Intermediate value theorem

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Statement

Full version

Suppose f is a continuous function and a closed interval [a,b] is contained in the domain of f (in particular, the restriction of f to the interval [a,b] is continuous). Then, for any t between the values f(a) and f(b) (see note below), there exists c \in [a,b] such that f(c) = t.

Note: When we say t is between f(a) and f(b), we mean t \in [f(a),f(b)] if f(a) \le f(b) and we mean that t \in [f(b),f(a)] if f(b) \le f(a).

Short version

Any continuous function on an interval satisfies the intermediate value property.

Caveats

The statement need not be true for a discontinuous function

It is possible for a function having a discontinuity to violate the intermediate value theorem. Below is an example, of the function f(x) := (x + 1)\operatorname{sgn}(x) where \operatorname{sgn} is the signum function and we define it to be zero at 0.

Intermediatevaluetheoremfailsfordiscontinuousfunction.png

Here, we consider the domain [a,b] = [-1,3], with f(-1) = 0 and f(3) = 4, but there is no c \in [-1,3] satisfying f(c) = 1/2, even though 1/2 \in [0,4].

The function needs to be defined throughout the domain

Consider the function f(x) := 1/x. We have f(-1) = -1 and f(1) = 1. However, there is no c \in [-1,1] such that f(c) = 1/2. The reason the theorem fails is that f is not defined at the point 0, and hence it is not defined on the domain [-1,1].

1byxviolatesintermediatevaluetheorem.png