Question

Let \( f(x) = \log x \) and \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] Then the domain of \( f \circ g \) is:

• A. \( \mathbb{R} \)
• B. \( (0, \infty) \)
• C. \( [0, \infty) \)
• D. \( [1, \infty) \)

Hint:

For composite functions, analyze the inner function's range and ensure it aligns with the domain of the outer function.

Answer 1

The Correct Option is A

To find the domain of \( f \circ g \), where \( f(x) = \log x \) and \( g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \), we need to determine the values of \( x \) for which \( g(x) > 0 \) because the natural logarithm \( f(x) = \log x \) is only defined for positive \( x \). 

Let's start by analyzing the expression for \( g(x) \): \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \]

**Step 1: Determine when the denominator is not zero.**

We need \( 2x^2 - 2x + 1 \neq 0 \). The quadratic equation: \[ 2x^2 - 2x + 1 = 0 \] has a discriminant \( \Delta = (-2)^2 - 4 \times 2 \times 1 = 4 - 8 = -4 \).

Since the discriminant is negative, the quadratic has no real roots. Thus, the denominator is never zero for any real \( x \).

**Step 2: Find where \( g(x) > 0 \).**

Since the denominator does not change sign, we only need to check when the numerator is positive: \[ x^4 - 2x^3 + 3x^2 - 2x + 2 > 0 \]

Notice that if we substitute \( x = 0 \): \[ g(0) = \frac{2}{1} = 2 \]

For large positive and negative \( x \), the \( x^4 \) term dominates, implying that the entire expression is positive.

Checking critical points or derivatives might be complex here due to the polynomial degrees, but observing the polynomial's end behavior suggests positivity over all reals.

Thus, the entire range of \( x \) satisfies \( g(x) > 0 \).

**Conclusion: The domain of the composition \( f \circ g \) is \(\mathbb{R}\) since \( g(x) > 0 \) for all real values of \( x \).

Answer 2

Step 1: Understanding domain constraints. The function \( f(x) = \log x \) requires \( x>0 \), so we must ensure \( g(x)>0 \) for \( f(g(x)) \) to be defined. 

Step 2: Finding domain of \( g(x) \). Given: \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \] The denominator is a quadratic equation: \[ 2x^2 - 2x + 1 \] Since the discriminant is negative, it is always positive. 

Step 3: Solving for \( g(x)>0 \). Setting the numerator \( x^4 - 2x^3 + 3x^2 - 2x + 2>0 \), we find that \( x<0 \) satisfies this condition. Hence, \( g(x) \) is always positive. Thus, \( g(x)>0 \) for all \( x \), meaning the domain of \( f \circ g \) is \( \mathbb{R} \).