1. **Problem statement:** Given two functions defined by points: $f = \{(3,4), (4,5), (5,6), (6,7)\}$ and $g = \{(5,3), (6,4), (7,-2), (8,0)\}$, find: a) $(f \cdot g)(x)$ b) $(g \cdot f)(x)$ 2. **Understanding the problem:** The notation $(f \cdot g)(x)$ means the product of the function values at $x$, i.e., $(f \cdot g)(x) = f(x) \times g(x)$, but since $f$ and $g$ are only defined at specific points, we can only compute the product at $x$ values where both $f(x)$ and $g(x)$ are defined. 3. **Find common domain points:** The $x$ values for $f$ are $3,4,5,6$ and for $g$ are $5,6,7,8$. The common $x$ values are $5$ and $6$. 4. **Calculate $(f \cdot g)(x)$ at common points:** - At $x=5$: $f(5) = 6$, $g(5) = 3$, so $(f \cdot g)(5) = 6 \times 3 = 18$ - At $x=6$: $f(6) = 7$, $g(6) = 4$, so $(f \cdot g)(6) = 7 \times 4 = 28$ Thus, $(f \cdot g)(x) = \{(5,18), (6,28)\}$. 5. **Calculate $(g \cdot f)(x)$ at common points:** Since multiplication is commutative, $(g \cdot f)(x) = (f \cdot g)(x)$, so the values are the same: $(g \cdot f)(x) = \{(5,18), (6,28)\}$. 6. **Summary:** - $(f \cdot g)(x) = \{(5,18), (6,28)\}$ - $(g \cdot f)(x) = \{(5,18), (6,28)\}$ --- 7. **Next problem:** Given $f(x) = 3x - 2$ and $g(x) = 3x + b$, find $b$ such that $(f \cdot g)(x) = (g \cdot f)(x)$ for all real $x$. 8. **Understanding:** Here, $(f \cdot g)(x)$ means function composition: $(f \cdot g)(x) = f(g(x))$ and $(g \cdot f)(x) = g(f(x))$. 9. **Compute $f(g(x))$:** $$f(g(x)) = f(3x + b) = 3(3x + b) - 2 = 9x + 3b - 2$$ 10. **Compute $g(f(x))$:** $$g(f(x)) = g(3x - 2) = 3(3x - 2) + b = 9x - 6 + b$$ 11. **Set equal for all $x$:** $$9x + 3b - 2 = 9x - 6 + b$$ 12. **Simplify by canceling $9x$ on both sides:** $$\cancel{9x} + 3b - 2 = \cancel{9x} - 6 + b$$ $$3b - 2 = b - 6$$ 13. **Solve for $b$:** $$3b - b = -6 + 2$$ $$2b = -4$$ $$b = \frac{-4}{2} = -2$$ 14. **Final answer:** $$b = -2$$