Lesson: Intermediate Value Theorem AP Calculus BC • AP Calculus AB • Precalculus • Sullivan • Tenth Edition • Precalculus • Ron Larson • Tenth Edition

The function satisfies and . But there is no between and 1 where . Why does this not violate the intermediate value theorem?

The figure shows the graph of the function on the interval together with the dashed line .

and , but anywhere on . Why does this not violate the intermediate value theorem?

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 .