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## Homework Statement

## 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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- Thread starter utkarsh009
- Start date

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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Lagrange's mean value theorem

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

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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

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Thanks mate! Solved it... But how did you think of that function h(x)??

- #4

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But your h(x) had solved the problem. What was the error??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.

- #5

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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.

- #6

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[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.

- #7

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Yah... Was just thinking the same and was about to post it... :Po_O said:

[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.

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.

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.

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.

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.

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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