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Lesson: The Mean Value Theorem and Its Interpretation

In this lesson, we will learn how to use the mean value theorem.

Worksheet • 4 Questions

Q1:

Madison is not convinced that the mean value theorem is true because, she says, the function 𝑓 ( π‘₯ ) = | π‘₯ | is certainly differentiable on ℝ βˆ’ { 0 } . But if we take π‘Ž = βˆ’ 1 and 𝑏 = 1 , we have 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž = 0 , and yet there is no point π‘₯ where 𝑓 ( π‘₯ ) = 0 β€² . What is her error?

  • AThe theorem requires the domain to be an interval, which ℝ βˆ’ { 0 } is not.
  • BThe function is not continuous, and the theorem requires continuity on an interval.
  • CThe theorem requires that the function be differentiable everywhere, which | π‘₯ | is not.
  • DThe function should be strictly increasing on the interval.
  • EThe function should be strictly decreasing on the interval.

Q2:

Consider the result: If 𝑓 is differentiable on an interval 𝐼 and 𝑓 ( π‘₯ ) = 0 β€² , then 𝑓 ( π‘₯ ) = 𝑐 , a constant, for all π‘₯ ∈ 𝐼 .

Which of the following statements says exactly the same thing as the constant function result?

  • AIf 𝑓 is differentiable on an interval 𝐼 and not a constant function, then 𝑓 β€² ( π‘Ž ) β‰  0 at some π‘Ž ∈ 𝐼 .
  • BIf 𝑓 is differentiable on an interval 𝐼 and 𝑓 β€² ( π‘₯ ) β‰  0 at each π‘₯ ∈ 𝐼 , then 𝑓 ( π‘₯ ) is not a constant function.
  • CIf 𝑓 is differentiable on an interval 𝐼 and not a constant function, then 𝑓 β€² ( π‘₯ ) β‰  0 at all π‘₯ ∈ 𝐼 .
  • DIf 𝑓 is differentiable on an interval 𝐼 and 𝑓 β€² ( π‘Ž ) β‰  0 at some π‘Ž ∈ 𝐼 , then 𝑓 ( π‘₯ ) is not a constant function.
  • EIf 𝑓 is a constant function on an interval 𝐼 , then 𝑓 is differentiable and 𝑓 β€² ( π‘₯ ) = 0 at all π‘₯ ∈ 𝐼 .

If 𝑓 is differentiable on an interval 𝐼 and not constant, we get points π‘Ž , 𝑏 ∈ 𝐼 with 𝑓 ( π‘Ž ) β‰  𝑓 ( 𝑏 ) . How does this show that 𝑓 β€² ( 𝑐 ) β‰  0 at some point 𝑐 ∈ 𝐼 ?

  • Abecause then 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž β‰  0 and by the mean value theorem, there is a point 𝑐 with 𝑓 β€² ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž
  • Bbecause then 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž β‰  0 and by the mean value theorem, there is a point 𝑐 with 𝑓 ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž )
  • Cbecause 𝑓 ( 𝑏 ) β‰  𝑓 ( π‘Ž ) means that one of 𝑓 β€² ( π‘Ž ) or 𝑓 β€² ( 𝑏 ) is not zero; we can take 𝑐 as this point
  • Dbecause if 𝑓 β€² ( π‘₯ ) = 0 on 𝐼 , then 𝑓 would be a constant function
  • Ebecause only constant functions has 𝑓 β€² ( π‘₯ ) = 0 everywhere

Q3:

Mason is not convinced that the mean value theorem is true because, he says, the function 𝑓 ( π‘₯ ) = | π‘₯ | has the property that if we take π‘Ž = βˆ’ 2 and 𝑏 = 2 , we have 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž = 0 , and yet there is no point π‘₯ where 𝑓 β€² ( π‘₯ ) = 0 . What is his error?

  • AThe function is not differentiable at π‘₯ = 0 . The theorem requires diferentiability on an interval.
  • BThe function is not continuous. The theorem requires continuity on an interval.
  • CThe theorem requires the domain to be an interval, which ℝ is not.
  • DThe function should be strictly increasing on the interval.
  • EThe function should be strictly decreasing on the interval.
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