Let f(x) = log_e x and g(x) = frac{x^4 - 2x^3 | vedclass

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(A) The domain of $f \circ g$ is the set of all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$. Given $f(x) = \log_e x$,

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$R$

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(A) The domain of $f \circ g$ is the set of all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$. Given $f(x) = \log_e x$,the domain of $f$ is $(0, \infty)$,so we require $g(x) > 0$. Given $g(x) = \frac{x^4 - 2x^3 + 3x^2 - 2x + 2}{2x^2 - 2x + 1}$. First,check the denominator: $2x^2 - 2x + 1 = 2(x^2 - x + \frac{1}{4}) + \frac{1}{2} = 2(x - \frac{1}{2})^2 + \frac{1}{2}$,which is always positive for all $x \in R$. Now,analyze the numerator: $x^4 - 2x^3 + 3x^2 - 2x + 2 = (x^4 - 2x^3 + x^2) + (2x^2 - 2x + 1) + 1 = x^2(x - 1)^2 + (2(x - \frac{1}{2})^2 + \frac{1}{2}) + 1$. Since $x^2(x - 1)^2 \ge 0$,$2(x - \frac{1}{2})^2 + \frac{1}{2} > 0$,and $1 > 0$,the numerator is always positive for all $x \in R$. Thus,$g(x) > 0$ for all $x \in R$. Therefore,the domain of $f \circ g$ is $R$.

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

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