WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, welcome back to AP Calculus.
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Today, we are going to be discussing the mean value theorem.
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Let us just jump right on in.
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Before we do the mean value, we are actually going to do a smaller version of the theorem called Rolle’s theorem.
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Let us begin with that.
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We begin to with Rolle’s theorem.
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Rolle’s theorem is the following.
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If my three hypotheses, if this function satisfies these three hypotheses then the conclusion we can draw.
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The first one is if y = f(x), if the function that we are dealing with is continuous on a closed interval ab.
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B, if y = f(x) is differentiable, in other words, smooth, no sharp corners.
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If it is differentiable all the open interval ab.
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The endpoints do not necessarily need to apply.
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Last, if f(a) is equal to f(b), in other words if the y value at the left endpoint
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happens to be the same as the y value at the right endpoint.
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If these three hypotheses are satisfied, then we can conclude the following.
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Then, there exists at least one number, we will call it c, between a and b, such that f’ at c is actually equal to 0.
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Geometrically, this means the following.
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Geometrically, we have the following situation.
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We got a little function like that.
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Let us go ahead and call this a, let us go ahead and call this b.
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If we have the function that satisfies these three hypotheses is continuous, is differential,
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and f(a) happens to equal f(b), it does.
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We can conclude that at least one number c, such that the derivative of that function at c is equal to 0.
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In other words, the slope is equal to 0 at the y value of that point.
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That number c happens to be right there.
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In this particular case, it is only one value of c.
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You already know this is true intuitively, but again, with mathematics, we need to be very precise.
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We need to state the hypotheses.
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If does hypotheses are satisfied, then we can draw the particular conclusion.
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If this than this.
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Let me see here, let me go ahead and move on to the next page.
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As we said, there maybe more than one c such that f’(c) is equal to 0.
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Geometrically, that just means the following.
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Something like that.
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Let us go ahead and call this a, let us call this b.
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In this particular case, the function is continuous between a and b.
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The function is differentiable, the open interval a and b.
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F(a) is equal to f(b).
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The number c such that f’ at c is equal to 0, the places where the slope is 0 is 1, 2, 3, 4.
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We have 5 values of c.
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This is c1, this is c2, this is c3, this is c4, and this is c5.
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The number of c actually have to do with the function itself.
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Rolle’s theorem guarantees the existence of at least one.
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That is it, that is all that is going on here.
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Let us go ahead and do an example.
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For the function f(x) = 1/3 x + 3 sin x on the closed interval -3.55 to 3.55.
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Decide if Rolle’s interval applies, if so, apply it to find all values of c, such that f’(c) is equal to 0.
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Let us go ahead and check the hypotheses to see if Rolle’s theorem actually applies.
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A, the question is, is f continuous on the closed interval -3.55 to 3.55, the answer is yes.
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Part b, is f differentiable on the open interval -3.55 to 3.55.
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Once again, the answer is yes.
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Part c, is f(- 3.55) equal to f(3.55).
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F(-3.55) and f(3.55) both = 0, yes.
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The three hypotheses are satisfied.
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Therefore, Rolle’s theorem does apply.
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There does exist at least one c, such that f’(c) is equal to 0.
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Let us find what these values of c are.
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F’(x) is equal to, I have 1/3 + the derivative of sin is cos, 3 cos x.
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I’m going to go ahead and set that equal to 0.
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I have 3 cos x is equal to -1/3.
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I have cos x is equal to -1/9.
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Therefore, x is the inv cos(-1/9).
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The values that I get are x = -1.682 and x = +1.682.
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These are my values of c.
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C is equal to -1.682, c is equal to 1.682.
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Here I have two values of c that satisfy this conclusion of Rolle’s theorem.
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Let us take a look at what this looks like graphically, geometrically, all that wonderful stuff.
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Here is our function y = 1/3 x +, let me go ahead and write that down.
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Y is equal to 1/3 x + 3 sin x, this is our original function f.
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From -3.55 to +3.55, that is here and here.
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We notice that f(a) = f(b), in this particular case, it happens to equal 0.
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We know that it is continuous, it is differentiable.
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There are two places where the slope is 0 here and here.
