Chapter 3, Problem 73RE

Solution

(a)

To calculate: The initial number of drug.

Number of drug administered is $90\text{ units}$

Given information: The function $f(t)=90-52\ln (1+t)$ where, $0\le t\le 4$ .

Calculation:

We let $t=0$

in equation given:

  $\begin{array}{c}f\left(0\right)=90-52\ln \left(1+0\right)\\ =90-52\left(0\right)\\ =90\text{ units}\end{array}$

(b)

To calculate: present value after 2hr.

Amount of drug left after two hour $32.872units$ .

Given information: the function $f(t)=90-52\ln (1+t)$ where $0\le t\le 4$ .

Calculation:

we let $t=2$ in equation given:

  $\begin{array}{c}f(2)=90-52\ln (1+2)\\ =90-52(3)\\ =32.872\text{units}\end{array}$

(c)

To graph: graph of $(f)$ .

Given information: the function $f(t)=90-52\ln (1+t)$ where $0\le t\le 4$ .

Graph:

  

Interpretation:

To draw graph we should start basic function $\ln t$

then transform as follow:

  $\ln t$ Basic function

  $\ln (t+1)$ Translate 1 unit left

  $52\ln (t+1)$ Vertically stretch by factor of 52

  $-52\ln (t+1)$ Reflect about x-axis

  $90-52\ln (t+1)$ Translate 90 units up

Remember to restrict domain $0\le t\le 4$ and to label axis.