   Chapter 10.5, Problem 7E Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203

Solutions

Chapter
Section Finite Mathematics and Applied Cal...

7th Edition
Stefan Waner + 1 other
ISBN: 9781337274203
Textbook Problem

Consider the functions in Exercises 5–8 as representing the value of an ounce of palladium in U.S. dollars as a function of the time t in days. Find the average rates of change of R ( t ) over the time intervals [ t , t + h ] , where t is as indicated and h = 0 , 0.1 , and 0.01 days. Hence, estimate the instantaneous rate of change of R at time t, specifying the units of measurement. (Use smaller values of h to check your estimates.) [HINT: See Example 1.] R ( t ) = 270 + 20 t 3 ; t = 1

To determine

To calculate: The average rate of change for the function R(t)=270+20t3 over the time interval [t,t+h], when t=1 and h=1,0.1 and 0.01 days. Also estimate the instantaneous rate of change of R at time t. The function R(t) represents the value of an ounce of palladium in U.S. dollars as a function of the time t in days.

Explanation

Given information:

The provided function is R(t)=270+20t3 which represents the value of an ounce of palladium in U.S. dollars as a function of the time t in days and evaluate the average rate of change when t=1.

Formula used:

The formula for the average rate of change of function f(x) over the interval [a,a+h] is as follows:

Average rate of change of f(x)=f(a+h)f(a)h.

Calculation:

Consider the function, R(t)=270+20t3 and calculate the average rate of change when t=1.

The average rate of change of function R(t) over the interval [t,t+h] is:

Average rate of change of R(t) is R(t+h)R(t)h.

Average rate of change of R(t) over [1,1+h] is R(1+h)R(1)h.

Now, calculate the average rate of change for different values of h which is follows:

For h=1,

Evaluate the values of R(1) and R(1+1) as:

R(1)=270+20(1)3=270+20=290

And,

R(1+1)=R(2)=270+20(2)3=270+160=430

Substitute the values of R(1) and R(1+1),

The average rate of change of function R(t)=R(1+1)R(1)h=4302901=140

For h=0.1,

Evaluate the values of R(1) and R(1+0

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