5. Express the domain of the following functions in interval notation. (a) \( f(x)=\frac{4}{2-\sqrt{x+3}} \). (b) \( h(x)=\log _{10} \frac{2 x+3}{x-5} \).

Below is the step‐by‐step solution for each part. --- ### (a) Function: \( f(x)=\frac{4}{2-\sqrt{x+3}} \) 1. **Domain from the Square Root:** The expression under the square root must be nonnegative: \[ x+3 \ge 0 \quad \Longrightarrow \quad x \ge -3. \] 2. **Denominator Constraint:** The denominator cannot be zero. Set the denominator equal to zero and solve: \[ 2-\sqrt{x+3} = 0 \quad \Longrightarrow \quad \sqrt{x+3} = 2. \] Square both sides: \[ x+3 = 4 \quad \Longrightarrow \quad x = 1. \] So, \( x=1 \) must be excluded from the domain. 3. **Combine the Conditions:** The domain requires that \( x \ge -3 \) and \( x \neq 1 \). **Domain in Interval Notation:** \[ [-3, 1) \cup (1, \infty). \] --- ### (b) Function: \( h(x)=\log_{10} \frac{2x+3}{x-5} \) 1. **Logarithm Argument Requirement:** The argument of the logarithm must be positive: \[ \frac{2x+3}{x-5} > 0. \] 2. **Find Critical Values:** Set the numerator and denominator equal to zero: - Numerator: \[ 2x+3=0 \quad \Longrightarrow \quad x=-\frac{3}{2}. \] - Denominator: \[ x-5=0 \quad \Longrightarrow \quad x=5. \] 3. **Test the Intervals:** These critical values divide the number line into three intervals: - **Interval I:** \( x < -\frac{3}{2} \) Choose \( x=-2 \): \[ 2(-2)+3 = -4+3 = -1 \quad \text{(negative)}, \] \[ -2-5 = -7 \quad \text{(negative)}. \] A negative divided by a negative is positive. - **Interval II:** \( -\frac{3}{2} < x < 5 \) Choose \( x=0 \): \[ 2(0)+3 = 3 \quad \text{(positive)}, \] \[ 0-5 = -5 \quad \text{(negative)}. \] A positive divided by a negative is negative. - **Interval III:** \( x > 5 \) Choose \( x=10 \): \[ 2(10)+3 = 20+3 = 23 \quad \text{(positive)}, \] \[ 10-5 = 5 \quad \text{(positive)}. \] A positive divided by a positive is positive. 4. **Conclusion for the Inequality:** The fraction \(\frac{2x+3}{x-5}\) is positive when: \[ x \in (-\infty, -\frac{3}{2}) \quad \text{or} \quad x \in (5, \infty). \] Also, note that \( x=-\frac{3}{2} \) causes the numerator to be zero (giving the value 0, which is not allowed since \( \log_{10}0 \) is undefined), and \( x=5 \) is excluded because it makes the denominator zero. **Domain in Interval Notation:** \[ (-\infty, -\frac{3}{2}) \cup (5, \infty). \] --- ### Final Answers - **(a)** The domain of \( f(x)=\frac{4}{2-\sqrt{x+3}} \) is: \[ [-3, 1) \cup (1, \infty). \] - **(b)** The domain of \( h(x)=\log_{10} \frac{2x+3}{x-5} \) is: \[ (-\infty, -\frac{3}{2}) \cup (5, \infty). \]