...Rate of Change...

• Dec 8th 2007, 09:13 AM
Super Mallow
...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!
• Dec 8th 2007, 10:47 AM
curvature
Quote:

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).
• Dec 8th 2007, 11:05 AM
topsquark
Quote:

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
• Dec 8th 2007, 01:13 PM
angel.white
That's a cute problem, I like how it works.
• Dec 8th 2007, 02:22 PM
Jhevon
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 :cool:, 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
• Dec 8th 2007, 04:36 PM
Super Mallow
Yeah, I figure dit out shortly after posting this...it was just a mental lapse..