Worksheet: The Mean Value Theorem

In this worksheet, we will practice interpreting and using the mean value theorem and Rolle's theorem.

Q1:

Mason is not convinced that the mean value theorem is true because, he says, the function ๐‘“(๐‘ฅ)=|๐‘ฅ| has the property that if we take ๐‘Ž=โˆ’2 and ๐‘=2, we have ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž=0, and yet there is no point ๐‘ฅ where ๐‘“โ€ฒ(๐‘ฅ)=0. What is his error?

  • AThe function is not differentiable at ๐‘ฅ=0. The theorem requires diferentiability on an interval.
  • BThe theorem requires the domain to be an interval, which โ„ is not.
  • CThe function is not continuous. The theorem requires continuity on an interval.
  • DThe function should be strictly increasing on the interval.
  • EThe function should be strictly decreasing on the interval.

Q2:

Madison is not convinced that the mean value theorem is true because, she says, the function ๐‘“(๐‘ฅ)=|๐‘ฅ| is certainly differentiable on โ„โˆ’{0}. But if we take ๐‘Ž=โˆ’1 and ๐‘=1, we have ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž=0, and yet there is no point ๐‘ฅ where ๐‘“โ€ฒ(๐‘ฅ)=0. What is her error?

  • AThe function should be continuous on the interval.
  • BThe theorem requires the domain to be an interval, which โ„โˆ’{0} is not.
  • CThe function should be strictly increasing on the interval.
  • DThe theorem requires that the function be differentiable on its domain.
  • EThe function should be strictly decreasing on the interval.

Q3:

Does the mean value theorem apply for the function ๐‘ฆ=2โˆš๐‘ฅโˆ’4๏Šจ over the interval [โˆ’2,2]?

  • ANo
  • BYes

Q4:

Does the mean value theorem apply to the function ๐‘ฆ=2|๐‘ฅ+2| over the interval [0,6]?

  • ANo
  • BYes

Q5:

Does the mean value theorem apply for the function ๐‘ฆ=2๐‘ฅโˆ’4๐‘ฅ+7๏Šฉ over the interval [0,5]?

  • AYes
  • BNo

Q6:

Does the mean value theorem apply for the function ๐‘ฆ=3+|2๐‘ฅ| over the interval [โˆ’2,2]?

  • AYes
  • BNo

Q7:

For the function ๐‘“(๐‘ฅ)=โˆš๐‘ฅ, find all the possible values of ๐‘ that satisfy the mean value theorem over the interval [4,9].

  • A ๏„ž 5 2
  • B 2 5 4
  • C 1 5
  • D0
  • E 5 2

Q8:

For the function ๐‘“(๐‘ฅ)=๐‘ฅ+2๐‘ฅ๏Šจ, find all the values of ๐‘ that satisfy the mean value theorem over the interval [โˆ’4,4].

Q9:

Does the mean value theorem apply for the function ๐‘ฆ=3๐‘ฅ2๐œ‹๐‘ฅsin over the interval [0,2]?

  • ANo
  • BYes

Q10:

For the function ๐‘“(๐‘ฅ)=3๐‘ฅ๏Šฉ, find all the possible values of ๐‘ that satisfy the mean value theorem over the interval [1,3].

  • A2
  • B โˆš 3 9 3
  • C12
  • D 3 โˆš 3 9
  • E0

Q11:

For the function ๐‘“(๐‘ฅ)=๐‘ฅโˆ’4๐‘ฅ๏Šฉ, find all the possible values of ๐‘ that satisfy the mean value theorem over the interval [โˆ’2,2].

  • A ๐‘ฅ = 0 , ๐‘ฅ = 2
  • B ๐‘ฅ = 2 , ๐‘ฅ = โˆ’ 2
  • C ๐‘ฅ = 2 โˆš 3 , ๐‘ฅ = โˆ’ 2 โˆš 3
  • D ๐‘ฅ = โˆ’ 2 โˆš 3
  • E ๐‘ฅ = 2 โˆš 3

Q12:

For the function ๐‘“(๐‘ฅ)=(๐‘ฅโˆ’1)๏Šฎ, find all the possible values of ๐‘ that satisfy the mean value theorem over the interval [0,2].

