The Intermediate Value Theorem (IVT) states that if a function $f$ is continuous on a closed interval $[a, b]$, and if $N$ is any number between $f(a)$ and $f(b)$, then there exists at least one $c$ in the interval $(a, b)$ such that $f(c) = N$.

Typical Formal Statement for the IVT:

Let $f$ be a function that is continuous on the closed interval $[a, b]$. Suppose that $f(a) \neq f(b)$, and let $N$ be a number such that: $f(a) < N < f(b) \quad \text{or} \quad f(b) < N < f(a).$ Then, there exists a number $c$ in the open interval $(a, b)$ such that: $f(c) = N.$

Proof Outline:

1. Continuity on the Closed Interval: Since $f$ is continuous on $[a, b]$, by the properties of continuous functions, there are no jumps or gaps in the function values.

2. $N$ is Between $f(a)$ and $f(b)$: Assume without loss of generality that $f(a) < N < f(b)$. This means $N$ is a value that lies between $f(a)$ and $f(b)$.

3. Existence of $c$: Since $f$ is continuous and takes all intermediate values between $f(a)$ and $f(b)$, there must be some $c \in (a, b)$ such that $f(c) = N$.

Thus, the IVT ensures that for any continuous function, if a value lies between $f(a)$ and $f(b)$, the function will take that value at least once within the interval.

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5 Related Questions:

1. What are the conditions for a function to be continuous on an interval?
2. How can the IVT be applied to prove the existence of roots in an equation?
3. Can the IVT be applied to a function with discontinuities? Why or why not?
4. How does the IVT relate to the concept of a root of a function?
5. What are practical applications of the Intermediate Value Theorem in real-world problems?

Tip:

For IVT to work, the function must be continuous over the entire closed interval; discontinuities (even at a single point) break the theorem's validity.