If the domain of the function f(x) = _ (0.6) (|(2x-5)/(x^2-4)|) is...

For the function $f(x) = \sqrt{\log_{0.6}\left(\left|\dfrac{2x-5}{x^2-4}\right|\right)}$$to be defined, two conditions must be satisfied:<br><br>1. The argument of the logarithm must be strictly positive:<br>$\left|\dfrac{2x-5}{x^2-4}\right| > 0 \implies 2x - 5 \neq 0 \implies x \neq \dfrac{5}{2}$<br>Also, the denominator cannot be zero:<br>$x^2 - 4 \neq 0 \implies x \neq \pm 2$<br><br>2. The expression inside the square root must be non-negative:<br>$\log_{0.6}\left(\left|\dfrac{2x-5}{x^2-4}\right|\right) \ge 0$<br><br>Since the base of the logarithm is$0.6 < 1$, the inequality sign reverses when removing the logarithm:<br>\left|\dfrac{2x-5}{x^2-4}\right| \le (0.6)^0$
\left|\dfrac{2x-5}{x^2-4}\right| \le 1$<br><br>Since both sides are non-negative, we can square both sides:<br>$(2x-5)^2 \le (x^2-4)^2$<br>$(x^2-4)^2 - (2x-5)^2 \ge 0$<br><br>Using the identity$A^2 - B^2 = (A-B)(A+B)$:<br>$(x^2 - 4 - 2x + 5)(x^2 - 4 + 2x - 5) \ge 0$<br>$(x^2 - 2x + 1)(x^2 + 2x - 9) \ge 0$<br>$(x - 1)^2 (x^2 + 2x - 9) \ge 0$<br><br>Since$(x - 1)^2 \ge 0$for all real$x$, the inequality holds if:<br>$x - 1 = 0 \implies x = 1$<br>or<br>$x^2 + 2x - 9 \ge 0$<br><br>Finding the roots of$x^2 + 2x - 9 = 0$using the quadratic formula:<br>$x = \dfrac{-2 \pm \sqrt{4 - 4(1)(-9)}}{2} = \dfrac{-2 \pm \sqrt{40}}{2} = -1 \pm \sqrt{10}$<br><br>Thus,$x^2 + 2x - 9 \ge 0$gives:<br>$x \in (-\infty, -1 - \sqrt{10}] \cup [-1 + \sqrt{10}, \infty)$<br><br>Combining this with$x = 1$, the solution to the inequality is:<br>$x \in (-\infty, -1 - \sqrt{10}] \cup \{1\} \cup [-1 + \sqrt{10}, \infty)$<br><br>Now, we must exclude the restricted values$x = \pm 2 and x = \dfrac{5}{2}$.<br>Note that$-1 - \sqrt{10} \approx -4.16$and$-1 + \sqrt{10} \approx 2.16$.<br>The values$-2 and 2$do not fall in the above intervals, so they are already excluded.<br>However,$\dfrac{5}{2} = 2.5$, which lies in the interval$[-1 + \sqrt{10}, \infty)$. We must exclude it by splitting the interval:<br>$[-1 + \sqrt{10}, \dfrac{5}{2}) \cup (\dfrac{5}{2}, \infty)$<br><br>The final domain of the function is:<br>$(-\infty, -1 - \sqrt{10}] \cup \{1\} \cup [-1 + \sqrt{10}, \dfrac{5}{2}) \cup (\dfrac{5}{2}, \infty)$<br><br>Comparing this with the given domain$(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$, we get:<br>$a = -1 - \sqrt{10}$<br>$b = 1$<br>$c = -1 + \sqrt{10}$<br>$d = \dfrac{5}{2}$<br>$e = \dfrac{5}{2}$<br><br>We need to find the value of$a + b + c + d + e$:<br>$a + b + c + d + e = (-1 - \sqrt{10}) + 1 + (-1 + \sqrt{10}) + \dfrac{5}{2} + \dfrac{5}{2}$<br>$= -1 - \sqrt{10} + 1 - 1 + \sqrt{10} + 5$<br>$= 4$<br><br>Answer:$4