Question

1. the piecewise function f shown below models the insulin levels of a patient over time where x = 0 represents 8:00.

a. what is the domain in this context?

b. what is the range in this context?

c. find f(3). what does it mean in this context?

d. find average rate of change from 0, 3. explain the meaning in this context.

Explanation:

Step1: Determine the domain

The domain of a function is the set of all possible input values (x - values). Looking at the graph, the x - values (time since 8:00) start at \(x = 0\) and end at \(x=11\). So, the domain is \(0\leq x\leq11\).

Step2: Determine the range

The range of a function is the set of all possible output values (y - values). Looking at the graph, the y - values (insulin levels) start at \(y = 8\) and go up to \(y = 24\). So, the range is \(8\leq y\leq24\).

Step3: Find \(f(3)\)

To find \(f(3)\), we look at the value of the function when \(x = 3\). From the graph, when \(x = 3\), \(y=f(3)=24\). In the context of the problem, \(x = 3\) means 3 hours after 8:00 (i.e., 11:00), and \(y = 24\) is the insulin level at that time.

Step4: Calculate the average rate of change

The formula for the average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). Here, \(a = 0\), \(b = 3\), \(f(0)=8\) and \(f(3)=24\).

$$ \frac{f(3)-f(0)}{3-0}=\frac{24 - 8}{3}=\frac{16}{3}\approx5.33 $$

In the context of the problem, this represents the average increase in insulin level per hour from 8:00 (\(x = 0\)) to 11:00 (\(x = 3\)).

Answer:

a. The domain is \(0\leq x\leq11\).
b. The range is \(8\leq y\leq24\).
c. \(f(3) = 24\). It means that at 11:00 (3 hours after 8:00), the patient's insulin level is \(24\) mclU/ml.
d. The average rate of change is \(\frac{24 - 8}{3-0}=\frac{16}{3}\approx5.33\). It means that, on average, the patient's insulin level increases by approximately \(5.33\) mclU/ml per hour from 8:00 to 11:00.