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C value that is one of them and that is two of them.
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This was our -1.682, that is our c1, and this is our +1.682.
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These was our others c value.
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That is it, that is all that is going on here.
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You are finding the values of c, inside the domain that satisfy the conclusions of Rolle’s theorem which is of f’(c) is equal to 0.
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Let us go ahead and state the mean value theorem.
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The mean value theorem, the theorem itself geometrically, it makes total sense.
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It is completely intuitive but we need to state it explicitly and precisely.
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The mean value theorem is very important in calculus for theoretical reasons more than anything else.
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The mean value theorem says the following.
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My hypotheses are as follows, if a, my hypotheses are the same as Rolle’s theorem except for the third.
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It does away with that particular hypothesis.
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The first one is y = f(x) is continuous on the closed interval ab.
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If it is continuous and if y = f(x) is differentiable on the open interval ab.
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Then, I can conclude the following.
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Now, I'm no longer restricted to f(a) equaling f(b).
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Then, there is at least one number c between a and b, such that, the following is true.
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F’ at c is equal to f(b) - f(a)/ b – a.
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Geometrically, it means the following.
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I have got some function here.
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Let us go ahead and take a here, let us take b here, randomly.
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This is f(a) and this is f(b).
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I’m going to draw the cord, the line that connects them.
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If this continuous on the closed interval, if it is differentiable on the open interval, there is some number c between them.
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Such that f’ at that c, in other words, the slope where it actually touches the graph is equal to f(b) - f(a)/ b – a.
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F(b) – f(a)/ b – a is the slope of this cord.
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Because this point is a, f(a).
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This point is b, f(b).
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F(b) – f(a)/ b - a is just the change in y/ the change in x, that is the slope here.
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They are telling me that there is some number c such that it actually has, the f' at that c has the same slope as this.
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They end up being parallel, makes total sense.
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The slope has to pass through, there is going to be a point where the slope along the curve is going to equal the slope between the 2 points, at the endpoints.
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That is all this is saying.
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This slope, this slope, it has to pass through this, in order for it to go that down.
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This is the mean value theorem.
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Very important theorem.
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Sometimes, you will see it stated this way.
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F’(c) × b - a is equal to f(b) – f(a).
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Sometimes, they write it this way.
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I actually prefer this way.
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The expression for the slope is equal to the expression of a slope, the derivative at some point.
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That is the mean value theorem, nothing particularly strange.
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It is nice that we do not have to worry about that third hypothesis.
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Let us go ahead and do an example.
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For the function f(x) = 4 - x²/ the interval – 2 to 0, decide if the mean value theorem applies.
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If so, apply it to find all values of c such that f’(c) = f(b) – f(a)/ b – a, which is the conclusion of the mean value theorem.
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Let us go ahead and do this.
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Part a, we need to check, is f continuous on the closed interval from -2 to 0?
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The answer is yes, it is continuous.
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In case you are wondering, you are more that welcome to use a graph to decide whether it is continuous, differentiable.
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You want to use all the resources at your disposal.
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This is just the equation y = √4 - x², which is nothing more than x² + y² = 4, except for -2 to 0.
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This is nothing more than, in the equation of the circle of radius 2 going from here to here, including the endpoints.
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Here is -2 and here is 0, that is all.
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This graph, that is all this is.
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Certainly, it is continuous and it is also differentiable.
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Is f differentiable, it is differentiable on the open interval from -2 to 0.
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Yes, it is perfectly smooth.
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Was it differentiable, yes.
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Since that is the case, therefore, there exists some c such that f’(c) is actually equal to f(b) – f(a)/b – a.
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Let us see what is that we have got.
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Let us go ahead and find f’ first.
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F(x) = this.
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I will go ahead and draw a little line here.
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Let me rewrite f(x), let me actually do this in red.
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I have f(x), I’m going to write it as 4 - x² ^½.
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It makes differentiation a little bit easier.
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F'(x) is going to equal ½ × 4 - x²⁻¹/2 × -2x.
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That is going to equal equaling, this 2 and this 2 cancel.
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We are going to end up with –x/ √4 – x².
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This is our f'.
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We want to set this equal to 0.
