Application of Derivatives Part 3

Concept:The largest decreasing interval of a piecewise function determines the parameter $\alpha$, which is then used to find the local maximum of another function by analyzing its derivative.Explanation:First, consider $f(t)=\frac{|t+1|}{t^2}$ for $t<0$.The critical point is $t=-1$ (where $|t+1|=0$).Write piecewise form: for $t \le -1$, $f(t)=-\frac{t+1}{t^2}$; for $-1 < t<0$, $f(t)=\frac{t+1}{t^2}$.Derivatives: for $t<-1$, $f'(t)=\frac{t+2}{t^3}$; for $-1 < t<0$, $f'(t)=-\frac{t+2}{t^3}$.For $t<-1$, $f'(t)=0$ at $t=-2$.On $(-\infty,-2)$, $f'(t)>0$ (increasing). On $(-2,-1)$, $f'(t)<0$ (decreasing).On $(-1,0)$, $f'(t)>0$ (increasing).Thus the largest decreasing interval is $(-2,-1)$.Given that $(2\alpha,\alpha)$ is this interval, we have $2\alpha=-2$ and $\alpha=-1$.Now $g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha$, $x>2$.Substitute $\alpha=-1$: $g(x)=2\ln(x-2)-x^2+4x+1$.Find derivative: $g'(x)=\frac{2}{x-2}-2x+4 = \frac{-2x^2+8x-6}{x-2} = \frac{-2(x-3)(x-1)}{x-2}$.For $x>2$, the only critical point is $x=3$ (since $x-1>0$).Sign of $g'(x)$: left of 3, positive; right of 3, negative. So $x=3$ is a local maximum.Compute $g(3)=2\ln(1)-9+12+1 = 0+4=4$.Answer:4 (Option A)