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Worksheet: The Mean Value Theorem and Its Interpretation

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 function should be strictly decreasing on the interval.
  • BThe theorem requires that the function be differentiable everywhere, which | π‘₯ | is not.
  • CThe function should be strictly increasing on the interval.
  • DThe theorem requires the domain to be an interval, which ℝ βˆ’ { 0 } is not.
  • EThe function is not continuous, and the theorem requires continuity on an 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 a constant function on an interval 𝐼 , then 𝑓 is differentiable and 𝑓 β€² ( π‘₯ ) = 0 at all π‘₯ ∈ 𝐼 .
  • BIf 𝑓 is differentiable on an interval 𝐼 and not a constant function, then 𝑓 β€² ( π‘₯ ) β‰  0 at all π‘₯ ∈ 𝐼 .
  • CIf 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.
  • DIf 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 .
  • EIf 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.

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 if 𝑓 β€² ( π‘₯ ) = 0 on 𝐼 , then 𝑓 would be a constant function
  • Cbecause only constant functions has 𝑓 β€² ( π‘₯ ) = 0 everywhere
  • Dbecause 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
  • Ebecause 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 )

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

Q4:

Consider the statement that if 𝑓 is a differentiable function on an interval 𝐼 and 𝑓 β€² ( π‘₯ ) > 0 there, then 𝑓 is strictly increasing on 𝐼 .

Which of the following statements is equivalent to the above?

  • AIf 𝑓 is differentiable on an interval 𝐼 and 𝑓 β€² ( π‘₯ ) ≀ 0 there, then 𝑓 is not strictly increasing on 𝐼 .
  • BIf 𝑓 is differentiable on an interval 𝐼 and not strictly increasing there, then 𝑓 β€² ( π‘₯ ) ≀ 0 at all points π‘₯ ∈ 𝐼 .
  • CIf 𝑓 is differentiable on an interval 𝐼 and strictly increasing there, then 𝑓 β€² ( π‘₯ ) > 0 at all points π‘₯ ∈ 𝐼 .
  • DIf 𝑓 is differentiable on an interval 𝐼 and not strictly increasing there, then 𝑓 β€² ( π‘Ž ) ≀ 0 at some point π‘Ž ∈ 𝐼 .
  • EIf 𝑓 is differentiable on an interval 𝐼 and strictly increasing there, then 𝑓 β€² ( π‘Ž ) > 0 at some point π‘Ž ∈ 𝐼 .

What does it mean for a function 𝑓 to not be strictly increasing on the interval 𝐼 ?

  • AThere are points π‘Ž , 𝑏 ∈ 𝐼 with π‘Ž < 𝑏 but 𝑓 ( 𝑏 ) ≀ 𝑓 ( π‘Ž ) .
  • BThere are points π‘Ž , 𝑏 ∈ 𝐼 with π‘Ž < 𝑏 but 𝑓 ( 𝑏 ) = 𝑓 ( π‘Ž ) .
  • CThere are points π‘Ž , 𝑏 ∈ 𝐼 with π‘Ž < 𝑏 but 𝑓 ( 𝑏 ) < 𝑓 ( π‘Ž ) .
  • DThere is a point π‘Ž ∈ 𝐼 where 𝑓 β€² ( π‘Ž ) ≀ 0 .
  • EWhenever π‘Ž , 𝑏 ∈ 𝐼 satisfy π‘Ž < 𝑏 , then 𝑓 ( 𝑏 ) ≀ 𝑓 ( π‘Ž ) .

Using the equivalent statement to the main result, how can you use the mean value theorem to prove the equivalent statement?

  • AIf 𝑓 is differentiable on 𝐼 and not strictly increasing, then take π‘Ž < 𝑏 with 𝑓 ( π‘Ž ) ≀ 𝑓 ( 𝑏 ) . By the mean value theorem, we get 𝑐 between π‘Ž and 𝑏 where 𝑓 β€² ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž , so 𝑓 β€² ( 𝑐 ) ≀ 0 .
  • BIt is not possible to prove the statement using the mean value theorem.
  • CIf 𝑓 is differentiable on 𝐼 and not strictly increasing, then take π‘Ž < 𝑏 with 𝑓 ( π‘Ž ) β‰₯ 𝑓 ( 𝑏 ) . By the mean value theorem, we get 𝑐 between π‘Ž and 𝑏 and where 𝑓 β€² ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž , so 𝑓 β€² ( 𝑐 ) ≀ 0 .
  • DIf 𝑓 is differentiable on 𝐼 and not strictly increasing, then take π‘Ž < 𝑏 with 𝑓 ( π‘Ž ) β‰₯ 𝑓 ( 𝑏 ) . By the mean value theorem, we get 𝑐 between π‘Ž and 𝑏 where 𝑓 β€² ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž , so 𝑓 β€² ( 𝑐 ) 0 > .
  • EIf 𝑓 is differentiable on 𝐼 and not strictly increasing, then take π‘Ž < 𝑏 with 𝑓 ( π‘Ž ) β‰₯ 𝑓 ( 𝑏 ) . By the mean value theorem, we get 𝑐 between π‘Ž and 𝑏 where 𝑓 β€² ( 𝑐 ) = 𝑓 ( 𝑏 ) βˆ’ 𝑓 ( π‘Ž ) 𝑏 βˆ’ π‘Ž , so 𝑓 β€² ( 𝑐 ) = 0 .