The domain of the real-valued function
\[
f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}}
\]
is:
\[ f(x) = \sqrt{\frac{2x^2 - 7x + 5}{3x^2 - 5x - 2}} \] is:
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- \( (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty) \)
- \( (-\infty, 1) \cup (2, \infty) \)
- \( (-\frac{1}{3}, \frac{5}{2}) \)
- \( (-\infty, -\frac{1}{3}] \cup [\frac{5}{2}, \infty) \)
The Correct Option is A
Solution and Explanation
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: \[ (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty). \]
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