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Let $(2\alpha,\alpha)$ 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 $g(x)=2\log(x-2)+\alpha x^2+4x-\alpha,; x>2$ is ______.





Solution



From graph ⇒ decreasing on $(-2,-1)$

$\alpha=-1$

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

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

$=\frac{1-(x-2)^2}{x-2}$

Critical point ⇒ $x=3$

$g(3)=2\log1-9+12+1=4$


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