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Worksheet: The Intermediate Value Theorem

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

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

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

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 differentiable, because the parts shown look like that
  • Bthe function is continuous, because the parts shown look like that
  • C for and for , because of the way the graph is drawn
  • Dthe function is not continuous, because of what the intermediate value theorem states
  • ENo conclusion is possible with the information given.

Q5:

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

We know that the function has the following properties: 𝑓 ∢ [ 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?

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