# Mean value thereom application

• Arnoldjavs3
In summary: And as such this proves that since 48|y-x| will be greater since y is greater than c? So the side with c will always be smaller.No, it just states that if y is greater than c, then 48|y-x| will be greater.
Arnoldjavs3

## Homework Statement

Suppose that x,y are in (0,0.5). Show that:
$$\left|\frac{1}{x^3}-\frac{1}{y^3}\right| ≥ 48|x-y|$$

## The Attempt at a Solution

I think that we need a function ##f(t) = t^{-3}## & ##f'(t) = -3t^{-4}##. Now we use the MVT here to show that there is some c that makes the expression in the problem correct right? My question is how do I incorporate the absolute value notations in the statement here?
$$-3c^{-4} = \frac{f(y) - f(x)}{y-x}$$
I think we have to assume that either x or y is greater than the other(because if they were equal then the above owuld be proven). Do I just input c for 0.5? I'm unsure.

Arnoldjavs3 said:

## Homework Statement

Suppose that x,y are in (0,0.5). Show that:
$$\left|\frac{1}{x^3}-\frac{1}{y^3}\right| ≥ 48|x-y|$$

## The Attempt at a Solution

I think that we need a function ##f(t) = t^{-3}## & ##f'(t) = -3t^{-4}##. Now we use the MVT here to show that there is some c that makes the expression in the problem correct right? My question is how do I incorporate the absolute value notations in the statement here?
$$-3c^{-4} = \frac{f(y) - f(x)}{y-x}$$
I think we have to assume that either x or y is greater than the other(because if they were equal then the above owuld be proven). Do I just input c for 0.5? I'm unsure.

It's not clear why the MVT would help here.

PeroK said:
It's not clear why the MVT would help here.

I forgot to add that they are differentiable on the interval (x,y)

Arnoldjavs3 said:
I forgot to add that they are differentiable on the interval (x,y)

I still can't see why you think the MVT helps you.

Arnoldjavs3 said:

## Homework Statement

Suppose that x,y are in (0,0.5). Show that:
$$\left|\frac{1}{x^3}-\frac{1}{y^3}\right| ≥ 48|x-y|$$

## The Attempt at a Solution

I think that we need a function ##f(t) = t^{-3}## & ##f'(t) = -3t^{-4}##. Now we use the MVT here to show that there is some c that makes the expression in the problem correct right? My question is how do I incorporate the absolute value notations in the statement here?
$$-3c^{-4} = \frac{f(y) - f(x)}{y-x}$$
I think we have to assume that either x or y is greater than the other(because if they were equal then the above owuld be proven). Do I just input c for 0.5? I'm unsure.
Remember that all your variables are in ##(0,\frac 1 2)##. If you take the absolute value of both sides of what you have written you get$$\frac 3 {c^4} =\frac{|f(y) - f(x)|}{|y-x |}$$which can be rewritten$$\frac 3 {c^4} |y-x| =\left | \frac 1 {x^3} - \frac 1 {y^3}\right |$$Think about how large or small the left side can be.

Arnoldjavs3
LCKurtz said:
Remember that all your variables are in ##(0,\frac 1 2)##. If you take the absolute value of both sides of what you have written you get$$\frac 3 {c^4} =\frac{|f(y) - f(x)|}{|y-x |}$$which can be rewritten$$\frac 3 {c^4} |y-x| =\left | \frac 1 {x^3} - \frac 1 {y^3}\right |$$Think about how large or small the left side can be.

Since c is an element of x,y then it can't be larger than 0.5 or smaller than 0? And as such this proves that since 48|y-x| will be greater since y is greater than c? So the side with c will always be smaller.

Last edited:
Arnoldjavs3 said:
Since c is an element of x,y

That doesn't make any sense. ##c## is a number in ##(0,.5)##, not an "element" of x,y, whatever that means.

then it can't be larger than 0.5 or smaller than 0? And as such this proves that since 48|y-x| will be greater since y is greater than c? So the side with c will always be smaller.

Stating the result doesn't prove anything. You need to show some inequalities.

## 1. What is the Mean Value Theorem and how is it applied?

The Mean Value Theorem is a fundamental theorem in calculus that states that for a continuous and differentiable function, there exists at least one point in the interval where the slope of the tangent line is equal to the average rate of change of the function. This theorem is commonly used to solve optimization problems and find the average value of a function over a given interval.

## 2. How is the Mean Value Theorem used to prove the existence of solutions to equations?

The Mean Value Theorem can be used to prove the existence of solutions to equations by showing that the function has at least one root in the given interval. If the average rate of change of the function is equal to zero at a point, then by the Mean Value Theorem, there must be a point where the slope of the tangent line is also equal to zero, indicating a root of the function.

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

No, the Mean Value Theorem can only be applied to continuous and differentiable functions. This means that the function must have no breaks or jumps in its graph and it must have a well-defined slope at every point in the interval.

## 4. How can the Mean Value Theorem be used to find the average rate of change of a function?

The Mean Value Theorem can be used to find the average rate of change of a function by first finding a point in the interval where the slope of the tangent line is equal to the average rate of change. This point can then be substituted into the derivative of the function to find the slope of the tangent line, which is also equal to the average rate of change.

## 5. Can the Mean Value Theorem be extended to higher dimensions?

Yes, the Mean Value Theorem can be extended to higher dimensions through the use of partial derivatives. In this case, the average rate of change of a multivariable function is equal to the slope of the tangent plane at a point in the given interval. This extension is known as the Multivariable Mean Value Theorem.

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