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Sorry, we are not setting it equal to 0.
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We are setting it equal to f(b) – f(a)/ b – a.
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Let us go ahead and find f(b).
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F(b) = the f(0) which = 4 - 0², all under the radical, that is equal to 2.
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F(a) is equal to f(-2) which is equal to 4, - and -2², under the radical, that = 0.
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We have f(b) - f(a) is 2 – 0/ b - a which is 0 -2.
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Remember, this is a and this is b.
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F(b) is 2, f(a) is 0.
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B is 0, a is -2.
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We end up with 2/2 which is equal to 1.
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We need to solve –x/4 – x² is equal to 1.
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Is that correct? Yes, that was correct.
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Let us go ahead and do that.
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We get - x, I will just go ahead and call it c.
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Let us go ahead and go √–c/4 – c² = 1.
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That becomes -c = 4 - c².
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Square both sides, I end up with c² = 4 - c².
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Let me go to the next page here.
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I have got c² = 4 - c².
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I have got 2 c² = 4, c² = 2, c = + or –√2.
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I'm going to go ahead and take the -√2 because that is the one that is actually in the domain between -2 and 0, which is equal to -1.414.
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That is it, we just found it, c is equal to –√2.
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Let us go ahead and see what that is geometrically.
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Here we have our function y is equal to 4 - x², all under the radical.
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Our c was –√2, it is -1.414.
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We come up there, the slope of this line which is f' at c, f’(-√2) that happens to equal the slope of this line.
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We found the value of c that satisfies this relationship.
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The value of c such that when I go up there, the slope of the graph at that c
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happens to be the same as the slope between the two endpoints.
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Here and here.
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That is it, nice and simple, nothing very strange about this at all.
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Let us go ahead.
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Let us try another example here, see how that one goes.
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For the function f(x) = x²/5 + cos(x) on the interval from -1 to 1, whose graph is shown below.
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This graph right here.
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In case it is not too clear, here is the -1, here is the 1.
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Decide if the mean value theorem applies.
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If so, apply it to find all the values of c such that f’(c) = that right there.
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Part a, is f continuous between -1 and 1?
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Yes, it is continuous.
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B, is it differentiable on the open interval between -1 and 1?
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It is not, there is a sharp point there.
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It is not smooth there, it is not differentiable.
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No, it is not differentiable.
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Therefore, the mean value theorem does not apply.
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I cannot do this.
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In this particular case, we have the graph to let us know that something is differentiable or not differentiable.
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We just look and we found that this low cusp.
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It is not differentiable.
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What if you did not have the graph?
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What if you needed to actually decide whether this thing was going to be differentiable or not?
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Something not being differentiable means the derivative does not exist.
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Let us go ahead and do this analytically.
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Let me actually work in blue.
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If we did not have a graph to decide differentiability, then, we would just do it analytically.
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Let us go ahead and find f', f’(x).
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When I take the derivative of this, it is going to be 2/5 × x⁻³/5 – sin(x).
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I'm going to rewrite this as 2/5, I’m just going to bring this down.
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X³/5 - sin(x), watch what happens as x gets close to 0.
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It is not a problem, if x is any number, and this is certainly valid.
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It is certainly differentiable everywhere.
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But what happens when x gets close to 0?
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As x gets close to 0, if this goes to 0, 0.5, 0.4, 0.3, 0.2, 0.1, the denominator is going to go to 0 also.
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As this goes to 0, the denominator goes to 0.
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As the denominator goes to 0, the fraction, the first term, 2/0 is going to go to infinity.
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The first term, this whole thing is going to end up going to positive or negative infinity.
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Depending on if we approach 0 from the right or from the left.
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Therefore analytically, f’ at 0 does not exist.
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We can conclude it is not differentiable.
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We do not need a graph where we see a cusp, clearly it is not differentiable.
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Clearly, the mean value theorem does not apply.
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If we did have a graph, we will just go ahead and take the derivative
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and see if there is any point where the derivative does not exist.
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Sure enough the derivative does not exist at 0.
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0 happens to be right in the middle of this domain.
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Therefore, the mean value theorem does not apply.
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That is it, nice and straightforward.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time, bye.