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

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To verify the hypotheses of the Mean Value Theorem (MVT) for the function f(x)=1/(x-2) on the interval [1, 4], it is essential to check continuity and differentiability. The function is not continuous on [1, 4] due to a vertical asymptote at x=2, which means it fails the first condition of the MVT. Consequently, since f(1) and f(4) yield different values (-1 and 1/2), there is no c that satisfies the conclusion of the MVT for this interval. In contrast, the function is continuous and differentiable on the interval [3, 6], allowing for the application of the MVT there. Thus, verifying continuity involves checking for points of discontinuity, which in this case confirms that the MVT does not apply for the first interval.
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Homework Statement


Find all the numbers c that satisfy the conclusion of the Mean Value Theorem for the functions

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

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

I don't need help solving for c, I just want to know how I can verify that the hypotheses of the mean Value Theorem are satisfied by the function f(x) on the given interval. I know the first one, there is no such number c 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 f be a function that satisfies the following three hypotheses:

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

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

The Attempt at a Solution



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

f(1)=\dfrac{1}{1-2}=-1

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

So for the first interval of [1, 4] there is no such c 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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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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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?
 
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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Alright, thank you!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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