Chapter 9.6, Problem 7CYU

Solution

(a)

What value for the domain and range are reasonable in the context of this situation, $f\left(t\right)=100{\left(1.05\right)}^t$ .

  $\begin{array}{l}\text{Domain:}\left(0,\infty \right)\\ \text{Range}:\left(100,\infty \right)\end{array}$

Given information:

  $f\left(t\right)=100{\left(1.05\right)}^t$

f(t) is the number of flies and t is time in days.

Concept used:

• Domain of a function $y=f\left(x\right)$ is the set of x values for which the function is defined.
• Range of a function $y=f\left(x\right)$ is the set of y -values that the function can take.
• x -intercept is the point where the graph of the function crosses/touches the x -axis.
• y -intercept is the point where the graph of the function crosses/touches the y -axis.

Calculation:

In order to get the graph of $f\left(t\right)=100{\left(1.05\right)}^t$ , first make a table by taking some random values of x and find the corresponding values of y . And then plot the ordered pairs and connect the points with a smooth curve, so the table is:

X	$f\left(t\right)=100{\left(1.05\right)}^t$	Y
- 2	$f\left(t\right)=100{\left(1.05\right)}^{-2}$	90.7
- 1	$f\left(t\right)=100{\left(1.05\right)}^{-1}$	95.23
0	$f\left(t\right)=100{\left(1.05\right)}^0$	100
1	$f\left(t\right)=100{\left(1.05\right)}^1$	105
2	$f\left(t\right)=100{\left(1.05\right)}^2$	110.25

Graph of the function is shown below:

  

From the above graph it is clear that the curve intersecting the y -axis at the point (0, 100). So, the y -intercept of the function is at $y=100$ .

Since the t represents the time in days, so it can’t be negative. Thus, the domain of the function is all real numbers greater than 0, that is, $\text{Domain}=\left(0,\infty \right)$ .

The range of the function will be all real number greater than 100, that is, $\text{Range}=\left(100,\infty \right)$ .

(b)

After two weeks, approximately how many flies are in this population.

  $f\left(t\right)=198$

Given information:

  $f\left(t\right)=100{\left(1.05\right)}^t$

f(t) is the number of flies and t is time in days.

t is equal to 2 weeks that is 14 days

Explanation:

In order to find how many flies are there in this population after two weeks , putting the value of t in the function $f\left(t\right)=100{\left(1.05\right)}^t$ and solving it as shown below:

  $\begin{array}{l}f\left(t\right)=100{\left(1.05\right)}^t\\ t=14\\ f\left(t\right)=100{\left(1.05\right)}^{14}\\ f\left(t\right)=197.99\\ f\left(t\right)\approx 198\end{array}$

So, after two weeks, there will be about 198 fruit flies in this population.