1. **State the problem:** Given the temperature function during illness $$T(t) = -0.6t^2 + 0.67t + 37$$ where $T$ is temperature in degrees Celsius and $t$ is time in days, we need to find: (a) The rate of change of temperature with respect to time, $T'(t)$. (b) The rate of change at $t=3$ days. (c) When the temperature begins to fall. 2. **Formula and rules:** The rate of change of a function is its derivative. For a polynomial $$f(t) = at^2 + bt + c$$, the derivative is $$f'(t) = 2at + b$$. 3. **Find $T'(t)$:** $$T(t) = -0.6t^2 + 0.67t + 37$$ Differentiate term-by-term: $$T'(t) = \frac{d}{dt}(-0.6t^2) + \frac{d}{dt}(0.67t) + \frac{d}{dt}(37)$$ $$= -0.6 \times 2t + 0.67 + 0$$ $$= -1.2t + 0.67$$ 4. **Evaluate $T'(3)$:** Substitute $t=3$: $$T'(3) = -1.2 \times 3 + 0.67 = -3.6 + 0.67 = -2.93$$ 5. **Find when temperature begins to fall:** Temperature begins to fall when rate of change becomes negative, i.e., when $T'(t) < 0$. Solve for $t$: $$-1.2t + 0.67 < 0$$ Add $1.2t$ to both sides: $$0.67 < 1.2t$$ Divide both sides by 1.2: $$\frac{0.67}{1.2} < t$$ Show cancellation: $$t > \cancel{\frac{0.67}{\cancel{1.2}}}$$ Calculate: $$t > 0.5583...$$ days Convert to hours: $$0.5583... \times 24 = 13.4$$ hours **Final answers:** (a) $$T'(t) = -1.2t + 0.67$$ (b) $$T'(3) = -2.93$$ degrees Celsius per day (c) Temperature begins to fall after approximately 13 hours from onset of illness.