 In previous videos for lecture 32 in our course, which is about the mean value theorem, I've said that the mean value theorem is essentially the most important result for differential calculus, that is the calculus of derivatives. Why do I put so much emphasis on the mean value theorem? Well, let me, in this video as we end lecture 32, mention two important consequences of the mean value theorem, which we will use later on in our lecture series when we reverse the process. Let me start considering anti-derivatives. The mean value theorem will tell us that if you know a function's derivative, you basically know the function minus one bit of information. And that's what we're gonna talk about in this video here. So first, kind of like when we prove the mean value theorem, we first prove Rohl's theorem, and then we prove the mean value theorem, that where Rohl's theorem had sort of this assumption that the starting and end point were the same, which tells us the average rate change was zero, right? So I kind of wanna make that same assumption. Let's first assume that the derivative is zero. So if a function's derivative is zero on some open interval a to b, that means the function must be constant. And so we do know that constant functions have zero derivatives. This theorem gives us the converse of that statement. If your derivative is zero, then you must have been a constant, okay? And this is how we're gonna do it. So we're gonna choose just arbitrarily numbers x1, x2, which are inside this interval. So x1, x2, there's two numbers between a and b. And without the loss of generality, let's assume that x2 is bigger than x1, okay? Since the derivative of f is equal to zero on the domain a to b, f is differentiable on a to b, right? The differential just means that the derivative exists, okay? And so in particular, if f is differentiable on the interval a to b, then it'll be differentiable on the smaller interval x1, x2. And this is why we chose two arbitrary numbers, x1, x2, is why we can't guarantee differentiability on the points a and b. We can, we can, well, sorry, we can guarantee a and b here, but we want to also guarantee the endpoints x1, x2 are gonna be continuous here. We're gonna start again in three, two, one. Since f is differentiable on the entire interval a to b, in particular, it'll be differentiable at the point x1 and x2, for which if you're differentiable, then you are also continuous. So we know that f will be differentiable on the open interval x1, x2. It'll be continuous likewise on the closed interval x1, x2, because we have a stronger understanding of this function given it's differentiable on this interval. The reason I have to make these comments here though is that in order to invoke the mean value theorem, I need to check that its conditions are satisfied, the assumptions are satisfied. So the mean value theorem applies to this interval x1 to x2. The mean value theorem guarantees there exists some point C that sits between x1 and x2, such that f prime of C is equal to f of x2 minus f of x1 over x2 minus x1. If you clear the denominators, you get this equation right here. But the derivative f prime of C is equal to zero. So if you plug in a zero right here, we're going to get that f of x2 minus f of x1 is equal to zero. Add f of x1 to both sides. We end up with f of x2 is equal to f of x1. So this told us that the function evaluated at x2 is the same as the, wait a second, x1 and x2 were chosen arbitrarily. I just picked two numbers between A and B. I don't have any specifications on them except there's two numbers in the domain. So given any two arbitrary numbers, if their function evaluation is the same and those two numbers were chosen arbitrarily, that implies the function's constant because it applies to any two numbers x1 and x2. f of x is always the same and therefore is a constant function. This is an important consequence of the mean value theorem. If a function's derivative is zero, it must have been a constant function. That's kind of like proving roles theorem that we did before. How do we get the more general argument? Well, in this situation, we're going to say that, assume that two functions have the same derivative. So f prime of x equals g prime of x for all x on the interval A to B. If two functions have the same, then it turns out those two functions only differ by a constant. That is, f of x equals g of x plus some constant number. The reason behind this basically comes from the fact that if we take the difference of the derivatives, we're going to get zero. So let's take capital f of x to be the difference of f of x and g of x. We're going to prove that capital F is constant. But because if capital F is constant, we're going to get that c equals f of x minus g of x. Add g of x to both sides. You get f of x equals g of x plus c. So the difference of the functions is going to be constant. Well, if capital F is little f minus little g, then the derivative of capital F will be the derivative of little f minus the derivative of little g, which by assumption f prime of x is equal. So the difference here is going to equal zero. Then we can invoke the previous theorem, right? We now have a function whose derivative is equal to zero. By the previous theorem, we just proved on the previous slide, that shows us that capital F of x is a constant function. Therefore, we then get the remainder of our argument right here. We get that since the difference between f and g is constant, then f and g only differ by constant. That is f equals g of x plus a constant right there. This is an extremely important result. And at the end of this lecture series, particularly when we get to chapter five, we start talking about antiderivatives and integrals, time this plus c. Because as we take an antiderivative, something we haven't talked about yet, but basically we're reversing the derivative process. We only, if you know a function's derivative, you know that function up to some vertical translation, up to some plus c. Basically, if you know the derivative of a function and you know the y-intercept of the function, you know the function. Let me show you one example before we close lecture 32. So let's prove that the sum of the two functions, arc tangent of x plus cotangent of x, excuse me. Let's show that arc tangent of x plus arc cotangent of x. That is, if you take the sum of the inverse of tangent and cotangent, you always equal pi halves. This number, the sum of these two functions is always pi halves, no matter what. And the way we're gonna prove this is to show that the derivative of arc tangent plus arc cotangent is zero. Take the function f of x to be the sum of arc tangent plus arc cotangent. Take its derivative. As we've seen previously, the derivative of arc tangent is one over one plus x squared. The derivative of arc cotangent is gonna be negative one over one plus x squared. So when you add those two functions together, you're gonna get zero. Because the function, because the function's derivative is zero, like we saw previously, this means that the function f is a constant. The arc tangent plus arc cotangent is a constant, but what's that constant? Well, it doesn't matter where you evaluate the function. It's always gonna be the same. So let's take, for example, arc tangent, excuse me. Let's evaluate the function at one. So arc tangent at one is gonna be pi force because we're looking for the angle which tangent's equal to one and it's tangent sine over cosine. We're looking for an angle where sine and cosine are the same. That happens at pi force or a 45 degree angle. Cotangent inverse, you're gonna get the same thing if we've got a way to have one. Well, cotangent is cosine over sine. So if we want cosine over sine to equal one, we need to find an angle where that happens that likewise happens at pi force. Pi force plus pi force is equal to pi halves. And so since our function f is constant, if once it's equal to pi halves, that means it's always equal to pi halves. And so this then proves by the mean value theorem that the sum of arc tangent with arc cotangent is always equal to pi halves. And so this lecture from a calculus one perspective is to try to impress upon us the power behind the mean value theorem. We will return to these results later on in chapter five of this lecture series when we talk about integrals which is essentially by the fundamental theorem calculus the opposite of differential calculus. But the mean value theorem is really showing us like in these examples, why knowing something about the derivative of the function tells us so much about the function. Later on in chapter four, as we continue to do applications of derivatives, one in particular one we're gonna do is curve sketching. If we know information about the function's derivative and second derivative, we can basically figure out stuff about the original function. Knowing about the derivative of the function tells you about the function because of what we learned from the mean value theorem. I hope you learned something in these videos from lecture 32. If you did, please hit the like button here. 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