Let's break down the problem step by step: ### Problem Restatement The temperature \( T \) in degrees Fahrenheit at time \( x \) hours after midnight is given by: \[ T(x) = 73 - 9 \cos \left(\frac{\pi}{12}(x - 2)\right) \] We are asked to **find the rate at which the temperature is changing at 11 AM** (i.e., find \( \frac{dT}{dx} \) at \( x = 11 \)). Round to one decimal place. --- ### Step 1: Find \( \frac{dT}{dx} \) Given: \[ T(x) = 73 - 9 \cos \left(\frac{\pi}{12}(x - 2)\right) \] The derivative with respect to \( x \) is: \[ \frac{dT}{dx} = -9 \cdot \frac{d}{dx}\left[\cos\left(\frac{\pi}{12}(x-2)\right)\right] \] Recall the chain rule: \[ \frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx} \] where \( u = \frac{\pi}{12}(x-2) \) and \( \frac{du}{dx} = \frac{\pi}{12} \). So, \[ \frac{dT}{dx} = -9 \left[ -\sin\left(\frac{\pi}{12}(x-2)\right) \cdot \frac{\pi}{12} \right] = 9 \cdot \frac{\pi}{12} \cdot \sin\left(\frac{\pi}{12}(x-2)\right) \] \[ \frac{dT}{dx} = \frac{9\pi}{12} \sin\left(\frac{\pi}{12}(x-2)\right) \] \[ \frac{dT}{dx} = \frac{3\pi}{4} \sin\left(\frac{\pi}{12}(x-2)\right) \] --- ### Step 2: Plug in \( x = 11 \) (11 AM) \[ \frac{dT}{dx}\Bigg|_{x=11} = \frac{3\pi}{4} \sin\left(\frac{\pi}{12}(11-2)\right) \] \[ = \frac{3\pi}{4} \sin\left(\frac{\pi}{12} \times 9\right) \] \[ = \frac{3\pi}{4} \sin\left(\frac{9\pi}{12}\right) \] \[ = \frac{3\pi}{4} \sin\left(\frac{3\pi}{4}\right) \] --- ### Step 3: Calculate the Values - \(\sin\left(\frac{3\pi}{4}\right) = \sin(135^\circ) = \frac{\sqrt{2}}{2}\) - \(\pi \approx 3.1416\) So, \[ \frac{dT}{dx}\Bigg|_{x=11} = \frac{3 \times 3.1416}{4} \times \frac{\sqrt{2}}{2} \] \[ = \frac{9.4248}{4} \times 0.7071 \] \[ = 2.3562 \times 0.7071 \] \[ \approx 1.666 \] --- ### Step 4: Round to One Decimal Place \[ \boxed{1.7} \] **Final Answer:** At 11 AM, the temperature is increasing at a rate of **1.7°F per hour**.

The rate of change of temperature at a given time is found by differentiating the temperature function with respect to time and evaluating it at that specific time.