Given that \( f \) is the function detine by \( f(x)=\sqrt{\frac{2-x}{3+x}} \) Determine whe the functian \( f(x) \) is continuous ano whether the is a left or. right continuen of the interval of Continuity

To determine the continuity of the function \( f(x) = \sqrt{\frac{2-x}{3+x}} \), we need to analyze the expression under the square root and the domain of the function. 1. **Identify the domain**: The function \( f(x) \) is defined when the expression inside the square root is non-negative, i.e., \[ \frac{2-x}{3+x} \geq 0. \] This inequality holds when both the numerator and denominator are either both positive or both negative. - **Numerator**: \( 2 - x \geq 0 \) implies \( x \leq 2 \). - **Denominator**: \( 3 + x > 0 \) implies \( x > -3 \). Now we need to find the intervals where the fraction is non-negative. We can analyze the critical points: - The numerator \( 2 - x = 0 \) at \( x = 2 \). - The denominator \( 3 + x = 0 \) at \( x = -3 \). We can test the intervals determined by these points: \( (-\infty, -3) \), \( (-3, 2) \), and \( (2, \infty) \). - For \( x < -3 \) (e.g., \( x = -4 \)): \[ \frac{2 - (-4)}{3 + (-4)} = \frac{6}{-1} < 0. \] - For \( -3 < x < 2 \) (e.g., \( x = 0 \)): \[ \frac{2 - 0}{3 + 0} = \frac{2}{3} > 0. \] - For \( x > 2 \) (e.g., \( x = 3 \)): \[ \frac{2 - 3}{3 + 3} = \frac{-1}{6} < 0. \] Thus, the function \( f(x) \) is defined and non-negative in the interval \( (-3, 2] \). 2. **Check continuity**: The function \( f(x) \) is continuous wherever it is defined, as it is a composition of continuous functions (the square root and the rational function) in the interval \( (-3, 2] \). 3. **Endpoints**: We need to check the behavior at the endpoints: - At \( x = -3 \), the function is not defined (the denominator becomes zero). - At \( x = 2 \), the function is defined and \( f(2) = \sqrt{\frac{2-2}{3+2}} = \sqrt{0} = 0 \). 4. **Conclusion**: The function \( f(x) \) is continuous on the interval \( (-3, 2] \). It is not continuous at \( x = -3 \) (not defined) and is continuous from the right at \( x = -3 \) (as we approach from the right, the function is defined). The left endpoint \( -3 \) is not included in the interval of continuity. Thus, the interval of continuity for \( f(x) \) is \( (-3, 2] \).