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 [- ∞, + ∞]. 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:
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.
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.
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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