Chapter 6.4, Problem 19Q

a.

To determine

To find: Whether the function is of exponential growth or decay

Answer to Problem 19Q

The function is of exponential growth.

Explanation of Solution

Given:

  $f(t)=5{(4)}^t$

Calculation:

The function is in the form of $y=a{b}^x$

And since $b=4>1$ , it is an exponential growth function

b.

To determine

To graph: The function $f(t)=5{(4)}^t$ and find its domain and range.

Answer to Problem 19Q

From the graph, the domain is all real numbers and the range is $y>0$ .

Explanation of Solution

Given:

  $f(t)=5{(4)}^t$

Calculation:

From the function solve and plot some points on the graph.

  

The domain of a function is the set of points for which the value of function is defined and real. Thus, here the domain of the function is set of all real numbers. The range is $y>0$ as for all values of x the value of y is $y>0$ .

c.

To determine

To find: The yearly percent change and approximate monthly percent change.

Answer to Problem 19Q

The yearly percent change is 300% and monthly is 12% approximately.

Explanation of Solution

Given:

  $f(t)=5{(4)}^t$

Formula used: $\text{Percent change= }\frac{Difference}{Original}\times 100$

Calculation:

For 0 year, $f(0)=5{(4)}^0=5$

Now for 1 year, $f(1)=5{(4)}^1=20$

  $\text{Percent change= }\frac{20-5}{5}\times 100=300\%$

Thus the yearly percent change is 300%

For monthly, Use $t=\frac{1}{12}(12t)$

Therefore the equation becomes

  $\begin{array}{}$ f(t)=5{(4)}^{\frac{1}{12}(12t)}$$ f(t)=5{(4)}^{\frac{1}{12}(12t)}$$ f(t)=5{(1.122)}^{(12t)}$\end{array}$

Now use the growth factor $\left(1+r\right)$ to find the growth rate of $\approx$ 0.122

Therefore, the monthly percent increase is about 12%

d.

To determine

To find: The number of frogs after 4 years.

Answer to Problem 19Q

The number of frogs in the pond after 4 years is 1280.

Explanation of Solution

Given:

  $f(t)=5{(4)}^t$

Calculation:

For 4 year, $f(4)=5{(4)}^4=1280$

Thus, number of frogs in the pond after 4 years is 1280.