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

Please verify your account before proceeding.

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

Q1:

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

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 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 .

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?

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?

Donβt have an account? Sign Up