Question:medium

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

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For rational functions inside a square root, ensure the numerator is non-negative while avoiding zeros of the denominator.
Updated On: Nov 26, 2025
  • \( (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty) \)
  • \( (-\infty, 1) \cup (2, \infty) \)
  • \( (-\frac{1}{3}, \frac{5}{2}) \)
  • \( (-\infty, -\frac{1}{3}] \cup [\frac{5}{2}, \infty) \)
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The Correct Option is A

Solution and Explanation

Step 1: Determine domain restrictions. For \( f(x) \) to be defined, the radicand must be non-negative: \[ \frac{2x^2 - 7x + 5}{3x^2 - 5x - 2} \geq 0. \] The denominator also cannot be zero: \[ 3x^2 - 5x - 2 eq 0. \]
Step 2: Find numerator roots. Solve \( 2x^2 - 7x + 5 = 0 \). Factoring yields \( (2x - 5)(x - 1) = 0 \), so \( x = \frac{5}{2}, 1 \).
Step 3: Find denominator roots. Solve \( 3x^2 - 5x - 2 = 0 \). Factoring yields \( (3x + 1)(x - 2) = 0 \), so \( x = -\frac{1}{3}, 2 \).
Step 4: Analyze sign changes using a number line. Critical points are \( -\frac{1}{3}, 1, 2, \frac{5}{2} \). These define intervals: \( (-\infty, -\frac{1}{3}), (-\frac{1}{3}, 1), (1,2), (2, \frac{5}{2}), (\frac{5}{2}, \infty) \). Testing values reveals the function is non-negative in \( (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty) \).
Step 5: Conclusion. The domain is \( (-\infty, -\frac{1}{3}) \cup [1,2) \cup [\frac{5}{2}, \infty) \).

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