1. **Problem Statement:** Given the functions: $$f(x) = x + 5, \quad g(x) = 2x - 1, \quad h(x) = 2x^2 + 9x - 5$$ We need to find: a. $(f + g)(x)$ b. $(f - g)(x)$ c. $(f \cdot g)(x)$ d. $\left(\frac{h}{g}\right)(x)$ e. $(f + g)(3)$ f. $(f - g)(3)$ g. $(f \cdot g)(3)$ h. $\left(\frac{h}{g}\right)(3)$ --- 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: $(f \cdot g)(x) = f(x) \times g(x)$ - Quotient of functions: $\left(\frac{h}{g}\right)(x) = \frac{h(x)}{g(x)}$, provided $g(x) \neq 0$ - To evaluate at a point, substitute the value into the function. --- 3. **Calculations:** **a.** $$(f + g)(x) = (x + 5) + (2x - 1) = x + 5 + 2x - 1 = 3x + 4$$ **b.** $$(f - g)(x) = (x + 5) - (2x - 1) = x + 5 - 2x + 1 = -x + 6$$ **c.** $$(f \cdot g)(x) = (x + 5)(2x - 1) = 2x^2 - x + 10x - 5 = 2x^2 + 9x - 5$$ **d.** $$\left(\frac{h}{g}\right)(x) = \frac{2x^2 + 9x - 5}{2x - 1}$$ **e.** $$(f + g)(3) = 3(3) + 4 = 9 + 4 = 13$$ **f.** $$(f - g)(3) = -3 + 6 = 3$$ **g.** $$(f \cdot g)(3) = 2(3)^2 + 9(3) - 5 = 2(9) + 27 - 5 = 18 + 27 - 5 = 40$$ **h.** $$g(3) = 2(3) - 1 = 6 - 1 = 5 \neq 0$$ $$h(3) = 2(3)^2 + 9(3) - 5 = 18 + 27 - 5 = 40$$ $$\left(\frac{h}{g}\right)(3) = \frac{40}{5} = 8$$ --- 4. **Final answers:** - $(f + g)(x) = 3x + 4$ - $(f - g)(x) = -x + 6$ - $(f \cdot g)(x) = 2x^2 + 9x - 5$ - $\left(\frac{h}{g}\right)(x) = \frac{2x^2 + 9x - 5}{2x - 1}$ - $(f + g)(3) = 13$ - $(f - g)(3) = 3$ - $(f \cdot g)(3) = 40$ - $\left(\frac{h}{g}\right)(3) = 8$