Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.

Lesson: Intermediate Value Theorem

In this lesson, we will learn how to apply the intermediate value theorem.

Sample Question Videos

04:45

Worksheet • 5 Questions • 1 Video

Q1:

The function 𝐹 ( π‘₯ ) = 1 π‘₯ + 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 polynomial functions
  • Dbecause the intermediate value theorem only applies to cases where 𝐹 ( π‘₯ ) = 0 , not 𝐹 ( π‘₯ ) = 3
  • Ebecause the function 𝐹 is not continuous over its domain

Q2:

The figure shows the graph of the function 𝑓 on the interval [ 0 , 1 6 ] together with the dashed line 𝑦 = 3 0 .

𝑓 ( 0 ) < 3 0 and 𝑓 ( 1 6 ) > 3 0 , but 𝑓 ( π‘₯ ) β‰  3 0 anywhere on [ 0 , 1 6 ] . Why does this not violate the intermediate value theorem?

  • Abecause the function is not continuous at π‘₯ = 8
  • Bbecause the intermediate value theorem only applies to functions with 𝑓 ( π‘₯ ) < 0 at some value
  • Cbecause the function is not defined on the entire interval [ 0 , 1 6 ]
  • Dbecause the intermediate value theorem only applies to polynomial functions
  • Ebecause the intermediate value theorem only applies to cases where 𝑓 ( π‘₯ ) = 0 not where 𝑓 ( π‘₯ ) = 3 0

Q3:

The function is defined on the interval and is continuous there. It is known that and , and these are the only values of with . It is also known that . Explain why .

  • Abecause if , then would equal 3 at some point between and by the intermediate value theorem
  • Bbecause is an increasing function
  • Cbecause should be greater than or equal to
  • Dbecause 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 , which is defined on all of .

If we say that for every , what can we conclude about ? Why?

  • Athe function is not continuous, because of what the intermediate value theorem states
  • BNo conclusion is possible with the information given.
  • Cthe function is continuous, because the parts shown look like that
  • D for and for , because of the way the graph is drawn
  • Ethe function is differentiable, because the parts shown look like that

Q5:

The figure shows only parts of the curve .

We know that the function has the following properties: , is continuous, , and . By considering the difference , what can you conclude about this function?

  • AThere must exist a point such that .
  • BThe function is zero at some .
  • CThere is no conclusion possible.
  • DThe function has an inflection point somewhere.
  • EThe function takes the value 0.4 at some point.
Preview

Related Lessons