Finding the Interval for Strictly Decreasing Function

We need to find the interval where the function $f(t) = \frac{|t+1|}{t^2}$ for $t < 0$ is strictly decreasing. This requires finding the derivative $f'(t)$ and determining where $f'(t) < 0$. We consider two cases based on the sign of $t+1$.

Case 1: $-1 < t < 0$

In this interval, $t+1 > 0$, so $|t+1| = t+1$. The function becomes $f(t) = \frac{t+1}{t^2} = t^{-1} + t^{-2}$.

The derivative is $f'(t) = -t^{-2} - 2t^{-3} = -\frac{1}{t^2} - \frac{2}{t^3} = \frac{-t-2}{t^3}$.

For $f(t)$ to be strictly decreasing, $f'(t) < 0$. So, $\frac{-t-2}{t^3} < 0$. Since $-1 < t < 0$, $t^3$ is negative. Thus, we need $-t-2 > 0$, which implies $t < -2$. This condition contradicts the interval $-1 < t < 0$. Therefore, the function is not strictly decreasing in this range.

Case 2: $t \le -1$

In this interval, $t+1 \le 0$, so $|t+1| = -(t+1)$. The function becomes $f(t) = \frac{-(t+1)}{t^2} = -t^{-1} - t^{-2}$.

The derivative is $f'(t) = t^{-2} + 2t^{-3} = \frac{1}{t^2} + \frac{2}{t^3} = \frac{t+2}{t^3}$.

For $f(t)$ to be strictly decreasing, $f'(t) < 0$. So, $\frac{t+2}{t^3} < 0$. Since $t \le -1$, $t^3$ is negative. Thus, we need $t+2 > 0$, which implies $t > -2$. Combining $t \le -1$ and $t > -2$, the interval where $f(t)$ is strictly decreasing is $(-2, -1)$.

Determining the value of $\alpha$

The problem states that the largest interval is $(2\alpha, \alpha)$. Comparing this with the derived interval $(-2, -1)$, we have:

• $2\alpha = -2 \implies \alpha = -1$
• $\alpha = -1$

Both conditions consistently give $\alpha = -1$.

Calculating the Local Maximum Value of $g(x)$

Substitute $\alpha = -1$ into the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$. The function becomes $g(x) = 2\log_e(x - 2) - x^2 + 4x + 1$, for $x > 2$.

Finding Critical Points

Calculate the first derivative $g'(x)$:

$g'(x) = \frac{d}{dx} (2\log_e(x - 2) - x^2 + 4x + 1) = \frac{2}{x-2} - 2x + 4$

Set $g'(x) = 0$ to find critical points:

$\frac{2}{x-2} - 2x + 4 = 0$

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

$1 = (x-2)^2$

$x-2 = \pm 1$

This gives $x = 3$ or $x = 1$. Since the domain is $x > 2$, the only relevant critical point is $x=3$.

Confirming Local Maximum

Calculate the second derivative $g''(x)$:

$g''(x) = \frac{d}{dx} (\frac{2}{x-2} - 2x + 4) = \frac{d}{dx} (2(x-2)^{-1} - 2x + 4) = -2(x-2)^{-2} - 2 = -\frac{2}{(x-2)^2} - 2$

Evaluate $g''(x)$ at the critical point $x=3$:

$g''(3) = -\frac{2}{(3-2)^2} - 2 = -\frac{2}{1^2} - 2 = -2 - 2 = -4$

Since $g''(3) < 0$, the function $g(x)$ has a local maximum at $x=3$.

Calculating the Local Maximum Value

Substitute $x=3$ into $g(x)$:

$g(3) = 2\log_e(3 - 2) - (3)^2 + 4(3) + 1$

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

$g(3) = 2(0) + 3 + 1 = 4$

The local maximum value of the function $g(x)$ is 4.