Show that cosx=x has a solution in the interval [0,1] . Hint: Show that f(x)=x-cosx has a zero i

Show that cosx=x has a solution in the interval [0,1] ....

Key Concepts

Continuous Functions

A function is continuous if small changes in the input result in small changes in the output, meaning there are no abrupt jumps or breaks in the graph. In the context of this problem, the cosine function and the linear function are both continuous, which makes the combination f(x) = x - cos x continuous over the interval [0, 1]. This property is essential because it ensures that all intermediate values between f(0) and f(1) are attained, a condition needed for applying the Intermediate Value Theorem.

Intermediate Value Theorem

The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b] and takes different signs at the endpoints, then there exists at least one point c in the interval where the function equals zero. In this problem, by showing that f(0) and f(1) have opposite signs for f(x) = x - cos x, one can conclude that f must cross zero at some point between 0 and 1, thereby proving that cos x = x for some x in [0, 1].

Transcript

00:01 This problem they've given us an equation, and they're basically saying that we're supposed to prove that this equation actually has a solution somewhere on this zero to one interval.

00:10 Well, there's actually lots of ways that you can approach this, but if we slightly rearrange this equation, it might make things a little bit easier for us.

00:18 Okay, so we now have x equal to cosine of x.

00:20 If we move everything to be on one side, we'll have zero on one of the sides instead.

00:25 So what i'll do is subtract out.

00:27 Let's see, you can do either side here really, but let's just sort of.

00:30 Subtract out the cosine of x and move it to this right -hand side, perhaps.

00:36 So we'll end up with a zero now on this left -hand side then, equal to x minus cosine of x.

00:42 This might not seem like a big deal, but it's actually going to be quite helpful.

00:46 It's really the hint that they provide here.

00:48 And what we're going to be doing with this new expression that we've just created here, which is now equal to zero, is we're going to allow it to be a function, particularly a function of notice that x is kind of our variable of choice we're using here we're going to let it be the independent variable for this f of x function that we're creating here and what we're doing is kind of adding a visual nature to what we're trying to accomplish.

01:09 Remember earlier we have x equal to cosine of x.

01:12 We're trying to see when these two functions actually equal each other on the interval.

01:16 Well if you move one to the other side here you now end up with this expression.

01:20 We're trying to make sure that this expression equals zero sometime there in this interval.

01:25 Same problem, just slightly rearranged to give us a little bit more clarity about things.

01:30 And if you let this expression become a function, we'll call it f of x, then we're really asking ourselves if this function actually equals zero sometime on this interval zero to one.

01:42 Or another way to think about it is, does this function actually have an x intercept or a zero of the function somewhere on the zero to one interval? that's really the kind of the part that we really want to focus our attention on.

01:56 That's going to be a little bit easier.

01:57 It kind of adds a visual sense to what we're trying to accomplish.

02:00 So how do we do this? well, the intermediate value theorem is our go -to -here.

02:04 Okay, so the intermediate value theorem requires us to have a continuous function on the interval that we're working with zero to one.

02:12 Well, let's check this out.

02:14 X right here, this little guy is a polynomial.

02:16 And so we know from previous sections that we've worked with that this is definitely continuous everywhere.

02:22 And so let's try out cosine of x.

02:23 Well, cosine of x, as you all know, is continuous everywhere as well.

02:28 If these two guys are being subtracted one from another, that should not change anything.

02:33 We learned that in section 2 .4, that if you subtract a continuous function from another continuous function like we're doing here, that will not change its continuity.

02:41 So in fact, we actually have an f of x function that's continuous not only on this interval, but everywhere.

02:47 All we needed it to be continuous on was just the interval.

02:49 We kind of got a little bit more bang for our buck there.

02:51 So it's certainly continuous on this interval for sure.

02:55 Okay, so now that we have that established, which was everything hinged on, let's take a look at some values here on each endpoint.

03:02 Let's just see kind of what we're working with here as far as context.

03:06 So let's try out zero and one into this function to see what the y values look like at those places.

03:12 So we'll have f of zero that we're going to try out first here.

03:15 When you plug in zero, for all these x's, you'll have zero minus, cosine of zero is just one.

03:21 So you'll have 0 minus 1, which is just negative 1.

03:25 Notice how this value was less than 0.

03:27 We're trying to make sure that this function actually equals 0 at some point in time in this interval.

03:32 This guy that we came up with on the left endpoint of the interval is less than 0.

03:37 That's important to note here, less than 0...

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