1. **Problem 4:** Calculate the total weekly earnings $f(x)$ for $x=40$ and $x=60$ deliveries using the piecewise function: $$f(x) = \begin{cases} 12x, & 0 \leq x \leq 50 \\ 18 (x - 50) + 600, & x > 50 \end{cases}$$ 2. For $f(40)$, since $40 \leq 50$, use the first case: $$f(40) = 12 \times 40 = 480$$ 3. For $f(60)$, since $60 > 50$, use the second case: $$f(60) = 18 \times (60 - 50) + 600 = 18 \times 10 + 600 = 180 + 600 = 780$$ --- 1. **Problem 5:** Given $f(x) = 3x - 2$ and $g(x) = x^2$, find: (a) $f \circ g (x) = f(g(x)) = f(x^2) = 3x^2 - 2$ (b) $g \circ f (x) = g(f(x)) = g(3x - 2) = (3x - 2)^2 = 9x^2 - 12x + 4$ (c) $f \circ f (x) = f(f(x)) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8$ (d) $g \circ g (x) = g(g(x)) = g(x^2) = (x^2)^2 = x^4$ (e) Find $f^{-1}(x)$: Start with $y = 3x - 2$ Swap $x$ and $y$: $$x = 3y - 2$$ Solve for $y$: $$x + 2 = 3y$$ $$y = \frac{x + 2}{3}$$ So, $$f^{-1}(x) = \frac{x + 2}{3}$$ (f) Find $g^{-1}(x)$: Since $g(x) = x^2$, the inverse is $g^{-1}(x) = \pm \sqrt{x}$ (defined for $x \geq 0$). --- 1. **Problem 6:** The mass $M$ of a cube is directly proportional to the cube of its side length $L$: $$M = kL^3$$ Given $M = 500$ g when $L = 2$ cm, find $k$: $$500 = k \times 2^3 = 8k$$ $$k = \frac{500}{8} = 62.5$$ Now, find $L$ when $M = 13,500$ g: $$13,500 = 62.5 L^3$$ Divide both sides by 62.5: $$\cancel{62.5} \times L^3 = \frac{13,500}{\cancel{62.5}} = 216$$ So, $$L^3 = 216$$ Take the cube root: $$L = \sqrt[3]{216} = 6$$ **Final answers:** - $f(40) = 480$ - $f(60) = 780$ - (a) $f \circ g (x) = 3x^2 - 2$ - (b) $g \circ f (x) = 9x^2 - 12x + 4$ - (c) $f \circ f (x) = 9x - 8$ - (d) $g \circ g (x) = x^4$ - (e) $f^{-1}(x) = \frac{x + 2}{3}$ - (f) $g^{-1}(x) = \pm \sqrt{x}$ - Side length for mass 13,500 g is $6$ cm.