1. **Problem Statement:** We have a piecewise function describing the distance $f(x)$ in kilometers Michelle and Rashid travel after $x$ hours of biking: $$ f(x) = \begin{cases} 16x & 0 \leq x \leq 1 \\ 10(x - 1) + 16 & 1 < x \leq 2 \\ 14(x - 2) + 26 & 2 < x \leq 3 \\ 12(x - 3) + 40 & 3 < x \leq 4 \end{cases} $$ We will answer the following: - When are they going fastest and slowest? - What is the domain of $f(x)$? - Find $f(2)$ and interpret it. - Find $f(3)$ and interpret it. - Find the total distance traveled. 2. **Finding the speeds (rates) in each interval:** Speed is the rate of change of distance with respect to time, i.e., the slope of each piece. - For $0 \leq x \leq 1$, $f(x) = 16x$, speed = 16 km/h. - For $1 < x \leq 2$, $f(x) = 10(x-1) + 16 = 10x + 6$, speed = 10 km/h. - For $2 < x \leq 3$, $f(x) = 14(x-2) + 26 = 14x - 2$, speed = 14 km/h. - For $3 < x \leq 4$, $f(x) = 12(x-3) + 40 = 12x + 4$, speed = 12 km/h. 3. **Fastest and slowest parts:** - Fastest speed is $16$ km/h during $0 \leq x \leq 1$ hour. - Slowest speed is $10$ km/h during $1 < x \leq 2$ hours. 4. **Domain of $f(x)$:** The function is defined for $x$ from 0 to 4 hours, so the domain is: $$0 \leq x \leq 4$$ 5. **Find $f(2)$:** Since $2$ is in the interval $1 < x \leq 2$, use the second piece: $$f(2) = 10(2 - 1) + 16 = 10(1) + 16 = 26$$ This means after 2 hours, they have traveled 26 kilometers. 6. **Find $f(3)$:** Since $3$ is in $2 < x \leq 3$, use the third piece: $$f(3) = 14(3 - 2) + 26 = 14(1) + 26 = 40$$ After 3 hours, they have traveled 40 kilometers. 7. **Total distance traveled:** At $x=4$ (end of ride), use the last piece: $$f(4) = 12(4 - 3) + 40 = 12(1) + 40 = 52$$ So, total distance traveled is 52 kilometers. **Summary:** - Fastest: first hour at 16 km/h. - Slowest: second hour at 10 km/h. - Domain: $0 \leq x \leq 4$. - $f(2) = 26$ km means distance after 2 hours. - $f(3) = 40$ km means distance after 3 hours. - Total distance $f(4) = 52$ km.