Karl's Calculus Tutor - Box 3.0d: Solution to Continuity Proof Problem

Box 3.0d: Solution to Continuity Proof Problem

The problem is to prove the following: Suppose f(x) is continuous on the closed interval a £ x £ b, and suppose that there is a point, c, on that interval for which f(c) > 0. Prove that there exists a d such that f(x) > 0 for all x in the interval, c - d < x < c + d.

Step 1: Apply the definition of continuity. The function, f(x), is continuous on the entire interval and the point, c is on that interval, so f(x) is continuous at c. What does this mean according to the definition of continuity? It means that f(c) exists and


   f(c)  =   lim   f(x)
            x  > c

This statement has a delta-epsilon contract that goes along with it. Before continuing reading this, try to write out that contract for yourself.

The contract goes like this: For every e > 0, there is a d > 0 such that

   |f(x) - f(c)|  £  e

whenever

   |x - c|  £  d

In other words, if you tell me how close you need f(x) to be to f(c), I can tell you how close to make x to c in order to make it so.

We know that f(c) > 0 because that's given in the problem. That means that it falls on the positive side of zero. The problem asks us to show that if x is within some d of c, no f(x) falls on the other side of zero. You already have the hint that you should try setting e to something less than f(c). Before continuing reading this, see if you can't complete the problem with the information you have so far.

Step 2: Apply the contract. We set e to something less than f(c) but still positive. Whenever f(x) £ 0, we know that |f(x) - f(c)| ³ f(c). Can you see why? If not, try plugging in a positive number for f(c) and a negative number for f(x) and see what happens to the above inequality. And that means that if |f(x) - f(c)| £ e, then f(x) cannot be negative or zero. So it must be positive.

Remember the condition given in the problem? It is that for some d it should be true that f(x) is positive whenever c - d < x < c + d. Isn't that the same as saying |x - c| < d? Isn't that the same wording as our contract? And does't the contract stipulate that if |f(x) - f(c)| £ e then such a d > 0 must exist that always makes that happen? And we know that when it happens, f(x) must be positive. That completes the proof.

If you are asked to do a proof like this on an exam, you don't have to be as wordy as I have been here. The main points are: 1) that if you choose e < f(c), that implies that a nonpositive f(x) fails the e test, and 2) that the epsilon-delta contract implied by continuity guarantees that a d exists that causes every x in the d-interval to meet the e test.

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