# Theorems involving Continuous Functions

Intermediate Value Theorem. Suppose that f is continuous on closed interval [a,b] and let N is any number between f(a) and f(b) (or f(b) and f(a); depends what is bigger). Then there exists number c in (a,b) such that f(c)=N.

The Intermediate Value Theorem states that a continuous function takes on every intermediate value between the function values f(a) and f(b).

Note that the value N can be taken one or more times.

In geometric terms Intermediate Value Theorem states that if any horizontal line y= is given between f(a) and f(b) then the graph of f should intersect y=N somewhere.

Corollary. Suppose that f is continuous on closed interval [a,b] such that f(a) and f(b) have different signs. Then there exists number c in (a,b) such that f(c)=0.

Proof of this fact is straightforward. Since f(a) and f(b) have different signs then either f(a)<0<f(b) or f(b)<0<f(a), so, we take N=0 in intermediate value theorem.

Note, that Intermediate Value Theorem is general false for discontinuous functions.

One use of the Intermediate Value Theorem is in locating roots of equations.

Example 1. Function 2^x-4x=0 clearly has root 4, but it is harder to locate another root. However, for function f(x)=2^x-4x f(0)=2^0-4*0=1>0 and f(1/2)=2^(1/2)-4*2=sqrt(2)-8<0.

Thus, by corollary (because function f(x)=2^x-4x is continuous) there exists number c in interval (0,1/2) such that f(c)=0.

Example 2. Show that the root of equation cos(x)-x=0 is within interval [0,1].

Let f(x)=cos(x)-x. Clearly it is continuous on [0,1].

We are looking for the solution of the equation, that is a number c between 0 and 1 such that f(c)=0.

Since f(0)=cos(0)-0=1>0 and f(1)=cos(1)-1<0 then f(1)<0<f(0).

Thus, according to corollary there exists number c on interval (0,1) such that f(c)=0. Therefore, f(x) has at least one root in (0,1).

Boundedness of Function. Suppose that function f is continuous on closed interval [a,b] then exist such constant finite numbers m and M that m<=f(x)<=M for all a<=x<=b.

Note that this theorem doesn't hold if interval is not closed.

For example, let f(x)={(1/x if 0<x<=1),(0 if x=0):}.

This function can take only finite values, but it grows without a bound when x approaches 0.