Domain of Logarithmic Function with Absolute Value Fraction

To find the domain of the function $f(x) = \sqrt{\log_{(0.6)} \left| \frac{2x-5}{x^2-4} \right|}$ we must satisfy two primary conditions:

1. Logarithmic Argument Condition: The argument of the logarithm must be strictly positive: $\left| \frac{2x-5}{x^2-4} \right| > 0$ This requires that the numerator is not zero and the denominator is defined and non-zero: $2x - 5 \neq 0 \implies x \neq \frac{5}{2}$ $x^2 - 4 \neq 0 \implies x \neq \pm 2$

2. Square Root Condition: The quantity under the square root must be non-negative: $\log_{(0.6)} \left| \frac{2x-5}{x^2-4} \right| \ge 0$

   Since the base of the logarithm is $0.6$ (which is strictly between $0$ and $1$), applying the exponential function flips the inequality direction: $\left| \frac{2x-5}{x^2-4} \right| \le (0.6)^0 = 1$

Combining these conditions gives: $0 < \left| \frac{2x-5}{x^2-4} \right| \le 1$

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Solving the Inequality $\left| \frac{2x-5}{x^2-4} \right| \le 1$:

This absolute value inequality is equivalent to the double inequality: $-1 \le \frac{2x-5}{x^2-4} \le 1$

Case 1: $\frac{2x-5}{x^2-4} \le 1$

$\frac{2x-5}{x^2-4} - 1 \le 0$ $\frac{2x-5 - (x^2-4)}{x^2-4} \le 0$ $\frac{-(x^2 - 2x + 1)}{x^2-4} \le 0$ $\frac{(x-1)^2}{x^2-4} \ge 0$

Since $(x-1)^2 \ge 0$ for all real $x$:

• The numerator is $0$ at $x = 1$, which satisfies the inequality.
• For $x \neq 1$, $(x-1)^2 > 0$, so we require the denominator to be strictly positive: $x^2 - 4 > 0 \implies x \in (-\infty, -2) \cup (2, \infty)$

Thus, the solution set for Case 1 is: $S_1 = (-\infty, -2) \cup \{1\} \cup (2, \infty)$

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Case 2: $\frac{2x-5}{x^2-4} \ge -1$

$\frac{2x-5}{x^2-4} + 1 \ge 0$ $\frac{2x-5 + x^2-4}{x^2-4} \ge 0$ $\frac{x^2 + 2x - 9}{(x-2)(x+2)} \ge 0$

Finding the roots of the numerator $x^2 + 2x - 9 = 0$: $x = \frac{-2 \pm \sqrt{4 - 4(1)(-9)}}{2} = -1 \pm \sqrt{10}$

Using the sign scheme (wavy curve method) with critical points $-1-\sqrt{10} \approx -4.16$, $-2$, $2$, and $-1+\sqrt{10} \approx 2.16$: $S_2 = (-\infty, -1-\sqrt{10}] \cup (-2, 2) \cup [-1+\sqrt{10}, \infty)$

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Finding the Intersection $S_1 \cap S_2$:

1. For $x < -2$: $(-\infty, -2) \cap (-\infty, -1-\sqrt{10}] = (-\infty, -1-\sqrt{10}]$

2. For $-2 < x < 2$: $\{1\} \cap (-2, 2) = \{1\}$

3. For $x > 2$: $(2, \infty) \cap [-1+\sqrt{10}, \infty) = [-1+\sqrt{10}, \infty)$

Finally, we must exclude $x = \frac{5}{2} = 2.5$, which lies inside $[-1+\sqrt{10}, \infty)$: $[-1+\sqrt{10}, \infty) \setminus \left\{\frac{5}{2}\right\} = \left[-1+\sqrt{10}, \frac{5}{2}\right) \cup \left(\frac{5}{2}, \infty\right)$

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Determining the Domain and Value of $a+b+c+d+e$:

The complete domain of $f(x)$ is: $(-\infty, -1-\sqrt{10}] \cup \{1\} \cup \left[-1+\sqrt{10}, \frac{5}{2}\right) \cup \left(\frac{5}{2}, \infty\right)$

Comparing this with the given form $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$, we get:

• $a = -1 - \sqrt{10}$
• $b = 1$
• $c = -1 + \sqrt{10}$
• $d = \frac{5}{2}$
• $e = \frac{5}{2}$

Calculating the required sum: $a + b + c + d + e = (-1 - \sqrt{10}) + 1 + (-1 + \sqrt{10}) + \frac{5}{2} + \frac{5}{2}$ $a + b + c + d + e = -2 + 1 + 5 = 4$