Q13:

A rock is dropped from a height of 81 ft. Its position ๐‘ก seconds after it is dropped until it hits the ground is given by the function ๐‘ (๐‘ก)=โˆ’16๐‘ก+81๏Šจ.

Determine how long it will take for the rock to hit the ground.

  • A0 s
  • B 3 2 s
  • C 8 1 1 6 s
  • D 9 2 โˆš 2 s
  • E 9 4 s

Find the average velocity, ๐‘ฃavg, of the rock from the point of release until it hits the ground.

Find the time ๐‘ก according to the mean value theorem when the instantaneous velocity of the rock is ๐‘ฃavg.

  • A 9 4 s
  • B 8 9 s
  • C 4 9 s
  • D0 s
  • E 9 8 s

Q14:

Does the mean value theorem apply for the function ๐‘ฆ=๐‘ฅ๐‘ฅ+1cos๏Ž„๏Šจ over the interval [0,1]?

  • AYes
  • BNo

Q15:

Does the mean value theorem apply for the function ๐‘ฆ=23๐œ‹๐‘ฅtan over the interval [1,3]?

  • AYes
  • BNo

Q16:

Consider the statement that if ๐‘“ is a differentiable function on an interval ๐ผ and ๐‘“โ€ฒ(๐‘ฅ)>0 there, then ๐‘“ is strictly increasing on ๐ผ.

Which of the following statements is equivalent to the above?

  • AIf ๐‘“ is differentiable on an interval ๐ผ and not strictly increasing there, then ๐‘“โ€ฒ(๐‘Ž)โ‰ค0 at some point ๐‘Žโˆˆ๐ผ.
  • BIf ๐‘“ is differentiable on an interval ๐ผ and not strictly increasing there, then ๐‘“โ€ฒ(๐‘ฅ)โ‰ค0 at all points ๐‘ฅโˆˆ๐ผ.
  • CIf ๐‘“ is differentiable on an interval ๐ผ and strictly increasing there, then ๐‘“โ€ฒ(๐‘ฅ)>0 at all points ๐‘ฅโˆˆ๐ผ.
  • DIf ๐‘“ is differentiable on an interval ๐ผ and ๐‘“โ€ฒ(๐‘ฅ)โ‰ค0 there, then ๐‘“ is not strictly increasing on ๐ผ.
  • EIf ๐‘“ is differentiable on an interval ๐ผ and strictly increasing there, then ๐‘“โ€ฒ(๐‘Ž)>0 at some point ๐‘Žโˆˆ๐ผ.

What does it mean for a function ๐‘“ to not be strictly increasing on the interval ๐ผ?

  • AThere are points ๐‘Ž,๐‘โˆˆ๐ผ with ๐‘Ž<๐‘ but ๐‘“(๐‘)<๐‘“(๐‘Ž).
  • BThere are points ๐‘Ž,๐‘โˆˆ๐ผ with ๐‘Ž<๐‘ but ๐‘“(๐‘)โ‰ค๐‘“(๐‘Ž).
  • CThere is a point ๐‘Žโˆˆ๐ผ where ๐‘“โ€ฒ(๐‘Ž)โ‰ค0.
  • DWhenever ๐‘Ž,๐‘โˆˆ๐ผ satisfy ๐‘Ž<๐‘, then ๐‘“(๐‘)โ‰ค๐‘“(๐‘Ž).
  • EThere are points ๐‘Ž,๐‘โˆˆ๐ผ with ๐‘Ž<๐‘ but ๐‘“(๐‘)=๐‘“(๐‘Ž).

