Calulus Help: Mean Value Theorem

In summary, the Mean Value Theorem tells us that there will be a point in the interval where the function's derivative is equal to the function itself. The point is f '(c)=[f(b)-f(a)]/(b-a). I found this point to be 0, and it has a slope of zero.
  • #1
buffgilville
91
0
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.

What am I doing wrong? Please help.
 
Last edited:
Physics news on Phys.org
  • #2
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
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. :confused:
 
  • #4
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.
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists a point within that interval where the slope of the tangent line is equal to the slope of the secant line connecting the endpoints of the interval.

2. How is the Mean Value Theorem used in calculus?

The Mean Value Theorem is used to prove other important theorems and is also used to solve optimization problems and find the root of equations. It is a crucial tool in understanding the behavior of functions and their derivatives.

3. What is the significance of the Mean Value Theorem?

The Mean Value Theorem is significant because it provides a relationship between the average rate of change of a function and the instantaneous rate of change at a specific point. This allows for a deeper understanding of the behavior of functions and is essential in many applications of calculus.

4. How is the Mean Value Theorem related to Rolle's Theorem?

Rolle's Theorem is a special case of the Mean Value Theorem where the endpoints of the interval have the same function value. This means that the slope of the secant line connecting the endpoints is equal to zero, making the tangent line at the point of intersection horizontal.

5. Can the Mean Value Theorem be applied to all functions?

No, the Mean Value Theorem can only be applied to functions that are continuous on a closed interval and differentiable on the open interval. If a function does not meet these criteria, the Mean Value Theorem cannot be used to find the point where the instantaneous rate of change is equal to the average rate of change.

Similar threads

  • Calculus
Replies
12
Views
371
Replies
2
Views
383
  • Introductory Physics Homework Help
Replies
2
Views
417
  • Introductory Physics Homework Help
Replies
2
Views
553
Replies
3
Views
1K
Replies
11
Views
990
  • Introductory Physics Homework Help
Replies
25
Views
211
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
177
Back
Top