Chapter 4.8, Problem 48E

To determine

To sketch: the graph of the function for the interval $0\le x\le 41$, and to describe the public school enrollment changes over this period of time.

Answer to Problem 48E

The public school enrollment increases in interval $o<t<11.66$ and $29.8<t<41$ . The public school enrollment decreases in interval $11.66<t<29.8$ . The maximum enrollment is 44970.93 and minimum is 40078.21.

Explanation of Solution

Given information:

The function,

  $S=1.64x^3-102x^2+1710x+36300$

The number $S$ of students enrolled in public schools in a certain country can be modeled, where $x$ is time.

Calculation:

Input function using key $1.64X\wedge 3-102X{x}^2+1710X+36300$ .

The display is shown below:

  

Adjust the WINDOW and hit GRAPH key.

  

Press 2^ndTRACE to go the CALCULATE menu and highlight the Maximum and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed Maximum location.

When asked for Guess, simply Hit ENTER key.

  

Therefore, local maximum is at approximately $\left(11.66,44970.93\right)$ .

Press 2^ndTRACE to go the CALCULATE menu and highlight the Minimum and press ENTER.

  

When asked for left bound and right bound, move cursor to the left or right of observed Maximum location.

When asked for Guess, simply Hit ENTER key.

  

Therefore, local minimum is at approximately $\left(29.8,40078.21\right)$ .

Therefore, the public school enrollment increases in interval $o<t<11.66$ and $29.8<t<41$ . The public school enrollment decreases in interval $11.66<t<29.8$ . The maximum enrollment is 44970.93 and minimum is 40078.21.