1. **Problem Statement:** Given functions $f(x) = 2x + 1$ and $g(x) = 3x + 1$, find the following operations: $\frac{f}{g}(x)$, $(fg)(x)$, $f \circ g(x)$, and $g \circ f(x)$. Then determine if $fg$, $f \circ g$, and $g \circ f$ are equal, and find the domain and range of each operation. 2. **Formulas and Rules:** - Function addition, subtraction, multiplication, and division are defined as: \[ (f+g)(x) = f(x) + g(x), \quad (f-g)(x) = f(x) - g(x), \quad (fg)(x) = f(x) \cdot g(x), \quad \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \text{ where } g(x) \neq 0 \] - Composition of functions is defined as: \[ (f \circ g)(x) = f(g(x)), \quad (g \circ f)(x) = g(f(x)) \] - Domain of $\frac{f}{g}$ excludes values where $g(x) = 0$. - Domain of compositions depends on the domain of the inner function and the domain of the outer function evaluated at the inner function's output. 3. **Calculations:** (i) Find each operation: - $\frac{f}{g}(x) = \frac{2x + 1}{3x + 1}$, domain excludes $x$ where $3x + 1 = 0 \Rightarrow x = -\frac{1}{3}$. - $(fg)(x) = f(x) \cdot g(x) = (2x + 1)(3x + 1) = 6x^2 + 2x + 3x + 1 = 6x^2 + 5x + 1$. - $f \circ g(x) = f(g(x)) = f(3x + 1) = 2(3x + 1) + 1 = 6x + 2 + 1 = 6x + 3$. - $g \circ f(x) = g(f(x)) = g(2x + 1) = 3(2x + 1) + 1 = 6x + 3 + 1 = 6x + 4$. (ii) Are $fg$, $f \circ g$, and $g \circ f$ equal? - $fg(x) = 6x^2 + 5x + 1$ is a quadratic function. - $f \circ g(x) = 6x + 3$ is linear. - $g \circ f(x) = 6x + 4$ is linear. Since their expressions differ, $fg \neq f \circ g \neq g \circ f$. (iii) Domain and range: - Domain of $f$ and $g$ is all real numbers $\mathbb{R}$. - Domain of $\frac{f}{g}$ is $\mathbb{R} \setminus \left\{-\frac{1}{3}\right\}$ because denominator cannot be zero. - Domain of $(fg)(x)$ is $\mathbb{R}$ since product of polynomials is defined everywhere. - Domain of $f \circ g$ is all $x$ such that $g(x)$ is in domain of $f$. Since both have domain $\mathbb{R}$, domain is $\mathbb{R}$. - Domain of $g \circ f$ similarly is $\mathbb{R}$. - Range of $f(x) = 2x + 1$ is $\mathbb{R}$ (linear, no restrictions). - Range of $g(x) = 3x + 1$ is $\mathbb{R}$. - Range of $\frac{f}{g}(x)$ excludes values where denominator zero; range is all real numbers except values that make the fraction undefined; more complex but generally all real numbers except possibly some values. - Range of $(fg)(x) = 6x^2 + 5x + 1$ is a parabola opening upwards; minimum at vertex: $$x = -\frac{b}{2a} = -\frac{5}{2 \cdot 6} = -\frac{5}{12}$$ Evaluate: $$y = 6\left(-\frac{5}{12}\right)^2 + 5\left(-\frac{5}{12}\right) + 1 = 6 \cdot \frac{25}{144} - \frac{25}{12} + 1 = \frac{150}{144} - \frac{25}{12} + 1 = \frac{25}{24} - \frac{50}{24} + \frac{24}{24} = -\frac{1}{24}$$ So range is $\left[-\frac{1}{24}, \infty\right)$. - Range of $f \circ g(x) = 6x + 3$ is $\mathbb{R}$. - Range of $g \circ f(x) = 6x + 4$ is $\mathbb{R}$. **Final answers:** - $\frac{f}{g}(x) = \frac{2x + 1}{3x + 1}, \quad x \neq -\frac{1}{3}$ - $(fg)(x) = 6x^2 + 5x + 1$ - $f \circ g(x) = 6x + 3$ - $g \circ f(x) = 6x + 4$ - $fg \neq f \circ g \neq g \circ f$ - Domains: $\frac{f}{g}$ domain is $\mathbb{R} \setminus \{-\frac{1}{3}\}$, others are $\mathbb{R}$ - Ranges: $\frac{f}{g}$ range excludes values causing division by zero, $(fg)$ range is $\left[-\frac{1}{24}, \infty\right)$, others are $\mathbb{R}$.