The Intermediate Value TheoremDate: 09/14/98 at 05:42:14 From: BOOK Subject: Intermediate Value Theorem I don't understand about the "Intermediate Value Theorem" at all. Can you please explain it to me? Also for the function: f(x) = 1/(x-3) + 2, x not equal to 3 verify that f(0) < f(5), find the value of f(x) such that there's no x value in the interval where x is greater or equal than 0, but less than or equal to 5 such that the function takes this value. For the last question, can you explain why this case appears to be a counterexample to the Intermediate Value Theorem. I really need your help. Thank you very much for your kindness. Date: 09/14/98 at 08:41:58 From: Doctor Anthony Subject: Re: Intermediate Value Theorem We will examine the Intermediate Value Theorem using the equation: f(x) = 1/(x-3) + 2 First check f(0) and f(5): f(0) = 1 and 2/3 f(5) = 2.5 So f(0) < f(5) as required. Now we need to find a value of f(x), say c, such that there's no x value in the interval [0,5] where the function takes this value. In other words, there is no value x in [0,5] such that f(x) = c. The curve has an asymptote at y = 2, and the graph does not cross the line y = 2. This is because there is no value of x such that 1/(x-3) = 0. Thus there is no value of x (except +/- infinity) where f(x) = 2. The Intermediate Value Theorem would suggest that if f(0) = 1.666 and f(5) = 2.5, then there should be a value of x between x = 0 and x = 5 for which f(x) equals 2, since 2 is a value between f(0) and f(5). But we have seen that this is not so. The reason is that f(x) is not continuous in the interval 0 < x < 5. Because one of the conditions before you apply the Intermediate Value Theorem is that the function is continuous in the interval concerned, the Intermediate Value Theorem does not apply to this case, and so, this example is not really a counterexample. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 09/14/98 at 12:43:27 From: Doctor Teeple Subject: Re: Intermediate Value Theorem Hi, I'd like to expand a bit on interpreting the Intermediate Value Theorem. According to the CRC Concise Encyclopedia of Mathematics, the Intermediate Value Theorem states that if a function f is continuous on a closed interval [a,b] and c is between f(a) and f(b) inclusive, then there exists at least one x in [a,b] such that f(x) = c. So another condition to check is that f(a) < f(b). Often, before trying to prove a theorem, it is a good idea to try examples to convince yourself that the theorem is true. For example, examine the following graph: We start by picking two points on the x-axis, a and b, and noting f(a) and f(b). Then pick a point of the curve in the closed interval. We could pick either of the endpoints or a point in between. We label this point c. Since we know that [a,b], [f(a),f(b)] are closed and that f is continuous, we know, by the Intermediate Value Theorem, that there exists a point x in [a,b] such that f(x) = c. If you examine the graph, you can find x quite easily. Try playing around with these conditions to see what happens if one or more of them fail. For example, what if we took an open interval (a,b)? Could you find a graph where the conclusion of the Intermediate Value Theorem fails? In your example, f(x) = 1/(x-3) + 2. Here is the graph: We can note right away that we are relaxing the condition that f(x) is continuous. So we know that the Intermediate Value Theorem will not apply because all of the conditions have not been met. Let's keep going to see how the theorem may fail. With your example, we start by picking our two points as 0 and 5, noting that f(0) = 1.6667 and f(5) = 2.5. From the graph, we can see that there is an asymptote at y = 2. Note that 2 is in [1.6667, 2.5], and we know that there is no x value in [0, 5] such that f(x) = 2. Thus the Intermediate Value Theorem fails with this example when we relax the condition of continuity. - Doctor Teeple, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/