Average Rate of Change

[john3j]

Hello,

I have a problem on my Calculus homework that says this:
Find the average rate of change for the function over the given interval.
y=8x^3-3x^2-5
[3,7]
I took the equations and plugged in each x value and ended up with this:
(2592-184)/(7-3) = 2408/4 = 602
My question is should I have take the derivative of each instance and then solve, or is the way I did it correct. I know that if I find the derivative of y=8x^3-3x^2-5 I get 24x^2-6x. Would I then just plug in my x values? Any information would be greatly appreciated.

Thanks,
John

 

[srmichael]

Hello,

I have a problem on my Calculus homework that says this:
Find the average rate of change for the function over the given interval.
y=8x^3-3x^2-5
[3,7]
I took the equations and plugged in each x value and ended up with this:
(2592-184)/(7-3) = 2408/4 = 602
My question is should I have take the derivative of each instance and then solve, or is the way I did it correct. I know that if I find the derivative of y=8x^3-3x^2-5 I get 24x^2-6x. Would I then just plug in my x values? Any information would be greatly appreciated.

Thanks,
John

Average rate of change IS NOT the same as derivative. Instantaneous rate is the same as derviative. Think of average rate of change as the slope between the points generated by the values given, in this instance [3,7].

In other words, for a interval [a.b], the Average Rate of Change = \(\displaystyle \frac{f(b)-f(a)}{b-a}\)