Let's solve each part of the problem step by step.

Part (a) - Temperature of the Refrigerator

The temperature of the refrigerator is the initial temperature of the roast before it starts heating up in the oven. This happens when $t = 0$.

Given: $R(t) = 375 - 334e^{-0.009t}$

Substitute $t = 0$ into the equation to find the initial temperature: $R(0) = 375 - 334e^{-0.009 \times 0}$ $R(0) = 375 - 334e^{0}$ Since $e^0 = 1$, we have: $R(0) = 375 - 334 \times 1 = 375 - 334 = 41 \text{ degrees Fahrenheit}$ So, the temperature of the refrigerator is 41 degrees Fahrenheit.

Part (b) - Temperature After 20 Minutes

We are asked to express the temperature of the roast 20 minutes after being put in the oven in functional notation and calculate its value.

The functional notation is: $R(20)$

Substitute $t = 20$ into the equation: $R(20) = 375 - 334e^{-0.009 \times 20}$

Let's calculate this value: $R(20) = 375 - 334e^{-0.18}$

Now, compute $e^{-0.18}$ and then find $R(20)$.

$e^{-0.18} \approx 0.83527$ $R(20) \approx 375 - 334 \times 0.83527 = 375 - 279.9878 \approx 95.01 \text{ degrees Fahrenheit}$

So, the temperature of the roast after 20 minutes is approximately 95.01 degrees Fahrenheit.

Part (c) - Temperature Increase in the First 14 Minutes

We need to find the temperature increase during the first 14 minutes. This can be calculated by finding the difference between the temperature at $t = 14$ and the initial temperature at $t = 0$.

Substitute $t = 14$ into the equation: $R(14) = 375 - 334e^{-0.009 \times 14}$

Let's calculate $R(14)$: $R(14) = 375 - 334e^{-0.126}$ $e^{-0.126} \approx 0.88106$ $R(14) \approx 375 - 334 \times 0.88106 = 375 - 294.273 \approx 80.73 \text{ degrees Fahrenheit}$

Now, subtract the initial temperature (41°F) from $R(14)$ to get the increase: $\text{Increase} = 80.73 - 41 = 39.73 \text{ degrees Fahrenheit}$

The temperature of the roast increased by approximately 39.73 degrees Fahrenheit during the first 14 minutes of cooking.

Part (d) - Temperature Increase from the First Hour to 14 Minutes After the First Hour

First, let's find the temperature at 1 hour (60 minutes): $R(60) = 375 - 334e^{-0.009 \times 60}$ $R(60) = 375 - 334e^{-0.54}$

Calculate $R(60)$: $e^{-0.54} \approx 0.58275$ $R(60) \approx 375 - 334 \times 0.58275 = 375 - 194.635 = 180.37 \text{ degrees Fahrenheit}$

Now, find the temperature at 74 minutes (1 hour + 14 minutes): $R(74) = 375 - 334e^{-0.009 \times 74}$ $R(74) = 375 - 334e^{-0.666}$ $e^{-0.666} \approx 0.51325$ $R(74) \approx 375 - 334 \times 0.51325 = 375 - 171.4265 = 203.57 \text{ degrees Fahrenheit}$

Now, find the increase in temperature from 60 minutes to 74 minutes: $\text{Increase} = 203.57 - 180.37 = 23.20 \text{ degrees Fahrenheit}$

So, the temperature of the roast increased by approximately 23.20 degrees Fahrenheit from the first hour to 14 minutes after the first hour of cooking.

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Do you want details on any of the steps? Here are five related questions:

1. How does the exponential decay factor in the equation affect the temperature change over time?
2. What would be the roast's temperature after 45 minutes?
3. How would the equation change if the roast's temperature increased faster?
4. Can we predict when the roast will reach 300°F using this equation?
5. How would changing the oven temperature affect the function?

Tip: Always check your function inputs, especially time variables, to ensure correct calculations when using exponential functions.