Table of Contents
- 1 What is the idea behind the Intermediate Value Theorem How do we state the theorem formally?
- 2 When can the Intermediate Value Theorem be used?
- 3 Is the converse of the Intermediate Value Theorem true?
- 4 Is the converse of the intermediate value theorem true?
- 5 Is converse of extreme value theorem true?
What is the idea behind the Intermediate Value Theorem How do we state the theorem formally?
In mathematical analysis, the intermediate value theorem states that if f is a continuous function whose domain contains the interval [a, b], then it takes on any given value between f(a) and f(b) at some point within the interval.
When can the Intermediate Value Theorem be used?
In other words, the Intermediate Value Theorem tells us that when a polynomial function changes from a negative value to a positive value, the function must cross the x-axis. Figure 17 shows that there is a zero between a and b. Figure 17. Using the Intermediate Value Theorem to show there exists a zero.
Is the converse of Intermediate Value Theorem true?
In general, the converse of a statement is not true. The converse of the Intermediate Value Theorem is: If there exists a value c∈[a,b] such that f(c)=u for every u between f(a) and f(b) then the function is continuous. This statement is false.
Is the converse of the Intermediate Value Theorem true?
In general, the converse of a statement is not true. The converse of the Intermediate Value Theorem is: If there exists a value c∈[a,b] such that f(c)=u for every u between f(a) and f(b) then the function is continuous.
Is the converse of the intermediate value theorem true?
Is the intermediate value theorem the same as the mean value theorem?
No, the mean value theorem is not the same as the intermediate value theorem. The mean value theorem is all about the differentiable functions and derivatives, whereas the intermediate theorem is about the continuous function.
Is converse of extreme value theorem true?
The converse of the Extreme Value Theorem is: If there is at least one maximum and one minimum in the closed interval [a,b] then the function is continuous on [a,b]. This statement is false. In order to show the statement is false, all you need is one counterexample.