Question

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

• A. \( (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty) \)
• B. \( (-\infty, 1) \cup (2, \infty) \)
• C. \( (-\frac{1}{3}, \frac{5}{2}) \)
• D. \( (-\infty, -\frac{1}{3}] \cup [\frac{5}{2}, \infty) \)

Hint:

For rational functions inside a square root, ensure the numerator is non-negative while avoiding zeros of the denominator.

Answer 1

The Correct Option is A

Step 1: {Find the domain restrictions} 
For the function \( f(x) \) to be defined, the expression inside the square root must be non-negative: \[ \frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0. \] Also, the denominator must not be zero, i.e., \[ 3x^2 - 5x - 2 \neq 0. \] 
Step 2: {Find the zeros of the numerator} 
Solving: \[ 2x^2 - 7x + 5 = 0. \] Factoring: \[ (2x - 5)(x - 1) = 0. \] \[ x = \frac{5}{2}, 1. \] 
Step 3: {Find the zeros of the denominator} 
Solving: \[ 3x^2 - 5x - 2 = 0. \] Factoring: \[ (3x + 1)(x - 2) = 0. \] \[ x = -\frac{1}{3}, 2. \] 
Step 4: {Analyze sign changes using a number line} 
The critical points partition the number line into intervals: \[ (-\infty, -\frac{1}{3}), (-\frac{1}{3}, 1), (1,2), (2, \frac{5}{2}), (\frac{5}{2}, \infty). \] By testing values in each interval, the function is non-negative in: \[ (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty). \] 
Step 5: {Conclusion} 
Thus, the domain is: \[ (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty). \]

Answer 2

Step 1: Analyze the given function
We are given the function: \[ f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}. \] To determine the domain, we need to ensure that the expression inside the square root is non-negative and that the denominator is not zero. That is, we require: \[ \frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0 \quad \text{and} \quad 3x^2 - 5x - 2 \neq 0. \]

Step 2: Solve the inequality
We first solve for the roots of the numerator and denominator.

Numerator: \( 2x^2 - 7x + 5 = 0 \)
Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -7, c = 5 \): \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(5)}}{2(2)} = \frac{7 \pm \sqrt{49 - 40}}{4} = \frac{7 \pm \sqrt{9}}{4} = \frac{7 \pm 3}{4}. \] So, the roots are: \[ x = \frac{7 + 3}{4} = \frac{10}{4} = \frac{5}{2}, \quad x = \frac{7 - 3}{4} = \frac{4}{4} = 1. \] Denominator: \( 3x^2 - 5x - 2 = 0 \)
Using the quadratic formula again: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6}. \] So, the roots are: \[ x = \frac{5 + 7}{6} = \frac{12}{6} = 2, \quad x = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3}. \] Step 3: Analyze the intervals
We now have the following critical points: \( x = -\frac{1}{3}, 1, 2, \frac{5}{2} \). The sign of the expression \( \frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \) changes across these points. By testing values in each interval formed by these critical points, we find the intervals where the expression is non-negative.

Step 4: Determine the domain
The function is defined and non-negative in the intervals where: \[ (-\infty, -\frac{1}{3}) \cup [1, 2) \cup [\frac{5}{2}, \infty). \] Thus, the domain of the function is:

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