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In this lesson, we will learn how to apply the intermediate value theorem.
Q1:
The function π is defined on the interval [ 2 , 7 ] and is continuous there. It is known that π ( 2 ) = 3 and π ( 4 ) = 3 , and these are the only values of π₯ β [ 2 , 7 ] with π ( π₯ ) = 3 . It is also known that π ( 5 ) = 4 . Explain why π ( 6 ) > 3 .
Q2:
The figure shows the graph of the function π on the interval [ 0 , 1 6 ] together with the dashed line π¦ = 3 0 .
π ( 0 ) < 3 0 and π ( 1 6 ) > 3 0 , but π ( π₯ ) β 3 0 anywhere on [ 0 , 1 6 ] . Why does this not violate the intermediate value theorem?
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?
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