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:
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The functions that are continuous at every point in their domain are continuous functions.
Let's see an example:
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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.
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We will now look at two important theorems pertaining to continuous functions: the intermediate value theorem and the extreme 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.
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Taking the example of the previous function,
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Between the range [0,4], there exists a value f(q) = 4 for q = 2, which can be illustrated like this:
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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.
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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:
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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:
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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:
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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:
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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:
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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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