Lagrange's mean value theorem problem

In summary, the conversation discusses the use of Lagrange's mean value theorem and Rolle's theorem to solve a problem involving a function h(x) that is defined in terms of a linear combination of f(x), f(0), and f(1). The ultimate goal is to find a c value that satisfies a given equation by applying these theorems. After some discussion and confusion, a relatively simple function, h(x) = xf(x), is found to solve the problem.
  • #1
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Homework Statement


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Homework Equations



Lagrange's mean value theorem

The Attempt at a Solution


Applying LMVT,
There exists c belonging to (0,1) which satisfies f'(c) = f(1)-f(0)/1 = -f(0)
But this gets me nowhere close to the options... :(
 
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  • #2
edit
Sorry the post I made contained an error. I'm not quite sure how to fix it yet but basically what I think you need to do is find a function h(x) and define it in terms of some linear combination of f(x), f(0), f(1) so that it equals zero at the end points and then apply Rolle's Theorem.
 
  • #3
Thanks mate! Solved it... But how did you think of that function h(x)??
 
  • #4
o_O said:
edit
Sorry the post I made contained an error. I'm not quite sure how to fix it yet but basically what I think you need to do is find a function h(x) and define it in terms of some linear combination of f(x), f(0), f(1) so that it equals zero at the end points and then apply Rolle's Theorem.
But your h(x) had solved the problem. What was the error??
 
  • #5
Hey sorry I edited my previous post. What I wrote doesn't work because the two c's might be different. The c that works for the equation you posted might be a different c from the one I used for h(x).

When trying to define an h(x) we want it to be 0 at the end points so we can apply Rolle's Theorem. It looks like Rolle's Theorem is needed because of the way the question is set up (i.e it looks like some derivative is equal to 0). And we want f(0) to disappear from the equation for the derivative because we don't know its value.
 
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  • #6
Actually a relatively simple function works lol. Sorry for all this confusion:
[tex]
h(x) = xf(x)[/tex]
There exists a c in (0, 1) such that [tex]
0 = cf'(c) + f(c)[/tex]

by Rolle's Theorem.
 
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  • #7
o_O said:
Actually a relatively simple function works lol. Sorry for all this confusion:
[tex]
h(x) = xf(x)[/tex]
There exists a c in (0, 1) such that [tex]
0 = cf'(c) + f(c)[/tex]

by Rolle's Theorem.
Yah... Was just thinking the same and was about to post it... :P
 

1. What is Lagrange's mean value theorem problem?

Lagrange's mean value theorem is a fundamental theorem in calculus that establishes a relationship between the derivative of a function and the slope of a secant line connecting two points on the function. It states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point within the interval where the slope of the tangent line is equal to the slope of the secant line connecting the two points.

2. What is the significance of Lagrange's mean value theorem?

Lagrange's mean value theorem is significant because it provides a way to guarantee that a function has a point where the instantaneous rate of change is equal to the average rate of change. This is useful in many applications, such as optimization problems and curve sketching.

3. How is Lagrange's mean value theorem applied in real-world problems?

Lagrange's mean value theorem is often used in real-world problems to find the maximum or minimum values of a function. For example, it can be used to find the minimum amount of time it takes for a car to reach a certain distance or the maximum profit a company can make with a given production rate.

4. What are the conditions for Lagrange's mean value theorem to be applicable?

The conditions for Lagrange's mean value theorem to be applicable are that the function must be continuous on a closed interval and differentiable on the open interval. Additionally, the endpoints of the interval must have the same function value.

5. Can Lagrange's mean value theorem be generalized to higher dimensions?

Yes, Lagrange's mean value theorem can be generalized to higher dimensions. In multivariable calculus, it is known as the mean value theorem for vector-valued functions. It states that if a vector-valued function is continuous and differentiable on a closed interval, then there exists a point within the interval where the derivative of the function is equal to the average rate of change of the function.

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