Domain of $f(x)=\sqrt{2-x}-\frac{1}{\sqrt{9-x^{2}}}$ is:

Q: Domain of $f(x)=\sqrt{2-x}-\frac{1}{\sqrt{9-x^{2}}}$ is:

To determine the domain of the function $ f(x) = \sqrt{2-x} - \frac{1}{\sqrt{9-x^2}} $, we need to consider the constraints imposed by each component of the function. 1. **Constraint from $\sqrt{2-x}$:** - The expression under the square root, $2-x$, must be non-negative for the square root to be defined in the real numbers. Therefore, we require: \[ 2-x \geq 0 \implies x \leq 2 \] 2. **Constraint from $\frac{1}{\sqrt{9-x^2}}$:** - The expression under the square root, $9-x^2$, must be positive for the square root to be defined and for the denominator to be non-zero. Therefore, we require: \[ 9-x^2 > 0 \implies x^2 < 9 \] - Solving $x^2 < 9$, we find: \[ -3 < x < 3 \] Combining these constraints: - From $\sqrt{2-x}$, we have $x \leq 2$. - From $\frac{1}{\sqrt{9-x^2}}$, we have $-3 < x < 3$. The intersection of these intervals is: - $-3 < x \leq 2$ Thus, the domain of $f(x)$ is $(-3, 2]$. **Explanation of why each option is incorrect except Option 3:** 1. **Option 1: (-3, 1)** - This interval is too restrictive because it does not include all values up to 2, which are allowed by the constraint $x \leq 2$. 2. **Option 2: (-3, 3)** - This interval includes values greater than 2, specifically up to 3, which violates the constraint $x \leq 2$. 3. **Option 4: (2, 3)** - This interval only includes values greater than 2, which violates the constraint $x \leq 2$. Additionally, it does not include any values less than or equal to 2, which are part of the domain. Therefore, the correct domain of the function is $(-3, 2]$, which is Option 3.

Question

Accepted Answer

To determine the domain of the function $ f(x) = \sqrt{2-x} - \frac{1}{\sqrt{9-x^2}} $, we need to consider the constraints imposed by each component of the function.

1. **Constraint from $\sqrt{2-x}$:**
   - The expression under the square root, $2-x$, must be non-negative for the square root to be defined in the real numbers. Therefore, we require:
     $$
     2-x \geq 0 \implies x \leq 2
     $$

2. **Constraint from $\frac{1}{\sqrt{9-x^2}}$:**
   - The expression under the square root, $9-x^2$, must be positive for the square root to be defined and for the denominator to be non-zero. Therefore, we require:
     $$
     9-x^2 > 0 \implies x^2 < 9
     $$
   - Solving $x^2 < 9$, we find:
     $$
     -3 < x < 3
     $$

Combining these constraints:
- From $\sqrt{2-x}$, we have $x \leq 2$.
- From $\frac{1}{\sqrt{9-x^2}}$, we have $-3 < x < 3$.

The intersection of these intervals is:
- $-3 < x \leq 2$

Thus, the domain of $f(x)$ is $(-3, 2]$.

**Explanation of why each option is incorrect except Option 3:**

1. **Option 1: (-3, 1)**
   - This interval is too restrictive because it does not include all values up to 2, which are allowed by the constraint $x \leq 2$.

2. **Option 2: (-3, 3)**
   - This interval includes values greater than 2, specifically up to 3, which violates the constraint $x \leq 2$.

3. **Option 4: (2, 3)**
   - This interval only includes values greater than 2, which violates the constraint $x \leq 2$. Additionally, it does not include any values less than or equal to 2, which are part of the domain.

Therefore, the correct domain of the function is $(-3, 2]$, which is Option 3.