Let `f(x) = log_(e) x and g(x) =(x^(4) -2x^(3) + 3x^(2) - 2x+2)/(2x^(2) - 2x + 1)`
Then , the domain of fog (x) is

Options

A. `R`
B. `[0, oo)`
C. `(0, oo)`
D. `[1, oo)`

Explanation

To find the domain of the composite function \( f(g(x)) \), where \( f(x) = \log_e x \) and \( g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \), we need to ensure that \( g(x) > 0 \) since the logarithm function is only defined for positive values.

### Step 1: Analyze the denominator of \( g(x) \)

The denominator of \( g(x) \) is \( 2x^2 - 2x + 1 \). We need to check if this expression is always positive.

1. Calculate the discriminant \( D \) of the quadratic:
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot 1 = 4 - 8 = -4 \]
Since the discriminant is negative, the quadratic does not have real roots and is always positive (as the coefficient of \( x^2 \) is positive).

### Step 2: Analyze the numerator of \( g(x) \)

Next, we need to analyze the numerator \( x^4 - 2x^3 + 3x^2 - 2x + 2 \) to determine when it is greater than zero.

1. We can factor or analyze the numerator. We can check for roots using the Rational Root Theorem or synthetic division, but a more straightforward approach is to evaluate the expression at various points.

2. Evaluate \( g(x) \) at specific points:
- At \( x = 0 \):
\[ g(0) = \frac{0^4 - 2 \cdot 0^3 + 3 \cdot 0^2 - 2 \cdot 0 + 2}{2 \cdot 0^2 - 2 \cdot 0 + 1} = \frac{2}{1} = 2 > 0 \]
- At \( x = 1 \):
\[ g(1) = \frac{1^4 - 2 \cdot 1^3 + 3 \cdot 1^2 - 2 \cdot 1 + 2}{2 \cdot 1^2 - 2 \cdot 1 + 1} = \frac{1 - 2 + 3 - 2 + 2}{2 - 2 + 1} = \frac{2}{1} = 2 > 0 \]
- At \( x = 2 \):
\[ g(2) = \frac{2^4 - 2 \cdot 2^3 + 3 \cdot 2^2 - 2 \cdot 2 + 2}{2 \cdot 2^2 - 2 \cdot 2 + 1} = \frac{16 - 16 + 12 - 4 + 2}{8 - 4 + 1} = \frac{10}{5} = 2 > 0 \]

3. Since \( g(x) \) is a polynomial of degree 4, and we have evaluated it at several points and found it to be positive, we can conclude that \( g(x) > 0 \) for all \( x \).

### Conclusion

Since \( g(x) > 0 \) for all \( x \), the domain of \( f(g(x)) \) is all real numbers.

Thus, the domain of \( f(g(x)) \) is:
\[ \text{Domain of } f(g(x)) = (-\infty, \infty) \]