Analyse $f$ on $t<0$:

For $t<-1$: $f(t)=-\frac{t+1}{t^2}=-\frac{1}{t}-\frac{1}{t^2}$, so $f^{\prime }(t)=\frac{1}{t^2}+\frac{2}{t^3}=\frac{t+2}{t^3}$.

On $(-2,-1)$: $t+2>0$, $t^3<0\Rightarrow f^{\prime }(t)<0$ (decreasing).

For $-1<t<0$: $f(t)=\frac{t+1}{t^2}$, and a similar check gives $f^{\prime }(t)>0$ (increasing).

Hence the largest decreasing interval is $(-2,-1)$, so $(2\alpha ,\alpha )=(-2,-1)\Rightarrow \alpha =-1$.

Now $g(x)=2\ln (x-2)-x^2+4x+1$, $x>2$:

$g^{\prime }(x)=\frac{2}{x-2}-2(x-2)=\frac{2[1-(x-2{)}^2]}{x-2}$.

$g^{\prime }(x)=0\Rightarrow (x-2{)}^2=1\Rightarrow x=3$ (taking $x>2$).

Sign of $g^{\prime }$: positive on $(2,3)$, negative on $(3,\infty )$, so $x=3$ is a local maximum.

$g(3)=2\ln 1-9+12+1=4$.