IDNLearn.com provides a collaborative environment for finding and sharing answers. Our platform provides accurate, detailed responses to help you navigate any topic with ease.
Sagot :
a. The average rate of change of the function f(x)= [tex]120(0.1)^x[/tex] from x = 0 to x = 2 is
b. The average rate of change of the function f(x)= [tex]4(2)^x[/tex] from x = 0 to x = 2 is
c. The average rate of change of the function f(x)= [tex]480(0.3)^x[/tex] from x = 1 to x = 5 is
The average rate of change of the function is given by-
[tex]\frac{f(1)-f(0)}{1-0}[/tex]
Now, apply the same to all the given functions to find the average rate of change of the functions as follows.
a. The given function is f(x) = [tex]120(0.1)^x[/tex]
We have to find the average rate of change of this function from x = 0 to x = 2
The average rate of change = [tex]\frac{f(2)-f(0)}{2-0}[/tex]
= (120 [tex](0.1)^2[/tex] - 120) / 2
= 60 (0.01 - 1)
= 60 x - 0.99 = - 59.4
Thus, the average rate of change = - 59.4
b. The given function is f(x) = [tex]4(2)^x[/tex]
We have to find the average rate of change of this function from x = 0 to x = 2
The average rate of change = [tex]\frac{f(2)-f(0)}{2-0}[/tex]
= ([tex]4(2)^2[/tex] - 4) / 2
= 2(4 - 1) = 2 x 3 = 6
Thus, the average rate of change = 6
c. The given function is f(x) = [tex]480(0.3)^x[/tex]
We have to find the average rate of change of this function from x = 1 to x = 5
The average rate of change = [tex]\frac{f(5)-f(1)}{5-1}[/tex]
= ([tex]480(0.3)^5 - 480(0.3)^1[/tex] ) / 4
= 480 (0.00243 - 0.3) /4
= 120 x -0.29757
Thus, the average rate of change = -35.7084
Read more about the Rate of change of function:
brainly.com/question/24313700
#SPJ4
The complete question is -
a. what is the average rate of change of the function f(x)= [tex]120(0.1)^x[/tex] from x = 0 to x = 2?
b. what is the average rate of change of the function f(x)= [tex]4(2)^x[/tex] from x = 0 to x = 2?
c. what is the average rate of change of the function f(x)= [tex]480(0.3)^x[/tex] from x = 1 to x = 5?
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
