Worksheet: Intermediate Value Theorem

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In this worksheet, we will practice interpreting the intermediate value theorem and using it to approximate a zero of a function.

Q1:

The function 𝐹(π‘₯)=1π‘₯+3 satisfies 𝐹(βˆ’1)<3 and 𝐹(1)>3. But there is no π‘₯ between βˆ’1 and 1 where 𝐹(π‘₯)=3. Why does this not violate the intermediate value theorem?

  • ABecause the function is not defined on the entire interval [βˆ’1,1].
  • BBecause the intermediate value theorem only applies on the interval (0,∞).
  • CBecause the intermediate value theorem only applies to cases where 𝐹(π‘₯)=0, not 𝐹(π‘₯)=3.
  • DBecause the function 𝐹 is not continuous over its domain.
  • EBecause the intermediate value theorem only applies to polynomial functions.

Q2:

The figure shows the graph of the function 𝑓 on the interval [0,16] together with the dashed line 𝑦=30.

𝑓 ( 0 ) < 3 0 and 𝑓(16)>30, but 𝑓(π‘₯)β‰ 30 anywhere on [0,16]. Why does this not violate the intermediate value theorem?

  • Abecause the intermediate value theorem only applies to polynomial functions
  • Bbecause the intermediate value theorem only applies to functions with 𝑓(π‘₯)<0 at some value
  • Cbecause the intermediate value theorem only applies to cases where 𝑓(π‘₯)=0 not where 𝑓(π‘₯)=30
  • Dbecause the function is not continuous at π‘₯=8
  • Ebecause the function is not defined on the entire interval [0,16]

Q3:

The function 𝑔 is defined on the interval [2,7] and is continuous there. It is known that 𝑔(2)=3 and 𝑔(4)=3, and these are the only values of π‘₯∈[2,7] with 𝑔(π‘₯)=3. It is also known that 𝑔(5)=4. Explain why 𝑔(6)>3.

  • ABecause 𝑔(6) should be greater than or equal to 𝑔(5)
  • BBecause if 𝑔(6)<3, then 𝑔 would equal 3 at some point between π‘₯=4 and π‘₯=6 by the intermediate value theorem
  • CBecause 𝑔 is an increasing function
  • DBecause 𝑔(6)β‰ 3 and we already know the two values where it is equal to 3

Q4:

The figure shows only a part of the graph of the function 𝑓, which is defined on all of [0,1].

If we say that 𝑓(π‘₯)β‰ π‘₯ for every π‘₯∈[0,1], what can we conclude about 𝑓? Why?

  • A 𝑓 ( π‘₯ ) > π‘₯ for π‘₯<0.5 and 𝑓(π‘₯)<π‘₯ for π‘₯>0.5, because of the way the graph is drawn
  • Bthe function 𝑓 is differentiable, because the parts shown look like that
  • CNo conclusion is possible with the information given.
  • Dthe function 𝑓 is continuous, because the parts shown look like that
  • Ethe function 𝑓 is not continuous, because of what the intermediate value theorem states

Q5:

The figure shows only parts of the curve 𝑦=𝑓(π‘₯).

We know that the function has the following properties: π‘“βˆΆ[0,1]β†’[0,1], 𝑓 is continuous, 𝑓(0)=0.6, and 𝑓(1)=0.2. By considering the difference 𝑓(π‘₯)βˆ’π‘₯, what can you conclude about this function?

  • AThere must exist a point π‘βˆˆ[0,1] such that 𝑓(𝑝)=𝑝.
  • BThe function is zero at some π‘βˆˆ[0,1].
  • CThe function takes the value 0.4 at some point.
  • DThere is no conclusion possible.
  • EThe function has an inflection point somewhere.

Q6:

A ball is thrown into the air. Its vertical position above the ground is given by π‘₯(𝑑)=βˆ’3𝑑+12𝑑+7. According to the intermediate value theorem, during which of the following intervals must the ball land on the ground?

  • A [ 4 , 5 ]
  • B [ 2 , 3 ]
  • C [ 5 , 6 ]
  • D [ 3 , 4 ]
  • E [ 1 , 2 ]

Q7:

Let 𝑓(π‘₯)=3βˆ’π‘₯ο—οŠ«. According to the intermediate value theorem, which of the following intervals must contain a solution to 𝑓(π‘₯)=0?

  • A [ 2 , 3 ]
  • B [ βˆ’ 3 , βˆ’ 2 ]
  • C [ 1 , 2 ]
  • D [ 0 , 1 ]
  • E [ βˆ’ 2 , βˆ’ 1 ]

Q8:

Let 𝑓(π‘₯)=π‘₯+π‘₯sin. According to the intermediate value theorem, which of the following must be true?

  • AThere exists at least one 𝑐, where βˆ’πœ‹2<𝑐<0, such that 𝑓(𝑐)=1.
  • BNone of the choices is correct.
  • CThere exists at least one 𝑐, where πœ‹2<𝑐<πœ‹, such that 𝑓(𝑐)=1.
  • DThere exists at least one 𝑐, where 0<𝑐<πœ‹2, such that 𝑓(𝑐)=1.
  • EThere exists at least one 𝑐, where βˆ’πœ‹<𝑐<βˆ’πœ‹2, such that 𝑓(𝑐)=1.

Q9:

Amelia is studying the function 𝑓(π‘₯)=16π‘₯βˆ’3π‘₯βˆ’10. First, she wants to prove that a root exists between 5 and 6. She knows that the function is continuous and has calculated that 𝑓(5)=βˆ’4.17, to two decimal places, and 𝑓(6)=8.

