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Lesson: The Intermediate Value Theorem

Sample Question Videos

Worksheet • 5 Questions • 1 Video

Q1:

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

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 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
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