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:
To determine the domain of the composite function \( f \circ g \), where \( f(x) = \log x \) and \( g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \), we must identify the values of \( x \) for which \( g(x)>0 \). This is because the natural logarithm function \( f(x) = \log x \) is defined only for positive arguments.
We begin by analyzing the function \( g(x) \): \[ g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1} \]
Step 1: Analyze the denominator.
We must ensure that the denominator, \( 2x^2 - 2x + 1 \), is not equal to zero. We examine the roots of the quadratic equation \( 2x^2 - 2x + 1 = 0 \). The discriminant is \( \Delta = (-2)^2 - 4 \times 2 \times 1 = 4 - 8 = -4 \).
As the discriminant is negative, the quadratic \( 2x^2 - 2x + 1 \) has no real roots and is therefore never zero for any real value of \( x \).
Step 2: Determine when \( g(x)>0 \).
Since the denominator is always positive (its leading coefficient is positive and it has no real roots), the sign of \( g(x) \) is determined by the sign of the numerator. We need to find when \( x^4 - 2x^3 + 3x^2 - 2x + 2>0 \).
Evaluating \( g(x) \) at \( x = 0 \) gives \( g(0) = \frac{2}{1} = 2 \), which is positive.
For extreme values of \( x \) (both large positive and large negative), the \( x^4 \) term in the numerator dominates, indicating that the numerator will be positive. This suggests that \( g(x) \) is positive for all real \( x \).
Considering the end behavior and the fact that the denominator is always positive, we conclude that \( g(x)>0 \) for all real numbers \( x \).
Conclusion: The domain of \( f \circ g \) is \(\mathbb{R}\) because \( g(x) \) is positive for all real values of \( x \).