
No, $M$ is not a function of $T$. The problem states that the temperature, $T$, remains a steady 77 degrees Fahrenheit for a few minutes, $M$. Thus, we know that there is more than one value of $M$, or more than one minute, associated with the temperature $T=77$. This means that if $M$ were a function of $T$ there would be more than one output M associated with the single input value of $T=77$. By definition of a function each input must be associated with exactly one output. Therefore, $M$ cannot be a function of $T$ because there would be more than one output associated with the input $T=77$.

In constrast the previous part, it is the case that for each number of minutes $M$ after noon, there is one and only one temperature $T$ in the room, so $T$ is a function of $M$. The quantity $g(0)$ is the value of $T$ when $M=0$. From the context given in the problem we know that $M$ denotes the time in minutes since noon. Thus, $M=0$ is exactly noon, so $g(0)$ is the temperature in the high school gym in degrees Fahrenheit at exactly noon.

Let’s look at the meaning of each equation:

$g(30)=5$ says the temperature at 12:30 p.m. is 5 degrees Fahrenheit, which is not what Becky said.

$\frac{g(30)g(0)}{30}=5$ says that the average rate of change of the temperature of the gym between noon and 12:30 p.m. is 5 degrees per minute, which is not what Becky said.

$g(0)g(30)=5$ says that the difference in the temperature of the gym at 12:30 and the temperature of the gym at noon is 5 degrees, which is the same thing as saying the temperature increased by 5 degrees Fahrenheit.

$T=g(30)5$ says that the temperature equals the temperature at 12:30 minus five degrees, which is not what Becky said.
As in the previous part, we recognize the quantity $g(75)g(60)$ as the change in the temperature in the room in the 15 minutes following the students' departure from the gym. We are told that the temperature decreases as the students begin to leave, so we know this quantity must be negative.
Next, we note that the quantity in question, namely $\frac{g(75)g(60)}{15}$, is precisely the average rate of change of the temperature over these first fifteen minutes, and so is also negative. This eliminates the first three choices for the value of this quantity immediately.
Finally, we note that the last option corresponds to an average decrease of $5$ degrees per minute, which would combine for a 75degree drop in this 15minutes span. Since this would leave the gym in a frozen state, we conclude that this was an implausibly fast rate of temperature decrease. This leaves only the value $0.2$, corresponding to a significantly more plausible temperature drop of 3 degrees over these 15 minutes.
Comments
Log in to commentJustin says:
about 2 yearsThere seems to be a mistake in the answer key. Part c iii in the question reads g(30)g(0), but in the answer key reads g(o)g(30).
Deepti Khare says:
over 5 yearsGreat task that builds students' estimation and reasoning skills without depending on a calculator!