Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Rolls theorem. Suppose that Y equal F of X is continuous at every point of the closed interval AB and differentiable at every interior point. Sometimes that's written as parenthesis a comma B. So if we include the endpoints, it's closed interval it's brackets. If it's only the interior points, we use parentheses. If F of A equal F of B, then there exists at least one number C where the first derivative at that number C is 0. So if we look at some examples, if we have F of A, our Y value equaling another Y value somewhere. So this is A and this is B. Somewhere within that curve we have to have a derivative of 0. If we have here being our F of A and here being our F of B, we could have a curve that looks like so. And here we can see we have two locations that have the first derivative equaling 0 or our horizontal lines. If F of A and F of B were on a horizontal line, then every single point on that would have a first derivative of 0. So Roll's theorem says if the two Y is equal, then somewhere in the interval and it's closed and continuous and differentiable that somewhere along the interval we have to have a slope of 0. The mean value theorem is very similar, but the mean value theorem changes to the point where it can be any Y values and that it's going to equal the slope. So let Y equal F of X be continuous on AB and differentiable on all the interior points. Then there exists at least one C in AB at which the slope or the 1st derivative at some Point C is just equal to the FB minus FA over B -, A which is our formula for slope. So if we thought about this, this point is AF of A and this point over here is BF of B. So if I find the slope of that, that would be this line here. Somewhere on this curve there has to be. The same parallel slope may occur more than one place, but it's got to occur at least in one place. So if F prime of X = 0, then we know F of X has to equal some constant C. If this is a horizontal line, we know that the original function has to be Y equaling some constant. Functions with the same derivatives differ by a constant. So if we started with F of X equaling two X + 2, the derivative is 2. If we had G of X equal two X + 3, the derivative is 2H of X two X + 4, the derivative is 2A of X2X the derivative is 2. So if F prime is greater than 0 then the original function is increasing over some interval. So what it really says is if my first derivative, IE my slope is positive and my slope is greater than 0 or a positive number, the slopes greater than 0, then the original function is increasing over some interval. If we look at a slope, if we look at an original function and we think about the slopes, my slope's positive, my slope's positive, the slope's positive, etcetera. If we look at it being this direction, my slope is positive, my slope is positive, my slope is positive. Both those functions would be increasing if the first derivative is less than 0. IE if the slope is negative then the original function is decreasing over some interval. So if we look at a function and now we look at our slope, there it's negative, here it's negative, here it's negative. So if the derivative is negative, if the slope is negative, the original function has to be decreasing. If we have two constants FSC 1 is less than 0 and FSC 2 is greater than 0, then somewhere in between FSC 3 has to equal 0. So this is the intermediate value theorem that says if I have a point here C1 FSC one where my Y is negative and C2F of C2 where my Y is positive, then somewhere in that interval, as long as we're being a continuous function, I have to have AC three that actually has zero. Let's look at some examples. So our first example is given a function and given an interval. So F of ** equal F of X equal X to the 2/3 on the interval zero to 1. So the first thing we want to do is we actually want to find the slope here. So if we have just F of X equaling X to the 2/3 and 01, given, we're going to say let's find F of 0 and let's find F of one. If we find F of 0, we get 0 to the 2/3, which is 0, and we get one to the 2/3 or one. So the slope here is the change of YS or the F of X2 minus the F of X1 over X 2 -, X one. So we're going to have Oh yeah, F of X2. So our Y value minus our Y value over our X value minus our X value. Come back here and think about putting them as points. We have a .00 and a .11. So the slope here is 1. And what it says is that somewhere in this interval 0 to one, I have to have another point that's going to have the slope of 1. So if I find my first derivative, I get 2/3 X to the -1 third, and then we're going to set that derivative for some constant. So F prime at some constant C has got to be 2/3 C to the negative 1/3. So now we're going to set the formula for the derivative equal to the slope that we just found. If we cross multiply, we'd get 2 equaling 3C to the 1/3 or 2/3 = C to the one third. And that's really the cube root. So the way we get rid of the cube root is we cube it. If I cube one side, I cube the other. So the constant 820 sevenths, which is definitely between zero and one, is going to have the same slope as the endpoints. So if we thought about this curve, we'd have 00, we'd have one, one and it's going to go like that. So at 820 sevenths, we have the same slope as those endpoints. So another example, this one, we want to just, we're given this original equation and an interval and we're asked to show that there's got to be a 0 somewhere between it. Well, first of all, this is going to be a smooth and continuous curve except when X is 0 and our interval didn't include zero. So we're OK there. It is differentiable at every location except at zero, and we're OK because