WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, and welcome to www.educator.com, welcome back to AP Calculus.
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Today, we are going to do some example problems for the max and min that we discussed in the last lesson.
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Let us jump right on in.
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Here, identify and estimate the absolute max and min, and the local max and min on the following graph.
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We have got this graph that looks like, in this particular case the domain is constrained.
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We definitely have an endpoint here which belongs to it and an endpoint here which belongs to it.
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This would be our a and this would be our b, something like that.
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In this particular case, the absolute max, yes, definitely here.
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The absolute max is this point right here.
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The absolute max, and it looks like it is equal to about 60, 70, 80.
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It is 80 and looks like it is achieved at x = 16.
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The absolute min is definitely this one over here.
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The absolute min, it is the endpoint, it looks like -0.5, -0.6, -0.7, and it looks like this is -5, 6, 7, 8.
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It looks like absolute min itself is actually equal to -85, 50, 60, 70, 80.
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It looks like it is -85 at x = -8.
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Let me see we have got a local max, we got a local min.
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Then, we have another local min.
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This is actually we can consider that a local max as well.
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That is about it, it just basically identify the points.
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This one, it looks like, we will put it at -4.5 and actual value is 10, 20, 35.
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Let us do this one over here.
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Let us go ahead and put it at 0, not -5, -6, let us go ahead and put it at -6.
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Here we have 5, 6, 7.
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It looks like this point is 7, the value itself was 0.
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This one over here, maybe 1.8.
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The value was at 10, 20, maybe 27, 28, something like that.
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There you go, that is it.
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Absolute, absolute, local, local, local, local, that is all.
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Nothing strange happening here.
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Sketch the graph of a continuous function having the following properties on the closed interval 0,5.
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It has an absolute min at 0, it has an absolute and a local max at 4.
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It has a local max at 2, a local min at 3.5.
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Let us go ahead and take a look.
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I will go ahead and draw myself a little,
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Let us do 1, 2, 3, 4, and 5.
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That is 5 and this is our 0, 1, 2, 3, 4.
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It says we have an absolute min at 0.
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I’m just going to go ahead and put the absolute min right there.
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It says we have a local max at 2, I’m just going to go ahead and put it there.
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We have a local min at 3.5.
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That is 3.5, I will go ahead and just put it there.
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It says we have an absolute max and a local max at 4.
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I will go ahead and put that there.
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There you go.
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That is it, just draw the graph, nice and simple.
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Sketch the following graphs.
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A function having a local min at 3 and differentiable at 3.
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Something that has a local minimum at 3 and is differentiable at 3.
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That is fine, that is just something like that.
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It has a local min and it is differentiable.
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In other words, the derivative is defined there.
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We have to label that a.
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A function having a local min at 3 but not differentiable at 3.
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That is going to look something like this.
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That is one possibility, this is 3.
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Local min, because to the left and to the right of it, function increases, function increases, but is not differentiable there.
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The derivative does not exist there, it is a cusp.
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That takes care of that.
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Of course, our last one, a function having a local min at 3 but not continuous at 3.
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I’m going to go ahead and do this.
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Here is our 3, it is a local min.
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In other words, to the left of 3 and to the right of 3, the function is above 3.
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This is definitely a local min but it is not continuous there.
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Very simple, very straightforward.
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Now the good stuff, find the critical values of 3x⁴ – 7x³ + 4x².
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Critical values, we say that we take the derivative and we set it equal to 0.
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We set it equal to 0, we solve for the x values.
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And then, we make sure that there is no point where it is not differentiable, just watching out for that.
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F’(x) = 12x³ - 21x² + 8x.
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We are going to go ahead and set that to 0.
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I’m going to factor an x and this is going to be 12x² - 21x + 8 is equal to 0.
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One of our solutions to this is x = 0.
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When I do this, either quadratic formula or calculator, however you want to do it,
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you are going to get x = 0.559 and x is equal to 1.190.
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There you go, those are our three critical values.
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This is differentiable everywhere.
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I do not have to worry about a critical value showing up at a point that is not differentiable.
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These are the only critical values for this particular function.
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Let us take a look at what it looks like.
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0.559 and 1.190, there you go, this is the function.
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We see it is continuous and differentiable everywhere.
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The critical values that we have, this is the original function, this is f(x).
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This is not the derivative of the function.
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We had 0, we had 0.559, and we had 1.190.
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These are critical values, they are the values in the domain along the x axis.
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Such that, the derivative of the function at that point is equal to 0.
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We can see that we have slope of 0, slope of 0, slope of 0.
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That is all that is going on here.
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Find the critical values of f(x) = absolute value of 2x – 5.
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The easiest way to do this is probably graphically.
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But, let us do it algebraically anyway because we are going to have to.
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We want to know first of all, where this is going to equal 0, where it is going to bounce.
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2x - 5 = 0 which implies that x = 5/2 or 2 ½.
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At 2 ½, we got a 1, 2, 3, at 2 ½, that is where it is going to bounce.
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When x is bigger than 5/2, f(x) is going to equal 2x – 5.
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They are all positive.
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In other words, when x is greater than 5/2, what is in here is a positive number.
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Therefore, I can drop the absolute value signs.
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That means f’(x) is equal to 2.
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2 does not equal 0 anywhere.
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To the right of 5/2, there are no critical values.
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When x is less than 5/2, what is inside here is less than 0 which means that f(x) is actually equal to negative of what is inside.
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-2x-5 which equals 5 - 2x, that is the definition of the absolute value.
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Therefore, f'(x) is equal to -2.
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-2 does not equal 0 anywhere.
