The function f(x) = 2log(x - 2) - x² + 4x + 1 increases on the interval

(2, 3)

Clearly, f(x) is defined for all x > 2.

We have,

$f(x)=2 \log (x-2)-x^2+4 x+1 $

$\Rightarrow f'(x)=\frac{2}{x-2}-2 x+4 $

$\Rightarrow f'(x)=2\left\{\frac{1-(x-2)^2}{x-2}\right\}=-\frac{2(x-1)(x-3)}{x-2} $

$\Rightarrow f'(x)=-\frac{2(x-1)(x-3)(x-2)}{(x-2)^2} $

∴ $f'(x)>0 $

$\Rightarrow -2(x-1)(x-3)(x-2)>0$

$\Rightarrow (x-1)(x-2)(x-3)<0 \Rightarrow x \in(-\infty, 1) \cup(2,3)$

Clearly, domain of $f(x)$ is $(2, \infty)$.

Thus, $f(x)$ is increasing on $(2,3)$.