we didn't include zero. So if we chose some big negative value, let's say -42 negative 42 ^3 is a big negative plus 4 / -42 ^2, that's going to be a positive number. But four over something fairly big is going to really be rather small, +7. So this -42 thirds is going to be way bigger. So without evaluating the number, we know that that F of -42 is less than 0. Now I'm going to choose a number close to the zero point, say maybe -1, and when I take -1 and cube it, I get -1 + 4 / -1 ^2 is 4 + 7. When I evaluate this, I get a positive. So the fact that at F and -42 I'm negative and at F and -1 I'm positive means there has to be a 0 somewhere between -42 and -1 and -42 to -1 is definitely on the given interval of negative Infinity to 0. A next example, they give us a bunch of derivatives and they want us to find the antiderivative. So what we're going to do is we're going to undo the derivative. We're going to find the original function. So what gives us out 2X as a derivative? And the way we're going to do this is we're going to take and we're going to add 1 to the exponent now. So add 1 to the exponent, IE 1 + 1 is the two, and then we're going to divide by the new exponent. So if we divide by the new exponent, 2 / 2 is really just one. And then we have to have a + C because the derivative of any constant is always 0. So Y prime giving us 2X is really just the function Y equal X ^2 + C If we think about the derivative of X ^2, it's 2X the derivative of constant 0. If we look at Y prime equal 2 X -1, we're going to add 1 to the exponent to get 2, and we're going to divide by the new exponent. So 2 / 2 is 1 derivative of negative or the antiderivative of -1 or what gives us out -1 for a derivative is just a linear. So it's going to be -1 X. If you wanted to, you could think of this with an X to the zero there. So if I added one to the exponent, I get X to the 1st. And then if I took and divided by that new exponent, and then we always have to have a plus Cy prime equal three X ^2 + 2 X -1, we're going to add 1 to the exponent to get 3 and divide by the new exponent. 3 / 3 is 1. We're going to add 1 to the new exponent. So 1 + 1 is 2 divided by the new exponent. 2 / 2 is one -1. That's just a linear coefficient. So we're going to have negative X in the original function plus some constant C. The next example, we're going to do the chain rule. Actually, we're going to undo the chain rule. So if we have the velocity equaling 2 / π cosine 2π two t / π, we want to find the distance. Well, what gave us out a derivative of cosine? That would be sine. And then we have to take the derivative of the inside by the chain rule. So we're going to take the derivative of the inside again, but instead of multiplying it this time we're going to divide by it, so the derivative of the inside. And then we still had the coefficient of 2 / π from a moment ago. So S is going to equal 2π sine 2T over π over the derivative 2T Pi is 2 / π and we're going to have a + C here because the derivative of any old constant is 0. When we simplify this, we get S equals sine of two t / π + C. Now this problem, they gave us some information. They told us that there was a location where S of π ^2 is 1. So now we're actually going to take this given information and stick it in. So we know 1 is going to equal sine of 2 * π ^2 / π + C What this allows us to do is it allows us to actually figure out what our C is going to be. So we get π ^2 / π which is going to be π and sine of 2π is 0 so our C is going to be 1. So that distance formula is really sine of two t / π + 1. We could always check this by taking the derivative the derivative of sine as cosine and then by the chain rule we multiply by the derivative of the inside and then the derivative of one is 0. So we would get a velocity of 2 / π cosine 2T over π which is indeed what we started with. Another example, we want to figure out what the original function is. If we know the derivative is secant T tangent t -, 1 at the .00. So our original function, what gives us out a derivative of secant tangent? And that's going to be secant. What gives us out a derivative of -1 that's just going to be negative T Whenever we see a constant, it's really just the coefficient of the linear term of the original function, and we have a + C there. So if we had the .00 we would put zero in for R of T and we would put zero in for R TS secant of 0 is 1, so in this case we get C equaling -1. So our RT would equal secant t -, t -, 1 for the original function going through 00. Now what if we change the equation to what we want it to be going through? What if we gave ourselves a new point? Let's say we gave ourselves Pi force, 3, root 2 - π force. So now we'd have 3 root 2 - π force equaling secant of π force -3 root 2 - π force plus some C SO3 root 2 - π force. The secant of π force is root 2 distributing. If we have a root 2 and a -3 root 2, that's going to give us, oh, let's go back and redo that. I put in the wrong thing there, didn't I? OK, the Y values here, the secant of π force minus π force plus C There we go. Have to pay attention. This was just my Y value, but this is my T value or my R value and my T value. So doing that again, 3 root 2 - π force. The secant and π force, hopefully you remember is root 2 - π force plus C taking everything to one side other than the C solving for C basically we get 2 root 2 for our C. So this equation would be R of T equaling secant t -, t +2 root 2. So this is the original function going through the given point P And if we think about the derivative of both of those functions, the derivative of both of those functions would come back with secant tangent, secant T, tangent t -, 1. Thank you and have a wonderful day.