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 δ such that  f(x) > 0  for all x in the interval,  c - δ < x < c + δ

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.

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

```   |f(x) - f(c)|  ≤  ε
```
whenever
```   |x - c|  ≤  δ
```
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 δ of c, no  f(x)  falls on the other side of zero. You already have the hint that you should try setting ε 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 ε to something less than  f(c)  but still positive. Whenever  f(x) ≤ 0,  we know that  |f(x) - f(c)| ≥ 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)| ≤ ε,  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 δ it should be true that  f(x)  is positive whenever  c - δ < x < c + δ.  Isn't that the same as saying  |x - c| < δ?  Isn't that the same wording as our contract? And doesn't the contract stipulate that if  |f(x) - f(c)| ≤ ε  then such a  δ > 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  ε < f(c),  that implies that a nonpositive  f(x)  fails the ε test, and 2) that the epsilon-delta contract implied by continuity guarantees that a δ exists that causes every x in the δ-interval to meet the ε test.

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