Explain how Amelia’s calculations can be used to prove that a root exists between 5 and 6.

  • AAs there is a sign change between the values of 𝑓(5) and 𝑓(6), and the function is continuous, there is at least one root between 5 and 6.
  • BAs there is a sign change between the values of 𝑓(5) and 𝑓(6), and the function is continuous, there is exactly one root between 5 and 6.
  • CAs there is a sign change between the values of 𝑓(5) and 𝑓(6), there is at least one root between 5 and 6 regardless of whether or not the function is continuous.
  • DAs the function increases between the values of 𝑓(5) and 𝑓(6), there is exactly one root between 5 and 6 regardless of whether or not the function is continuous.
  • EAs the function value decreases between the values of 𝑓(5) and 𝑓(6), and the function is continuous, there is at least one root between 5 and 6.

Amelia decides to approximate the root to 1 decimal place using linear interpolation, which uses properties of similar triangles. She draws the following diagram.

Amelia uses the diagram to form the equation π‘₯βˆ’5|𝑓(5)|=6βˆ’π‘₯|𝑓(6)| which she can solve to find a first approximation of the root. Calculate the value of π‘₯ to 3 decimal places.

Amelia then uses her first approximation to improve the interval in which the root lies and then repeats the process with this new interval. Continue this process to find the value of the root accurate to 1 decimal place.

Q10:

If 𝑓(π‘₯) is continuous over [0,3], 𝑓(0)>0, and 𝑓(3)>0, can we use the intermediate value theorem to conclude that 𝑓(π‘₯) has no zeros in the interval [0,3]?

  • ANo
  • BYes

Q11:

Consider a polynomial 𝑃(π‘₯).

Given that 𝑃(2)=22 and 𝑃(4)=βˆ’1, which of the following conclusions can you draw about the zeros of 𝑃?

  • A 𝑃 ( π‘₯ ) β‰  0 for all π‘₯≀2, so it cannot be zero there.
  • BThere is a zero in the interval (2,4).
  • CYou cannot draw any conclusion.
  • DThere is no zero in the interval [2,4].
  • E 𝑃 ( π‘₯ ) > 0 for π‘₯β‰₯2, so it cannot be zero there.

Given that limο—β†’οŠ±βˆžπ‘ƒ(π‘₯)=βˆ’βˆž, what can you conclude about the zeros of 𝑃 in the interval (βˆ’βˆž,2)?

  • AThere is exactly one zero in this interval.
  • BThere is nothing we can conclude.
  • CThere is at least one zero in this interval.
  • DThere are no zeros in this interval.
  • EThere are two zeros in this interval.

Q12:

According to the intermediate value theorem, in which of the following intervals must cossinπ‘₯+π‘₯+π‘₯=2 have a solution?

  • A [ 0 , 1 ]
  • B [ 2 , 3 ]
  • C [ 1 , 2 ]
  • D [ βˆ’ 3 , βˆ’ 2 ]
  • E [ βˆ’ 2 , βˆ’ 1 ]

Q13:

A particle moving along a line has at each time 𝑑 a position function 𝑠(𝑑), which is continuous. Assume 𝑠(2)=βˆ’5 and 𝑠(5)=βˆ’2. Another particle moves such that its position is given by β„Ž(𝑑)=𝑠(𝑑)+𝑑. Which of the following must be true?

  • AThere exists at least one 𝑐, where 2<𝑐<5, such that 𝑠(𝑐)=1.
  • B 𝑠 has at least one zero.
  • CThere exists at least one 𝑐, where 2<𝑐<5, such that β„Ž(𝑐)=5.
  • D β„Ž has at least one zero.
  • E βˆ’ 5 < 𝑠 ( 𝑑 ) < βˆ’ 2 for all 𝑑 between 2 and 5.

Q14:

Ethan is studying the function 𝑓(π‘₯)=3π‘₯+9π‘₯βˆ’1,000. First, he wants to prove that a root exists between 6 and 7. He knows that the function is continuous and has calculated that 𝑓(6)=βˆ’298 and 𝑓(7)=92.

Explain how Ethan’s calculations can be used to prove that a root exists between 6 and 7.

  • AAs there is a sign change between the values of 𝑓(6) and 𝑓(7) and the function is continuous, there is exactly one root between 6 and 7.
  • BAs the function value decreases between the values of 𝑓(6) and 𝑓(7) and the function is continuous, there is at least one root between 6 and 7.
  • CAs the function decreases between the values of 𝑓(6) and 𝑓(7), there is exactly one root between 6 and 7 regardless of whether or not the function is continuous.
  • DAs there is a sign change between the values of 𝑓(6) and 𝑓(7) and the function is continuous, there is at least one root between 6 and 7.
  • EAs there is a sign change between the values of 𝑓(6) and 𝑓(7),there is at least one root between 6 and 7 regardless of whether or not the function is continuous.

Ethan decides to approximate the root accurate to one decimal place using interval bisection. His method is as follows.

He takes the endpoints of the interval, finds the middle, and then substitutes this value into the function to check its sign. He uses this sign to revise his interval and then repeats the steps with the new interval.

He has so far completed the following table, where π‘Ž and 𝑏 are the endpoints of the interval.

π‘Ž 𝑓 ( π‘Ž ) 𝑏 𝑓 ( 𝑏 ) π‘Ž + 𝑏 2 𝑓 ο€½ π‘Ž + 𝑏 2 
6 βˆ’298 7 92 6.5 βˆ’117.625

Use the information in Ethan’s table to find the next interval that Ethan should consider.

  • A(6,6.5)
  • B(6.5,7)

Continue Ethan’s process to find the root accurate to 1 decimal place.

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