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Intermediate Value Theorem

The Intermediate Value Theorem states that a continuous function must pass through every value between its endpoints. If ff is continuous on [a,b][a,b] and NN is between f(a)f(a) and f(b)f(b), then some c∈(a,b)c \in (a,b) satisfies f(c)=Nf(c) = N.

The Intermediate Value Theorem

If a continuous function moves from one value to another, it cannot skip any value in between — like a person walking from 3 ft to 6 ft who must pass through every intermediate height. The Intermediate Value Theorem (IVT) formalizes this intuition:

If ff is continuous on [a,b][a,b], and NN is any number strictly between f(a)f(a) and f(b)f(b), then there exists at least one c∈(a,b)c \in (a,b) such that f(c)=Nf(c) = N.

Continuity is essential. A discontinuous function can "jump" over values instead of passing through them. For example, if ff jumps from f(a)=1f(a)=1 to f(b)=5f(b)=5 with a break in between, there may be no cc where f(c)=3f(c)=3. Continuity guarantees an unbroken graph from (a,f(a))(a, f(a)) to (b,f(b))(b, f(b)), so every horizontal line y=Ny = N between those heights must cross the graph at least once.

Graph illustrating Intermediate Value Theorem

Common Applications

The most frequent use of IVT is proving a root exists without finding it. Show that ff is continuous on [a,b][a,b], then verify f(a)f(a) and f(b)f(b) have opposite signs. Since 00 lies between them, the IVT guarantees some c∈(a,b)c \in (a,b) with f(c)=0f(c) = 0.

Example: For f(x)=x3−x−1f(x) = x^3 - x - 1 on [1,2][1,2]: f(1)=−1<0f(1) = -1 < 0 and f(2)=5>0f(2) = 5 > 0. Since ff is continuous (polynomial) and changes sign, a root exists in (1,2)(1,2).

For target values beyond just zero, define g(x)=f(x)−Ng(x) = f(x) - N and check if gg changes sign. If g(c)=0g(c) = 0, then f(c)=Nf(c) = N.

Key Limitation

The IVT guarantees existence — at least one solution exists — but says nothing about how many solutions exist or their exact values. It's a powerful existence tool, not a solution-finding method.

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