Verifying Hypotheses of the Mean Value Theorem for f(x)=1/(x-2) on [1,4]

In summary, the function f(x) on the interval [1, 4] is continuous but not differentiable, and the function f(x) on the interval [3, 6] is differentiable but not continuous.
  • #1
FritoTaco
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Homework Statement


Find all the numbers [itex]c[/itex] that satisfy the conclusion of the Mean Value Theorem for the functions

[itex]f(x)=\dfrac{1}{x-2}[/itex] on the interval [itex][1, 4][/itex]

[itex]f(x)=\dfrac{1}{x-2}[/itex] on the interval [itex][3, 6][/itex]

I don't need help solving for [itex]c[/itex], I just want to know how I can verify that the hypotheses of the mean Value Theorem are satisfied by the function [itex]f(x)[/itex] on the given interval. I know the first one, there is no such number [itex]c[/itex] that is guaranteed by the mean Value Theorem but there is for the second one. How can we verify the first one?

Homework Equations



Rolle's Theorem: Let [itex]f[/itex] be a function that satisfies the following three hypotheses:

1. [itex]f[/itex] is continuous on the closed interval [itex][a, b][/itex]
2. [itex]f[/itex] is differentiable on the open interval [itex](a, b)[/itex]
3. [itex]f(a)=f(b)[/itex]

Then there exists a number [itex]c[/itex] between [itex]a[/itex] and [itex]b[/itex] such that [itex]f'(c)=0[/itex]

The Attempt at a Solution



So I would test if it's differentiable for the first function.

[itex]f(1)=\dfrac{1}{1-2}=-1[/itex]

[itex]f(4)=\dfrac{1}{4-2}=1/2[/itex]

So for the first interval of [itex][1, 4][/itex] there is no such [itex]c[/itex] number because the inputs (-1 and 1/2) are not equal.
How do we know if it's continuous? If it's differentiable does that mean it has to be continuous?
 
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  • #2
The mean value theorem is more general than Rolle's theorem, it does not need the f(a)=f(b) condition.
FritoTaco said:
If it's differentiable does that mean it has to be continuous?
Yes - but keep in mind that this just helps directly with (a,b), not with [a,b].
 
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  • #3
So my interval is a closed interval for [a, b]. If the case doesn't apply, how would you go about verifying if it's continuous?
 
  • #4
At the time you get those questions, it should be fine to do that just by inspection.

But there is also a statement about the continuity of f(x)=c/g(x) with a constant c that depends on properties of g(x).
 
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  • #5
Alright, thank you!
 

What is the Mean Value Theorem?

The Mean Value Theorem is a mathematical theorem that states if a function is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there exists at least one point c in the open interval where the slope of the tangent line is equal to the average rate of change of the function over the closed interval.

What is the significance of the Mean Value Theorem?

The Mean Value Theorem is significant because it provides a way to connect the concepts of continuity and differentiability. It also allows us to find the instantaneous rate of change at a specific point on a curve by using the average rate of change over an interval.

How is the Mean Value Theorem applied in real life?

The Mean Value Theorem has various applications in real life, such as in physics, economics, and engineering. For example, it can be used to determine the maximum velocity of a moving object, the average rate of change of a stock price, or the maximum acceleration of a car.

Can the Mean Value Theorem be extended to higher dimensions?

Yes, the Mean Value Theorem can be extended to higher dimensions. In multivariable calculus, it is known as the Mean Value Theorem for Partial Derivatives and it states that if a function is continuous and differentiable on a closed interval in a multi-dimensional space, then there exists a point c within that interval where the partial derivatives of the function are equal to the average rates of change in each direction.

What are the conditions for the Mean Value Theorem to hold?

The Mean Value Theorem holds if the function is continuous on a closed interval and differentiable on the open interval. Additionally, the function must not have any vertical tangents or sharp turns within the interval.

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