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

  • AIf ๐‘“ is differentiable on ๐ผ and not strictly increasing, then take ๐‘Ž<๐‘ with ๐‘“(๐‘Ž)โ‰ฅ๐‘“(๐‘). By the mean value theorem, we get ๐‘ between ๐‘Ž and ๐‘ and where ๐‘“โ€ฒ(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž, so ๐‘“โ€ฒ(๐‘)โ‰ค0.
  • BIf ๐‘“ is differentiable on ๐ผ and not strictly increasing, then take ๐‘Ž<๐‘ with ๐‘“(๐‘Ž)โ‰ค๐‘“(๐‘). By the mean value theorem, we get ๐‘ between ๐‘Ž and ๐‘ where ๐‘“โ€ฒ(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž, so ๐‘“โ€ฒ(๐‘)โ‰ค0.
  • CIf ๐‘“ is differentiable on ๐ผ and not strictly increasing, then take ๐‘Ž<๐‘ with ๐‘“(๐‘Ž)โ‰ฅ๐‘“(๐‘). By the mean value theorem, we get ๐‘ between ๐‘Ž and ๐‘ where ๐‘“โ€ฒ(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž, so ๐‘“โ€ฒ(๐‘)=0.
  • DIf ๐‘“ is differentiable on ๐ผ and not strictly increasing, then take ๐‘Ž<๐‘ with ๐‘“(๐‘Ž)โ‰ฅ๐‘“(๐‘). By the mean value theorem, we get ๐‘ between ๐‘Ž and ๐‘ where ๐‘“โ€ฒ(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž, so ๐‘“โ€ฒ(๐‘)>0.
  • EIt is not possible to prove the statement using the mean value theorem.

Q17:

Consider the result: If ๐‘“ is differentiable on an interval ๐ผ and ๐‘“(๐‘ฅ)=0๏Ž˜, then ๐‘“(๐‘ฅ)=๐‘, a constant, for all ๐‘ฅโˆˆ๐ผ.

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

  • AIf ๐‘“ is differentiable on an interval ๐ผ and ๐‘“โ€ฒ(๐‘ฅ)โ‰ 0 at each ๐‘ฅโˆˆ๐ผ, then ๐‘“(๐‘ฅ) is not a constant function.
  • BIf ๐‘“ is a constant function on an interval ๐ผ, then ๐‘“ is differentiable and ๐‘“โ€ฒ(๐‘ฅ)=0 at all ๐‘ฅโˆˆ๐ผ.
  • CIf ๐‘“ is differentiable on an interval ๐ผ and ๐‘“โ€ฒ(๐‘Ž)โ‰ 0 at some ๐‘Žโˆˆ๐ผ, then ๐‘“(๐‘ฅ) is not a constant function.
  • DIf ๐‘“ is differentiable on an interval ๐ผ and not a constant function, then ๐‘“โ€ฒ(๐‘Ž)โ‰ 0 at some ๐‘Žโˆˆ๐ผ.
  • EIf ๐‘“ is differentiable on an interval ๐ผ and not a constant function, then ๐‘“โ€ฒ(๐‘ฅ)โ‰ 0 at all ๐‘ฅโˆˆ๐ผ.

If ๐‘“ is differentiable on an interval ๐ผ and not constant, we get points ๐‘Ž,๐‘โˆˆ๐ผ with ๐‘“(๐‘Ž)โ‰ ๐‘“(๐‘). How does this show that ๐‘“โ€ฒ(๐‘)โ‰ 0 at some point ๐‘โˆˆ๐ผ?

  • Abecause then ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Žโ‰ 0 and by the mean value theorem, there is a point ๐‘ with ๐‘“โ€ฒ(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Ž
  • Bbecause ๐‘“(๐‘)โ‰ ๐‘“(๐‘Ž) means that one of ๐‘“โ€ฒ(๐‘Ž) or ๐‘“โ€ฒ(๐‘) is not zero; we can take ๐‘ as this point
  • Cbecause only constant functions has ๐‘“โ€ฒ(๐‘ฅ)=0 everywhere
  • Dbecause then ๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)๐‘โˆ’๐‘Žโ‰ 0 and by the mean value theorem, there is a point ๐‘ with ๐‘“(๐‘)=๐‘“(๐‘)โˆ’๐‘“(๐‘Ž)
  • Ebecause if ๐‘“โ€ฒ(๐‘ฅ)=0 on ๐ผ, then ๐‘“ would be a constant function

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