31. Find $f \circ g$ and $g \circ f$ and their domains for $f(x)=x^2$ and $g(x)=\sqrt{1-x}$. 1. Compute $f \circ g$: \\ $f(g(x)) = (\sqrt{1-x})^2 = 1-x$. 2. Domain of $f \circ g$: The expression $\sqrt{1-x}$ requires $1-x \ge 0$ giving $x \le 1$. So, domain of $f \circ g$ is $(-\infty, 1]$. 3. Compute $g \circ f$: \\ $g(f(x)) = \sqrt{1 - x^2}$. 4. Domain of $g \circ f$: Inside the square root must be non-negative: $1 - x^2 \ge 0 \Rightarrow -1 \le x \le 1$. So, domain of $g \circ f$ is $[-1, 1]$. --- 32. Find $f \circ g$ and $g \circ f$ and their domains for $f(x)=\sqrt{x-3}$ and $g(x)=\sqrt{x^2+3}$. 1. Compute $f \circ g$: \\ $f(g(x)) = \sqrt{\sqrt{x^2+3} - 3}$. 2. Domain for $f \circ g$: Expression inside $f$ must be $\ge 3$ so that $x-3 \ge 0$. Thus, require $\sqrt{x^2+3} -3 \ge 0 \Rightarrow \sqrt{x^2+3} \ge 3$. 3. Square both sides: $x^2 + 3 \ge 9 \Rightarrow x^2 \ge 6 \Rightarrow x \le -\sqrt{6}$ or $x \ge \sqrt{6}$. 4. Compute $g \circ f$: \\ $g(f(x)) = \sqrt{ (\sqrt{x-3})^2 + 3 } = \sqrt{x-3 + 3} = \sqrt{x}$. 5. Domain of $g \circ f$: The expression $f(x)=\sqrt{x-3}$ requires $x-3 \ge 0 \Rightarrow x \ge 3$, so domain is $[3, \infty)$. Note: Also $g \circ f(x) = \sqrt{x}$ domain becomes $x \ge 0$ but domain restricted by $f$ to $x \ge 3$. --- 33. Find $f \circ g$ and $g \circ f$ and their domains for $f(x)=\frac{1+x}{1-x}$, $g(x)=\frac{x}{1-x}$. 1. Compute $f \circ g$: \\ $f(g(x))= \frac{1 + \frac{x}{1-x}}{1- \frac{x}{1-x}} = \frac{\frac{1-x+x}{1-x}}{\frac{1-x - x}{1-x}} = \frac{1}{1-2x}$. 2. Domain of $g$: Denominator $1-x \neq 0 \Rightarrow x \neq 1$. 3. Domain of $f \circ g$: Denominator of $f(g(x))$ is $1 - 2x \neq 0 \Rightarrow x \neq \frac{1}{2}$. Also $x \neq 1$ from $g$. So domain of $f \circ g$ is $\mathbb{R} \setminus \{ 1, \frac{1}{2} \}$. 4. Compute $g \circ f$: \\ $g(f(x)) = \frac{\frac{1+x}{1-x}}{1 - \frac{1+x}{1-x}} = \frac{\frac{1+x}{1-x}}{\frac{1-x - (1+x)}{1-x}} = \frac{1+x}{-2x} = - \frac{1+x}{2x}$. 5. Domain of $f$: Denominator $1-x \neq 0 \Rightarrow x \neq 1$. 6. Domain of $g \circ f$: Denominator $2x \neq 0 \Rightarrow x \neq 0$, also $x \neq 1$. So domain of $g \circ f$ is $\mathbb{R} \setminus \{0,1\}$. --- 34. Find $f \circ g$ and $g \circ f$ and their domains for $f(x) = \frac{x}{1+x^2}$ and $g(x) = \frac{1}{x}$. 1. Compute $f \circ g$: \\ $f(g(x)) = f\left(\frac{1}{x}\right) = \frac{\frac{1}{x}}{1 + \left(\frac{1}{x}\right)^2} = \frac{\frac{1}{x}}{1 + \frac{1}{x^2}} = \frac{\frac{1}{x}}{\frac{x^2 + 1}{x^2}} = \frac{1}{x} \cdot \frac{x^2}{x^2+1} = \frac{x}{x^2+1}$. 2. Domain of $g$: Denominator $x \neq 0$. 3. So domain of $f \circ g$ is $\mathbb{R} \setminus \{0\}$. 4. Compute $g \circ f$: \\ $g(f(x)) = \frac{1}{\frac{x}{1+x^2}} = \frac{1+x^2}{x}$. 5. Domain of $f$: Denominator $1 + x^2 \neq 0$ always true. 6. Domain of $g \circ f$: denominator $x \neq 0$. So domain is $\mathbb{R} \setminus \{0\}$. --- Final domain summary: - $f \circ g$ for 31: $(-\infty,1]$ - $g \circ f$ for 31: $[-1,1]$ - $f \circ g$ for 32: $(-\infty,-\sqrt{6}] \cup [\sqrt{6},\infty)$ - $g \circ f$ for 32: $[3,\infty)$ - $f \circ g$ for 33: $\mathbb{R} \setminus \{1, \frac{1}{2}\}$ - $g \circ f$ for 33: $\mathbb{R} \setminus \{0,1\}$ - $f \circ g$ for 34: $\mathbb{R} \setminus \{0\}$ - $g \circ f$ for 34: $\mathbb{R} \setminus \{0\}$