Discussion of Intermediate Value Theorem
The idea and use of theorems can be very confusing. I hope to clear up some points with this discussion.
First you need to understand what a theorem is. A theorem is made up of two parts. The first part are the hypotheses or the conditions under which the theorem can be applied. The second part is the conclusion or what the theorem implies. If the conditions of the theorem are satisfied, then the result is true.
Take the example of the Intermediate Value Theorem (IVT).
The IVT states that if f is a continuous function on [a,b] and N is a number between f(a) and f(b), then there is a number c in (a,b) such that f(c) = N.
The first part of the theorem gives us the conditions. We need to have a continuous function on a closed interval [a,b] and we need to pick a number N that is between f(a) and f(b). If we have these two conditions, then the second part of the theorem, the conclusion, tells us that there is a number c in (a,b) such that f(c) = N.
You can visualize this theorem as a machine. You give the IVT machine a continuous function on [a,b] and a number N between f(a) and f(b) and the machine gives you a number c in (a,b) such that f(c) = N.
Now, that's enough theoretical discussion. Let's try a couple examples.
Let's say that we want to show that there is a solution to f(x) = 5 where
f(x) = 2x 3 + x 2 + 5x. Suppose further that we want to show that there is a solution in the interval (-1, 1).
We would like to apply the intermediate value theorem. In order to do this, we need to make sure that the conditions of the IVT are satisfied. First, f(x) is a polynomial, so it is continuous. Second, we want f(x) = 5, so we want N = 5 in the statement of the theorem. So we need to make sure that 5 is between f(a) and f(b). In this case, a = -1 and b = 1. Thus, f(a) = f(-1) = -2 + 1 - 5 = -7 and f(b) = f(1) = 2 + 1 + 5 = 7. Then f(-1) < 5 < f(1), so we can apply the IVT. The IVT tells us that there is a c in (-1,1) such that f(c) = 5 and this completes the problem.
The above solution is somewhat lengthy, so that you understand what is happening. The following is a perfectly fine solution to the same problem:
f(x) is continuous and f(-1) = -7 < 5 < 7 = f(1), so the IVT says there is c in (-1,1) such that f(c) = 5.
Now suppose we want to show that there is a solution to x 2 -1 = 1/(x + 3) in the interval (-2, 1).
We first define a function f(x) = x 2 -1 - 1/(x+3). Now we want to find c in (-2,1) such that f(c) = 0. We check our conditions: f(x) is only discontinuous at -3, so f is continuous on [-2,1]. We let N = 0, a = -2, and b = 1. Then f(a) = f(-2) = 4 - 1 - 1 = 2 and f(b) = f(1) = 1 - 1 - 1/2 = -1/2, so f(b) < 0 < f(a), so IVT says that there is c in (-2, 1) such that f(c) = 0.
A more succint answer: f is continuous on [-2, 1] and f(1) = -1/2 < 0 < 2 = f(-2), so IVT says there is c in (-2, 1) such that f(c) = 0.