# 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.