Answer
The domain of \( f(x)=\frac{4}{2-\sqrt{x+3}} \) is \([-3, 1) \cup (1, \infty)\).
The domain of \( h(x)=\log_{10} \frac{2x+3}{x-5} \) is \((-\infty, -\frac{3}{2}) \cup (5, \infty)\).
Solution
Below is the step‐by‐step solution for each part.
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### (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:**
\[
(-\infty, -\frac{3}{2}) \cup (5, \infty).
\]
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### 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:
\[
(-\infty, -\frac{3}{2}) \cup (5, \infty).
\]
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