Mean value theorem, closed intervals

In summary, the Mean Value Theorem states that if a function is continuous on [a,b] and differentiable on (a,b), then there is at least one c in (a,b) such that f'(c) = (f(b) - f(a))/(b - a). Some may argue that it would be better to state that c is in [a,b] rather than (a,b), as this would account for the possibility of the function not being differentiable at the endpoints. However, this is not necessary as c being in (a,b) also implies that c is in [a,b]. Additionally, the theorem is stronger when stated in terms of (a,b) rather than [a,b]. It should be
  • #1
PFuser1232
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Mean Value Theorem

Suppose that ##f## is a function that is continuous on ##[a,b]## and differentiable on ##(a,b)##. Then there is at least one ##c## in ##(a,b)## such that:
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

My question is: wouldn't it be better to state that ##c## is in ##[a,b]## rather than ##(a,b)##? For example, if ##f(x) = 2## for ##1 \leq x \leq 3##, then:
$$f'(x) = 0 = \frac{f(3) - f(1)}{2}$$
For all ##x##, including 3, which is one of the endpoints of the interval.
 
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  • #2
If ##c## is in ##(a,b)##, then ##c## is in ##[a,b]##. Besides, ##f## might not be differentiable at ##a## or ##b##.
 
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  • #3
Because the theorem is a stronger result.
 
  • #4
micromass said:
If ##c## is in ##(a,b)##, then ##c## is in ##[a,b]##. Besides, ##f## might not be differentiable at ##a## or ##b##.
micromass said:
If ##c## is in ##(a,b)##, then ##c## is in ##[a,b]##. Besides, ##f## might not be differentiable at ##a## or ##b##.

Yeah, I think that's it. ##f## is differentiable on ##(a,b)##, not ##[a,b]##.
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that 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 (a,b) where the slope of the tangent line to the function at c is equal to the average rate of change of the function over the interval [a,b]. In other words, it guarantees the existence of a point where the instantaneous rate of change is equal to the average rate of change over a given interval.

2. How is the Mean Value Theorem used in calculus?

The Mean Value Theorem is used to prove other important theorems and results in calculus, such as the First and Second Derivative Tests, Rolle's Theorem, and the Fundamental Theorem of Calculus. It is also used to find critical points and to determine if a function has a maximum or minimum value on a given interval.

3. What is a closed interval?

A closed interval is a set of real numbers that includes its endpoints. For example, the interval [0,1] includes the numbers 0 and 1, whereas the interval (0,1) does not include these endpoints. In the Mean Value Theorem, the function must be continuous on the closed interval [a,b] in order for the theorem to hold.

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

No, the Mean Value Theorem can only be applied to continuous functions on a closed interval [a,b] and differentiable on the open interval (a,b). If a function is not continuous or differentiable on the given interval, the Mean Value Theorem cannot be used to find a point where the slope of the tangent line is equal to the average rate of change.

5. How is the Mean Value Theorem related to the concept of average rate of change?

The Mean Value Theorem states that at some point c in the open interval (a,b), the instantaneous rate of change of a function is equal to the average rate of change over the interval [a,b]. This means that the Mean Value Theorem connects the instantaneous rate of change, which is represented by the derivative of a function, to the average rate of change, which is found by calculating the slope of a secant line between two points on the function.

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