The domain of the real-valued function f(x) = | vedclass

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Question

(A) For the function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ to be defined, the expression inside the square root must be non-negative:$\f

Vedclass · Accepted Answer

$\left(-\infty, -\frac{1}{3}\right) \cup [1, 2) \cup \left[\frac{5}{2}, \infty\right)$

Vedclass · Answer

(A) For the function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ to be defined, the expression inside the square root must be non-negative:$\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0$Factorizing the numerator and denominator:Numerator: $2x^2 - 7x + 5 = 2x^2 - 2x - 5x + 5 = 2x(x - 1) - 5(x - 1) = (2x - 5)(x - 1)$Denominator: $3x^2 - 5x - 2 = 3x^2 - 6x + x - 2 = 3x(x - 2) + 1(x - 2) = (3x + 1)(x - 2)$So, the inequality is $\frac{(2x - 5)(x - 1)}{(3x + 1)(x - 2)} \geq 0$The critical points are $x = -\frac{1}{3}, 1, 2, \frac{5}{2}$.Using the wavy curve method (sign scheme) on the number line:- For $x > \frac{5}{2}$, the expression is positive.- For $2 - For $1 - For $-\frac{1}{3} - For $x Including the points where the numerator is zero $(x = 1, \frac{5}{2})$ and excluding points where the denominator is zero $(x = -\frac{1}{3}, 2)$:The domain is $\left(-\infty, -\frac{1}{3}\right) \cup [1, 2) \cup \left[\frac{5}{2}, \infty\right)$.Thus, option $A$ is correct.

The domain of the real-valued function $f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}$ is

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