# Homework Help: Calulus Help: Mean Value Theorem

1. Oct 21, 2004

### buffgilville

Calculus Help: Mean Value Theorem

Let

f(x) = x^3 - x

on the interval [2,3]. Find all numbers c in the interval (2,3) that satisfy the conclusion of the mean-value theorem.

Here's what I did:
f '(c) = f(b) - f(a) / (b - a)

f '(c) = f(3^3 - 3) - f(2^3 - 2) / (3-2)
= 18, but it's incorrect.

Last edited: Oct 21, 2004
2. Oct 21, 2004

### JasonRox

Why is it suppose to be in the interval?

It is differientiable in the interval (a,b). That means the function is continuous in the interval [a,b], or so that's what we have to assume.

All the mean value theorem says is that in the interval there will be a point f '(c)=[f(b)-f(a)]/(b-a).

It has a slope of 18 somewhere in the interval (2,3). If you draw the graph, you can probably find it.

The graph of a function and its derivative with not be within the same intervals.

I don't see anything wrong with 18.

Note: I have never used the Mean Value Theorem. I read it off mathworld.com, and personally it makes complete sense.

3. Oct 21, 2004

### buffgilville

I got the value of c=18 and the question asks to find all numbers c in the interval (2,3). Also, I entered 18 as the answer to this problem and I got it wrong.

4. Oct 21, 2004

### JasonRox

I wish I had a paint program on here so I can post pictures of graphs, but I can't.

Let's use this lame graph.

..........3
-------2------------------ <-Make this line function x.
.........1
-------0------------------ <-This is f '(x)

If we take the interval (-2,-1), and use the mean value theorem we should get the right value.

Let's try...

f(b)-f(a) = 2-2 (function x is a straight line).

b-a does equal zero because they are not equals, so let's pick -2, and -1, which are in the interval. Now, b-a=-1.

f '(c)=[f(b)-f(a)]/(b-a)=(2-2)/-1=0.

The slope is zero, and we know that. Also, 0 is not in the interval (-2,-1).

Every point has that slope, but if it were like this...

-.............-
..-..........-
....-.......-
......-...-
........-

See how it goes up and down? I hope so.

If we took the same intervals as above, we would still get zero. The mean value theorem says that atleast one point will be that slope, in the interval. Is there one? Yes, there is exactly one.

Try finding it.