How do you find the average rate of change for the function #s(t)=4.5t^2# on the indicated intervals [6,6+h]? Calculus Derivatives Average Rate of Change Over an Interval 1 Answer Konstantinos Michailidis Sep 13, 2015 It is #4.5*(h+12)# Explanation: It is #(Δs)/(Δt)=(s(6+h)-s(6))/(6+h-6)=4.5*((6+h)^2-6^2)/(6+h-6)= 4.5/h*(36+12h+h^2-36)=4.5*((h^2+12h)/h)=4.5*(h+12)# Answer link Related questions How do you find the average rate of change of a function from graph? How do you find the average rate of change of a function between two points? How do you find the average rate of change of #f(x) = sec(x)# from #x=0# to #x=pi/4#? How do you find the average rate of change of #f(x) = tan(x)# from #x=0# to #x=pi/4#? How do you find the rate of change of y with respect to x? How do you find the average rate of change of #y=x^3+1# from #x=1# to #x=3#? What is the relationship between the Average rate of change of a fuction and derivatives? What is the difference between Average rate of change and instantaneous rate of change? What does the Average rate of change of a linear function represent? What is the relationship between the Average rate of change of a function and a secant line? See all questions in Average Rate of Change Over an Interval Impact of this question 2104 views around the world You can reuse this answer Creative Commons License