Let $$(2a, a)$$ be the largest interval in which the function $$f(t)=\frac{|t+1|}{t^{2}},t < 0$$, is strictly decreasing. Then the local maximum value of the function $$g(x)=2\log_{e}(x-2)+a x^{2}+4x-a,x > 2$$, is______.

The function is $$f(t)=\dfrac{|t+1|}{t^{2}},\;t\lt 0$$.
To decide where $$f(t)$$ is strictly decreasing we must study its derivative. Because of the absolute value we treat the two sign-regions of $$t+1$$ separately.

$$t\lt -1\;(\;t+1\lt 0\;)$$ Here $$|t+1|=-(t+1)$$, so $$f(t)=\dfrac{-(t+1)}{t^{2}}=\dfrac{-t-1}{t^{2}}.$$

Using $$\dfrac{u}{v}\Rightarrow f'(t)=\dfrac{u'v-uv'}{v^{2}}$$ with $$u=-(t+1),\;u'=-1,\;v=t^{2},\;v'=2t,$$ we get $$f'(t)=\dfrac{(-1)t^{2}-\bigl(-(t+1)\bigr)2t}{t^{4}} =\dfrac{-t^{2}+2t(t+1)}{t^{4}} =\dfrac{t^{2}+2t}{t^{4}} =\dfrac{t(t+2)}{t^{4}} =\dfrac{t+2}{t^{3}}.$$

Since $$t^{3}\lt 0$$ for $$t\lt 0,$$ the sign of $$f'(t)$$ is opposite to the sign of $$t+2$$: • for $$t\lt -2$$, $$t+2\lt 0$$ ⇒ $$f'(t)\gt 0$$ (increasing) • for $$-2\lt t\lt -1$$, $$t+2\gt 0$$ ⇒ $$f'(t)\lt 0$$ (decreasing)

$$-1\lt t\lt 0\;(\;t+1\gt 0\;)$$ Now $$|t+1|=t+1,$$ so $$f(t)=\dfrac{t+1}{t^{2}}.$$

With $$u=t+1,\;u'=1,\;v=t^{2},\;v'=2t$$, $$f'(t)=\dfrac{1\cdot t^{2}-(t+1)2t}{t^{4}} =\dfrac{t^{2}-2t(t+1)}{t^{4}} =\dfrac{-t(t+2)}{t^{4}} =-\dfrac{t+2}{t^{3}}.$$

Again $$t^{3}\lt 0$$, so the sign of $$f'(t)$$ is the same as the sign of $$t+2$$: for every $$t\in(-1,0)$$ we have $$t+2\gt 0$$ ⇒ $$f'(t)\gt 0$$ (increasing).

Combining both cases, $$f(t)$$ is strictly decreasing only on the open interval $$(-2,-1).$$ The question states this interval as $$(2\alpha,\alpha),$$ hence $$2\alpha=-2,\;\alpha=-1.$$

Next, consider $$g(x)=2\log_{e}(x-2)+\alpha x^{2}+4x-\alpha,\;x\gt 2.$$ Substituting $$\alpha=-1$$ gives $$g(x)=2\ln(x-2)-x^{2}+4x+1.$$

Differentiate to locate critical points: $$g'(x)=\dfrac{2}{x-2}-2x+4.$$ Setting $$g'(x)=0$$, $$\dfrac{2}{x-2}-2x+4=0\quad\Rightarrow\quad \frac{1}{x-2}-x+2=0.$$ Multiply by $$x-2$$: $$1=(x-2)^{2}\;\Rightarrow\;x-2=\pm1.$$ Since $$x\gt 2,$$ take $$x-2=1$$ ⇒ $$x=3.$$

Second derivative: $$g''(x)=-\dfrac{2}{(x-2)^{2}}-2.$$ At $$x=3,$$ $$g''(3)=-2-2=-4\lt 0,$$ so $$x=3$$ is a point of local maximum.

Maximum value: $$g(3)=2\ln(3-2)-3^{2}+4\cdot3+1 =2\ln1-9+12+1 =0+4=4.$$

Therefore, the local maximum value of $$g(x)$$ is $$4$$.