Q. D - Calculus

Question

The limit of a rational function model at infinity: Upon further reflection, Leila decides that the quadratics used in the previous problem are unreasonable since the quadratic model for the relative maximum values could be interpreted as indicating that the eventual number of beavers would be unbounded. She decides to change her model for the relative maximum beaver populations to

$M\left(t\right)=\frac{40944t^2+454512t-1732032}{t^2+9t-36}.$

(a) Verify that this function does pass through the data points at $t=0,2,$and $4 .$Is this function continuous everywhere? (Hint: Consider $\underset{t\to 3}{lim}M\left(t\right)$)

(b) Compute $\underset{t\to \infty }{lim}M\left(t\right)$What is the significance of this number?

Step-by-Step Solution

Answer

(a) the function passes through the data points at $t=0,2,\& 4$because the function has specific values at $t=0,2,\& 4.$

(b) $\underset{t\to \infty }{\lim }M\left(t\right)=40944$state that the population will never go less than $40944$with increasing time.

Part (a) Step 1. Given information

Given function is $M\left(t\right)=\frac{40944t^2+454512t-1732032}{t^2+9t-36}.$

Given value of t is $t=0,2,\& 4.$

Part (a) Step 2. Verification.

Determine the value of the function at $t=0.$

$$\begin{gathered} M\left(0\right)=\frac{40944{\left(0\right)}^2+454512\left(0\right)-1732032}{{\left(0\right)}^2+9\left(0\right)-36} \\ M\left(0\right)=48112 \end{gathered}$$

Determine the value of the function at $t=2.$

$$\begin{gathered} M\left(2\right)=\frac{40944{\left(2\right)}^2+454512\left(2\right)-1732032}{{\left(2\right)}^2+9\left(2\right)-36} \\ M\left(0\right)=47088 \end{gathered}$$

Determine the value of the function at $t=4.$

$$\begin{gathered} M\left(4\right)=\frac{40944{\left(4\right)}^2+454512\left(4\right)-1732032}{{\left(4\right)}^2+9\left(4\right)-36} \\ M\left(4\right)=46320 \end{gathered}$$

The function has specific values at $t=0,2, \&4$so function pass through the data points at $t=0,2, \& 4.$

Part (b) Step 1. Given information.

Given function is $M\left(t\right)=\frac{40944t^2+454512t-1732032}{t^2+9t-36}.$

The Limit that needs to determine is $\underset{t\to \infty }{\lim }M\left(t\right).$

Part (b) Step 2. The limit value of the function.

Determine the limit $\underset{t\to \infty }{\lim }M\left(t\right).$

$$\begin{gathered} \underset{t\to \infty }{\lim }M\left(t\right)=\underset{t\to \infty }{\lim }\frac{40944t^2+454512t-1732032}{t^2+9t-36} \\ =\underset{t\to \infty }{\lim }\frac{40944+{\displaystyle \frac{454512}{t}}-{\displaystyle \frac{1732032}{t^2}}}{1+{\displaystyle \frac{9}{t}}-{\displaystyle \frac{36}{t^2}}} \\ \underset{t\to \infty }{\lim }M\left(t\right)==40944 \end{gathered}$$

$\underset{t\to \infty }{\lim }M\left(t\right)=40944$ state that the population will never go less than $40944$with increasing time.