1. **Problem statement:** Given two functions $f(x) = x^2$ and $g(x) = 2x + 1$, find the following combined functions: $(f+g)(x)$, $(f-g)(x)$, $(fg)(x)$, and $\frac{f}{g}(x)$. 2. **Formulas and rules:** - Sum of functions: $(f+g)(x) = f(x) + g(x)$ - Difference of functions: $(f-g)(x) = f(x) - g(x)$ - Product of functions: $(fg)(x) = f(x) \cdot g(x)$ - Quotient of functions: $\frac{f}{g}(x) = \frac{f(x)}{g(x)}$, with $g(x) \neq 0$ 3. **Calculate each:** - $(f+g)(x) = x^2 + (2x + 1) = x^2 + 2x + 1$ - $(f-g)(x) = x^2 - (2x + 1) = x^2 - 2x - 1$ - $(fg)(x) = x^2 \cdot (2x + 1) = 2x^3 + x^2$ - $\frac{f}{g}(x) = \frac{x^2}{2x + 1}$, where $2x + 1 \neq 0$ (i.e., $x \neq -\frac{1}{2}$) 4. **Explanation:** - For sum and difference, simply add or subtract the expressions. - For product, multiply each term carefully. - For quotient, divide the expressions but note the domain restriction where denominator is not zero. **Final answers:** $$(f+g)(x) = x^2 + 2x + 1$$ $$(f-g)(x) = x^2 - 2x - 1$$ $$(fg)(x) = 2x^3 + x^2$$ $$\frac{f}{g}(x) = \frac{x^2}{2x + 1}, \quad x \neq -\frac{1}{2}$$