Domain of the function f(x) = `(x+0. 5)^log(0. 5+x)((x^2+2x-3)/(4x^2-4x-3)))`

Explanation

To find the domain of the function \( f(x) = (x + 0.5)^{\log(0.5 + x) \left( \frac{x^2 + 2x - 3}{4x^2 - 4x - 3} \right)} \), we need to ensure that all components of the function are defined.

### Step 1: Determine the conditions for the logarithm
The logarithm function \( \log(0.5 + x) \) is defined when its argument is positive:
\[ 0.5 + x > 0 \implies x > -0.5 \]

### Step 2: Ensure the base of the logarithm is valid
The base of the logarithm, \( 0.5 + x \), must also be positive and cannot be equal to 1:
\[ 0.5 + x \neq 1 \implies x \neq 0.5 \]

### Step 3: Analyze the fraction
Next, we need to ensure that the fraction \( \frac{x^2 + 2x - 3}{4x^2 - 4x - 3} \) is defined and positive.

#### Step 3.1: Factor the numerator
The numerator \( x^2 + 2x - 3 \) can be factored:
\[ x^2 + 2x - 3 = (x - 1)(x + 3) \]

#### Step 3.2: Factor the denominator
The denominator \( 4x^2 - 4x - 3 \) can be factored:
\[ 4x^2 - 4x - 3 = (2x + 3)(2x - 1) \]

### Step 4: Set up the inequality
We need to find where the fraction is positive:
\[ \frac{(x - 1)(x + 3)}{(2x + 3)(2x - 1)} > 0 \]

### Step 5: Find critical points
The critical points from the factors are:
- From the numerator: \( x = 1 \) and \( x = -3 \)
- From the denominator: \( x = -\frac{3}{2} \) and \( x = \frac{1}{2} \)

### Step 6: Test intervals
We will test the sign of the fraction in the intervals determined by these critical points:
- Intervals: \( (-\infty, -3) \), \( (-3, -\frac{3}{2}) \), \( (-\frac{3}{2}, \frac{1}{2}) \), \( (\frac{1}{2}, 1) \), \( (1, \infty) \)

1. **For \( x < -3 \)**: Choose \( x = -4 \):
\[ \frac{(-4 - 1)(-4 + 3)}{(2(-4) + 3)(2(-4) - 1)} = \frac{(-5)(-1)}{(-5)(-9)} > 0 \]
Positive.

2. **For \( -3 < x < -\frac{3}{2} \)**: Choose \( x = -2 \):
\[ \frac{(-2 - 1)(-2 + 3)}{(2(-2) + 3)(2(-2) - 1)} = \frac{(-3)(1)}{(-1)(-5)} < 0 \]
Negative.

3. **For \( -\frac{3}{2} < x < \frac{1}{2} \)**: Choose \( x = 0 \):
\[ \frac{(0 - 1)(0 + 3)}{(2(0) + 3)(2(0) - 1)} = \frac{(-1)(3)}{(3)(-1)} > 0 \]
Positive.

4. **For \( \frac{1}{2} < x < 1 \)**: Choose \( x = 0.75 \):
\[ \frac{(0.75 - 1)(0.75 + 3)}{(2(0.75) + 3)(2(0.75) - 1)} = \frac{(-0.25)(3.75)}{(4.5)(-0.5)} < 0 \]
Negative.

5. **For \( x > 1 \)**: Choose \( x = 2 \):
\[ \frac{(2 - 1)(2 + 3)}{(2(2) + 3)(2(2) - 1)} = \frac{(1)(5)}{(7)(3)} > 0 \]
Positive.

### Step 7: Combine intervals
The fraction is positive in the intervals:
- \( (-\infty, -3) \)
- \( (-\frac{3}{2}, \frac{1}{2}) \)
- \( (1, \infty) \)

### Step 8: Combine with logarithm conditions
Now we combine the conditions:
1. From the logarithm: \( x > -0.5 \) and \( x \neq 0.5 \)
2. From the fraction: \( (-\infty, -3) \cup (-\frac{3}{2}, \frac{1}{2}) \cup (1, \infty) \)

### Final Domain
The final domain is:
\[ (-\frac{3}{2}, -0.5) \cup (1, \infty) \]