Question

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 the function \[ g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha,\; x>2 \] is
 

Hint:

First determine parameters using monotonicity, then substitute into the second function.

Answer 1

Correct Answer: 2

To determine the largest interval where the function \( f(t)=\frac{|t+1|}{t^2},\, t<0 \) is strictly decreasing, we first need to analyze the function. Consider the domain where \( t<0 \). For \( t<-1 \), \( |t+1|=-(t+1) \), simplifying \( f(t) \) to \( \frac{-(t+1)}{t^2} \). For \( -1<t<0 \), \( |t+1|=t+1 \), simplifying \( f(t) \) to \( \frac{t+1}{t^2} \).

The derivative \( f'(t) \) tells us about the monotonicity. Compute \( f'(t) \) for \( t<-1 \): \( f'(t)=\frac{d}{dt}\left(\frac{-(t+1)}{t^2}\right)=\frac{2t+t}{t^3}=\frac{2+t}{t^3} \). It's decreasing if \( 2+t<0 \) which gives \( t<-2 \). Repeat for \( -1<t<0 \) but since \( t<-1 \), concentrate on \( t<-2 \).

Thus, \( (2\alpha,\alpha)=(2(-2),-1)=(-4,-1) \) is the largest interval where \( f(t) \) decreases.

Next, investigate the function \( g(x)=2\log_e(x-2)+\alpha x^2+4x-\alpha \). Here, \( \alpha=-1 \), making \( g(x)=2\log_e(x-2)-x^2+4x+1 \).Taking derivative \( g'(x) = \frac{2}{x-2}-2x+4 \) and setting it to zero for local extrema: \( \frac{2}{x-2}=2x-4 \). Solving, \( 2=2x(x-2)-4(x-2) \Rightarrow 2=-4/x+2x\) which reduces to a quadratic form.

Find \( g''(x) \) for concavity: \( g''(x)=\frac{-2}{(x-2)^2}-2 \), negative confirming maximum. Solve \( 4x-8=2(x-2) \Rightarrow x^2-x-2=0 \) leaves singular maximum point within domain as valid interpreted within valid range of \( x>2 \).

Therefore, confirming range 2 meets ultimate elevation \( x=2 \) or reciprocal symmetry simplifies effort.