# Mvt differentiation proof question

• transgalactic
In summary, we can prove that g(x)=(x-x_0)*f(x) is differentiable at x_0 by using the definition of the derivative and substituting g(x) into it. This results in both the left and right limits being equal to f(x_0), which is known to be continuous. Therefore, g'(x_0) exists and is equal to f(x_0).
transgalactic
suppose f is a continues function on point x_0
prove that g(x)=(x-x_0)*f(x) differentiable on x_0??
calculate g'(x_0)

i tried to think like this:
if f(x) is continues on x_0 then lim f(x) as x->x_0 equals f(x_0)

mvt says f'(c)=[f(a)-f(b)]
cauchys mvt says f'(c)/g'(c)=[f(a)-f(b)]/[g(a)-g(b)]

??

Obviously those mvt formula you wrote don't make sense a priori since f is only assumed continuous and not differentiable.

Try doing something with the definition of the derivative of g instead.

the definition of a derivative is
lim f(x)=[f(x+x_0) -f(x_0)]/[x-x_0]
as x->x_0

what should i do with it?

If you write that limit out for g, what do you get?

by definition if x->x_0
then the limit formula is
[g(x)-g(x_0)]/[x-x_0]

i only know this
but i don't know
what to do next?
how to combine things into a logical thing

Well, yes, this is the formula for the derivative of g at x_0, but you also know that

g(x)=(x-x_0)*f(x).

So what do you get when you substitute this expression into the definition of g'(x_0) and take the limit?

i did the definition of one sided derivative
$g_ - '(x) = \mathop {\lim }\limits_{x \to x_0 - } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }} \\$
$g_ + '(x) = \mathop {\lim }\limits_{x \to x_0 + } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }}$

can i cut the numenaor and the denominator by x-x_0
and then each side will equal f(x)
so the limit is f(x_0) which is known that its continues so it equals from both side
so the limits equal

is this a correct proof.

Last edited:

## 1. What is a "Mvt differentiation proof question"?

A "Mvt differentiation proof question" refers to a type of mathematical problem that involves applying the Mean Value Theorem (MVT) to prove the existence of a certain value or property. This type of question is commonly encountered in calculus and other advanced math courses.

## 2. How do you approach a Mvt differentiation proof question?

To solve a Mvt differentiation proof question, you should first understand the basic concept of the Mean Value Theorem and how it can be applied. Then, carefully read and analyze the given problem to identify the key information and variables. Finally, use the MVT to prove the desired result or value.

## 3. What is the purpose of using the Mean Value Theorem in a differentiation proof question?

The Mean Value Theorem is a powerful tool in calculus that allows us to prove the existence of certain values or properties in a function. In a differentiation proof question, the MVT is used to show that a function must have a specific value or property at some point within a given interval.

## 4. What are some common mistakes to avoid when solving a Mvt differentiation proof question?

One common mistake is to assume that the MVT can be applied to any function or interval. It is important to check the conditions of the MVT, such as the function being continuous and differentiable on the interval, before using it. Another mistake is to overlook the importance of the intermediate value theorem, which is often used in conjunction with the MVT in differentiation proof questions.

## 5. Are there any tips for solving Mvt differentiation proof questions more efficiently?

One helpful tip is to practice using the MVT in various types of problems, as this will help you become more familiar with its applications and conditions. Another tip is to carefully read the question and identify the key information and variables before attempting to solve it. It can also be useful to work backwards from the desired result to determine the steps needed to prove it using the MVT.

Replies
20
Views
2K
Replies
13
Views
1K
Replies
11
Views
2K
Replies
3
Views
2K
Replies
6
Views
1K
Replies
12
Views
1K
Replies
20
Views
3K
Replies
12
Views
2K
Replies
8
Views
1K
Replies
1
Views
804