`f(x)=sgn(cot^(- 1)x),g(x)=sgn(x^2-4x+5)`

Answer

To determine if the functions \( f(x) = \text{sgn}(\cot^{-1} x) \) and \( g(x) = \text{sgn}(x^2 - 4x + 5) \) are identical, we need to check three conditions: their domains, ranges, and whether they yield the same output for every \( x \) in their common domain.

### Step 1: Determine the Domain of \( f(x) \)

The function \( f(x) = \text{sgn}(\cot^{-1} x) \):
- The cotangent inverse function, \( \cot^{-1} x \), is defined for all real numbers \( x \).
- The output of \( \cot^{-1} x \) lies in the interval \( (0, \pi) \) for all \( x \in \mathbb{R} \).

**Domain of \( f(x) \):** \( \mathbb{R} \)

### Step 2: Determine the Range of \( f(x) \)

The signum function \( \text{sgn}(y) \) is defined as:
- \( 1 \) if \( y > 0 \)
- \( 0 \) if \( y = 0 \)
- \( -1 \) if \( y < 0 \)

Since \( \cot^{-1} x \) is always positive for all \( x \in \mathbb{R} \):
- Therefore, \( f(x) = \text{sgn}(\cot^{-1} x) = 1 \) for all \( x \in \mathbb{R} \).

**Range of \( f(x) \):** \( \{1\} \)

### Step 3: Determine the Domain of \( g(x) \)

The function \( g(x) = \text{sgn}(x^2 - 4x + 5) \):
- First, we analyze the quadratic expression \( x^2 - 4x + 5 \).
- The discriminant \( D \) of the quadratic is calculated as:
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 \]
- Since the discriminant is less than zero, the quadratic has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive).

**Domain of \( g(x) \):** \( \mathbb{R} \)

### Step 4: Determine the Range of \( g(x) \)

Since \( x^2 - 4x + 5 \) is always positive (it has no real roots and opens upwards):
- Therefore, \( g(x) = \text{sgn}(x^2 - 4x + 5) = 1 \) for all \( x \in \mathbb{R} \).

**Range of \( g(x) \):** \( \{1\} \)

### Step 5: Compare Outputs for Common Domain

Since both functions have the same domain \( \mathbb{R} \) and both yield the same output:
- \( f(x) = 1 \) for all \( x \)
- \( g(x) = 1 \) for all \( x \)

### Conclusion

Since the domains are the same, the ranges are the same, and \( f(x) = g(x) \) for all \( x \) in their common domain, we conclude that:

**Final Result:** \( f(x) \) and \( g(x) \) are identical functions.

---