2) > 0$.Divide by $2$: $\frac{1}{x-2} - (x-2) > 0$.Let $u = x-2$. Since the domain of $\ln(x-2)$ is $x > 2$,we have $u > 0$.The inequality becomes $\frac{1}{u} - u > 0 \implies \frac{1-u^2}{u} > 0$.Since $u > 0$,we have $1-u^2 > 0 \implies u^2 Given $u > 0$,we have $0 Substituting $u = x-2$,we get $0 Thus,the function increases on the interval $(2, 3)$.

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

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