# Math Help - ...Rate of Change...

1. ## ...Rate of Change...

Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

This problem isn't hard, I just can't remember how to do it.

1) In exercise 1, find the average rate of change of the function over the given interval or intervals
f(x)=X^3+1
a. [2.3] b. [-1,1]

I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!

2. Originally Posted by Super Mallow
Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

This problem isn't hard, I just can't remember how to do it.

1) In exercise 1, find the average rate of change of the function over the given interval or intervals
f(x)=X^3+1
a. [2.3] b. [-1,1]

I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!
The average of a function f(x) on an interval [a, b] is defined as int(f(x),x=a..b)/(b-a).

3. Originally Posted by Super Mallow
Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

This problem isn't hard, I just can't remember how to do it.

1) In exercise 1, find the average rate of change of the function over the given interval or intervals
f(x)=X^3+1
a. [2.3] b. [-1,1]

I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!
Ah! Curvature gave it to me. He (I assume curvature is a he anyway) gave you the average of the function over an interval, but you are looking for the average of the rate of change. So we have for the first one:
$\text{Average of Rate of Change} = \frac{1}{3 - 2}\int_2^3 \frac{df}{dx}dx = \frac{1}{1}(f(3) - f(2)) = 19$

You do the second one.

-Dan

4. That's a cute problem, I like how it works.

5. Gentlemen, you are thinking too hard (wow, i almost never get to say that, i'm always the one doing things the hard way. i feel so cool now , i almost never get to use that emoticon either). The average rate of change of a function on an interval, is the slope of the secant line on that interval, which is simply $\frac {f(b) - f(a)}{b - a}$ (if our interval is [a,b] and our function is f(x) of course), which ends up being the same as topsquark's answer...but without entering the realm of integrals. But i guess it's nice to see how it all works out in the end and we can use an integral to confirm our answer

6. Yeah, I figure dit out shortly after posting this...it was just a mental lapse..