If the domain of the function f(x) = log_e le | vedclass

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(B) For the function to be defined,we need: $1$) $\frac{2x+3}{4x^2+x-3} > 0$ $2$) $-1 \leq \frac{2x-1}{x+2} \leq 1$ Step $1$: Solve $\frac{2x+3}{(4x-3

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$12$

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(B) For the function to be defined,we need: $1$) $\frac{2x+3}{4x^2+x-3} > 0$ $2$) $-1 \leq \frac{2x-1}{x+2} \leq 1$ Step $1$: Solve $\frac{2x+3}{(4x-3)(x+1)} > 0$. The critical points are $x = -3/2, -1, 3/4$. Using the wavy curve method,the solution is $x \in (-3/2, -1) \cup (3/4, \infty)$. Step $2$: Solve $-1 \leq \frac{2x-1}{x+2} \leq 1$. Part $A$: $\frac{2x-1}{x+2} + 1 \geq 0 \implies \frac{3x+1}{x+2} \geq 0$. Solution: $x \in (-\infty, -2) \cup [-1/3, \infty)$. Part $B$: $\frac{2x-1}{x+2} - 1 \leq 0 \implies \frac{x-3}{x+2} \leq 0$. Solution: $x \in (-2, 3]$. Intersection of Part $A$ and Part $B$: $x \in [-1/3, 3]$. Step $3$: Find the intersection of Step $1$ and Step $2$. $((-3/2, -1) \cup (3/4, \infty)) \cap [-1/3, 3] = (3/4, 3]$. Thus,$\alpha = 3/4$ and $\beta = 3$. Step $4$: Calculate $5\beta - 4\alpha$. $5(3) - 4(3/4) = 15 - 3 = 12$.

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

If the domain of the function $f(x) = \log_e \left( \frac{2x+3}{4x^2+x-3} \right) + \cos^{-1} \left( \frac{2x-1}{x+2} \right)$ is $(\alpha, \beta]$,then the value of $5\beta - 4\alpha$ is equal to

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