Question

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

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

Answer 1

Answer: (C)

$\mathbb{R}$

Answer 2

To find the domain of $f \circ g$, 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 determine the values of $x$ for which $g(x)$ is defined and $g(x) > 0$ since the domain of $f(x) = \log_e x$ is $x > 0$.

First, let's analyze $g(x)$. The denominator is $2x^2 - 2x + 1$. We can rewrite this as $2(x^2 - x) + 1 = 2(x^2 - x + \frac{1}{4}) + 1 - \frac{1}{2} = 2(x - \frac{1}{2})^2 + \frac{1}{2}$. Since $(x - \frac{1}{2})^2 \geq 0$ for all $x$, the denominator is always positive and never zero. Thus, $g(x)$ is defined for all real numbers.

Next, let's analyze the numerator $x^4 - 2x^3 + 3x^2 - 2x + 2$. We can rewrite this as $(x^4 - 2x^3 + x^2) + (2x^2 - 2x + 2) = x^2(x^2 - 2x + 1) + 2(x^2 - x + 1) = x^2(x-1)^2 + 2(x^2 - x + 1)$. Notice that $x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4} > 0$ for all $x$. Also $x^2(x-1)^2 \geq 0$. Another approach is to rewrite the numerator as $(x^2 - x + 1)^2 + 1$. $(x^2 - x + 1)^2 = x^4 + x^2 + 1 - 2x^3 - 2x + 2x^2 = x^4 - 2x^3 + 3x^2 - 2x + 1$. Thus, $x^4 - 2x^3 + 3x^2 - 2x + 2 = (x^2 - x + 1)^2 + 1$. Since $(x^2 - x + 1)^2 \geq 0$, the numerator is always greater than or equal to 1.

Since the numerator is always positive and the denominator is always positive, $g(x) > 0$ for all $x \in \mathbb{R}$. Therefore, the domain of $f(g(x))$ is all real numbers, $\mathbb{R}$.