Consider the function as representing the value of an ounce of palladium in U. S. dollars as a function of the time t in days.
R(t) = 30t − 3t^2; t = 3
a. Find the average rate of change of R(t) over the time intervals [t, t + h], where t is as indicated and h = 1, 0.1, and 0.01 days. (Use smaller values of h to check your estimates.)

the average rate of change rc is 13 , 12.1 and 12.01 for h=1 , 0.1 and 0.01 respectively

Step-by-step explanation:

for

R(t) = 30*t − 3*t² ; t = 3

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

rc= [R(t+h)-R(t)] / [(t+h) -t) = [R(t+h)-R(t)] /h = 1/h * [ 30*(t+h) − 3*(t+h)²  - (  30*t − 3*t² ) ] =  (1/h) * ( 30*h - (3*t² + 6*t*h + t²) +3*t² ) = 30 - 6*t + h

then

rc= 30 - 6*t + h

for t=3 and h=1

rc= 30 - 6*3 + 1 = 13

for t=3 and h=0.1

rc= 30 - 6*3 + 0.1 = 12.1

for t=3 and h=0.01

rc= 30 - 6*3 + 0.01 = 12.01

for t=3 and h=0.001

rc= 30 - 6*3 + 0.01 = 12.001

when h goes smaller , the average rate of change gets closer to the instantaneous rate of change of R(t) in t=3 (the derivative of R in t=3) , that is

R'(t)= 30 - 6*t

answer: i believe it’s 245.04

step-by-step explanation: