# Continuous Functions Theorems

## Continuous functions

A **continuous function** in mathematics is defined as a function that is defined at each point in its domain. Basically, a function whose graph is an unbroken curve in its domain is a continuous function.

In mathematical terms, a function *f(x)* is considered to be continuous at a point *c* if and only if it satisfies the following conditions:

The functions that are continuous at every point in their domain are continuous functions.

Let's see an example:

The domain of this function is all real numbers [- âˆž, + âˆž]. This function satisfies all three conditions, so it is a continuous function, as can be seen in its graph.

## Theorems of Continuous Functions

We will now look at two important theorems pertaining to continuous functions: the intermediate value theorem and the extreme value theorem.

### 1. Intermediate Value Theorem

For the **intermediate value theorem**, let's assume a function *f(x)* that is continuous between [*a,b*]. Suppose there is a number *p* between *f(a)* and *f(b)*. This theorem states that there must be at least one value *q* between *a* and *b*, such that *f(q) = p*.

Taking the example of the previous function,

Between the range [0,4], there exists a value *f(q)* = 4 for *q* = 2, which can be illustrated like this:

There could be functions that have more than one point between *a* and *b* with the same value. Let's consider an interval of [-4,4] for our previous example. In this case, we get the value *f(q)* = 4 at two points: *q* = -2 and *q* = +2, as we can see in this graph.

### 2. Extreme Value Theorem

Now, when it comes to the **extreme value theorem**, which states that for a continuous function *f(x)* in the interval [*a,b*], there will be an absolute maximum and an absolute minimum. For a function *f(x)* that is continuous in the interval [*a,b*], there will be two points *c* and *d* such that *f(c)* will be absolute maximum and *f(d)* will be absolute minimum over [*a,b*], as seen here:

Taking the same example as before, in the range [1,2], the absolute maximum and minimum for the function will be 4 and 1 at points 2 and 1, respectively which can be seen in this graph:

## Lesson Summary

In this lesson, you learned about the **continuous functions** and the conditions that a function has to fulfill in order to be one. It's basically a function that has an unbroken, continuous graph between two points. Then we looked at two important theorems pertaining to continuous functions:

**Intermediate value theorem**: states that for a function*f(x)*that is continuous between [*a,b*], and there is a number*p*between*f(a)*and*f(b)*, then there must be at least one value*q*between*a*and*b*, such that*f(q) = p*

**Extreme value theorem**: states that for a continuous function*f(x)*in the interval [*a,b*], there will be an absolute maximum and an absolute minimum

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## Continuous Functions Theorems - Examples and Counterexamples

In the following problems, students will apply their knowledge of the continuous function theorems to solve for the values satisfying the theorems and to construct counterexamples.

## Problems

1) Verify that f(x) = 3x^2 - 2x is continuous, and find a value (q) satisfying the Intermediate Value Theorem for f(q) = 5, using the fact that f(0) = 0 and f(2) = 8.

2) Graph the function f(x) = 3x + 2 on the interval [0,3] and verify that the Extreme Value Theorem is satisfied.

3) Show that the function f(x) = x^3 + x - 2 has a root in the interval [0,2].

4) Create an example of a discontinuous function which doesn't satisfy the Intermediate Value Theorem.

## Solutions

1) The function is continuous, as seen on the graph:

We know a value for q exists by the Intermediate Value Theorem. To find q, we need to solve the equation 3q^2 - 2q = 5 , or equivalently, 3q^2 - 2q - 5 = 0. Factoring gives us (q + 1) (3q - 5) = 0 and so q = -1 or q = 5/3. The value we want is q = 5/3 since 0 < 5/3 < 2.

2) The graph of the function on the interval is below:

The Extreme Value Theorem is satisfied - we have a continuous function on a closed interval and the absolute maximum is 11 at x = 3 and the absolute minimum is 2 at x = 0.

3) Using the end points, we see that f(0) = 0^3 + 0 - 2 = -2 and f(2) = 2^3 + 2 - 2 = 8. The Intermediate Value Theorem says that for any f(0)< p < f(2) there exists 0 < q < 2, such that f(q) = p. Since p = 0 is a value between f(0) and f(2), there exists a 'q' with f(q) = 0; that is, there is a root.

4) One possible example is the piece-wise function f(x) = 2 when x is in the interval [0,1] and f(x) = 0 when x is interval (1,2]. The function is graphed below and the Intermediate Value Theorem is not satisfied. For example, a value between f(0) and f(2) is 1, but there is no x value for which f(q) = 1.

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