Problem Analysis

To find the domain of the function $f(x)$, we need to find the values of $x$ for which both the logarithmic term and the inverse sine term are defined. The domain will be the intersection of these two sets of values.

Step 1: Domain of the Logarithmic Term

For $\log_e \left( \frac{2x-3}{5+4x} \right)$, the argument must be strictly positive:

$\frac{2x-3}{5+4x} > 0$

The critical points are $x = \frac{3}{2}$ and $x = -\frac{5}{4}$. Using the sign-interval method (wavy curve):

• For $x > \frac{3}{2}$, the expression is positive.
• For $-\frac{5}{4} < x < \frac{3}{2}$, the expression is negative.
• For $x < -\frac{5}{4}$, the expression is positive.

So, the domain $D_1 = (-\infty, -1.25) \cup (1.5, \infty)$.

Step 2: Domain of the Inverse Sine Term

For $\sin^{-1} \left( \frac{4+3x}{2-x} \right)$, the argument must lie in the range $[-1, 1]$:

$-1 \leq \frac{4+3x}{2-x} \leq 1$

Part A: $\frac{4+3x}{2-x} \geq -1$

$\frac{4+3x}{2-x} + 1 \geq 0 \implies \frac{4+3x + 2-x}{2-x} \geq 0 \implies \frac{2x+6}{2-x} \geq 0 \implies \frac{x+3}{x-2} \leq 0$

This gives the interval $x \in [-3, 2)$.

Part B: $\frac{4+3x}{2-x} \leq 1$

$\frac{4+3x}{2-x} - 1 \leq 0 \implies \frac{4+3x - (2-x)}{2-x} \leq 0 \implies \frac{4x+2}{2-x} \leq 0 \implies \frac{2x+1}{x-2} \geq 0$

This gives the interval $x \in (-\infty, -0.5] \cup (2, \infty)$.

The intersection of Part A and Part B is $D_2 = [-3, -0.5]$.

Step 3: Intersection of Domains

The total domain $D = D_1 \cap D_2$:

$D = \{(-\infty, -1.25) \cup (1.5, \infty)\} \cap [-3, -0.5]$

The overlapping interval is $D = [-3, -1.25)$, which can be written as $[-3, -\frac{5}{4})$.

Step 4: Solve for $\alpha^2 + 4\beta$

Comparing $D = [-3, -\frac{5}{4})$ with the given form $[\alpha, \beta)$:

• $\alpha = -3$
• $\beta = -\frac{5}{4}$

$\alpha^2 + 4\beta = (-3)^2 + 4(-\frac{5}{4})$

$\alpha^2 + 4\beta = 9 - 5 = 4$

Final Answer

The value of $\alpha^2 + 4\beta$ is 4.