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To the left of 5/2, there are no critical values.
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However, we know that at 5/2, where it bounces, it is not differentiable there.
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That is the only critical value, not differentiable at 5/2, at x = 5/2.
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X = 5/2 is the only critical value.
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The graph for this sure enough looks like that.
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There is no place where the derivative is 0, no place where the derivative is 0.
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But here it is not differentiable, so 2.5.
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5/2 is the critical value for that function.
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Find the critical values of f(x) = 3√2x² – x.
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We have f(x) is equal to this.
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Let us rewrite it in a way that it makes it a little bit more tractable for us.
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It is going to be 2x² - x¹/3.
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Therefore, f’(x) is going to equal 1/3 × 2x² – x⁻²/3 × the derivative of what is inside, which is going to be 4x – 1.
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We are going to go ahead and set that equal to 0.
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This is the same as, this is negative so I will bring it down below.
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This is 4x - 1 divided by 3 × 2x² – x²/3 = 0.
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This is equal to 0, when the numerator is equal to 0.
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Despite the fact that it looks complicated, it is only the numerator that I have to deal with.
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Therefore, 4x - 1 = 0.
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Therefore, x = ¼ or 0.
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I have taken care of the critical value and so far as the derivative setting it equal to 0.
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At 1/4, the slope is going to be 0.
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In other words, there is going to be a local max and a local min.
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I have to account for the possible places where this thing is not differentiable.
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Are there places where f(x) is defined.
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It is very important, a function has to be defined there, in order for it to have a critical value there.
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It was defined but are there places where f(x) is defined but f’(x) does not exist?
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Yes, f’(x) is this thing right here which is this thing right here.
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It fails to exist, it is not defined but the derivative fails to exist when that is equal to 0.
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This happens and the answer is yes.
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The denominator is 0 when x is equal to 0.
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In other words, if you put 0 into here, you get 0 – 0, the denominator is 0, x = 0, and x = 0.5.
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When you put 0.5 into here, you also get 0 in the denominator.
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At 0 and 0.5, the derivative does not exist.
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At these two points, the function exists.
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If I put them into the original function, I get f(0) = 0 and f(0.5) = 0.
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The function is defined but the derivative is not defined.
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The derivative does not exist.
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Do you want to call it a critical value?
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Basically, what happens is the slope actually ends up going to infinity.
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Is it a differentiable there?
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No, technically, these are critical values.
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I’m just going to say, so I suppose.
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It is not the end of the world if you do not call it critical values.
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I suppose, we can say that x = 0 and x = 0.5 are also critical values.
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There are places where their function is defined but the derivative does not exist are also critical values.
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Again, it is really not the end of the world.
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I myself, personally, I do not consider these critical values.
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Do they fit the definition?
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Yes, they do but for the most part they are irrelevant and insignificant.
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Let us take a look at what the graph looks like.
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It looks like this, we know that at 0.25, ¼, yes, it is a place where the derivative = 0, that is a critical value.
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Over here at 0 and.5, the function exists.
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At 0, it passes through there, the derivative does not exist.
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It is going to be a fully vertical slope.
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Technically, we consider them critical values.
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Let us see what we have got.
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Find the critical values of f(x) = cos² 2x on 0 to 2π.
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Let us go ahead and do f’(x) and set it equal to 0.
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F’(x) is equal to 2 × cos(2x) × -sin(2x) × 2.
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That is equal to -4, I’m going to put the sin 2x first, cos 2x.
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I’m going to rewrite that as -2 × 2 sin 2x cos 2x.
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Hopefully, you recall an identity, that says the sin(2θ) = 2 sin θ cos θ.
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We have 2 sin θ cos θ.
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Therefore, it is equal to sin 2 θ.
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This becomes -2 × sin(4x).
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Therefore, we take -2 × sin (4x).
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We set the derivative equal to 0, we have sin(4x) is equal to 0.
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We have 4x is equal to 0 + n π.
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Sin(0) = 0, sin(π) = 0, 2π, 3π, 4π.
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Therefore, x is equal to 0 + n × π/4.
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I just take n, 1, 2, 3, 4, 5, on this particular interval, here is what I have.
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On the interval from 0 to 2π, x is equal to 0, π/4, 2π/4, 3π/4, 4π/4, 5π/4, 6π/4, 7π/4, and 8π/4,which puts us at 2π.
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Of course, it is differentiable everywhere so I do not have to worry about points where the derivative does not exist.
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These are my critical values.
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There are places where the derivative = 0.
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Let us take a look at what it looks like.
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There you go, 2π is 6.28.
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It is going to be somewhere around there.
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Sorry, we are doing critical values.
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There, there, there, there, and there.
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There you go, that is it.
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Let us take a look at example 8.
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It says find the absolute max and min of this function on the closed interval 0 to π/3.
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We have a function defined at a closed interval.
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We know by the extreme value theorem that the absolute max and the absolutely min do exist.
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We have a procedure for finding them.
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We take the derivative, we find the critical values.
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We evaluate those critical values in the original equation f(x).
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And then, we evaluate the endpoints, in this case 0 to π/3.
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We have a bunch of numbers, we take the biggest of those numbers, that is the absolute max.
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We take the smallest of those numbers, that is the absolute min.
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Let us do it.
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We have f’(x) is equal to 2 cos x – 2 sin x.
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We set that derivative equal to 0, we are trying to find the critical values here.
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I’m going to rewrite this as 2 cos x = 2 sin x which gives me sin x/ cos x is equal to 1.
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That is the same as the tan(x) equaling 1.
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If the interval from 0 to π/3, x is equal to π/4 or 45°.
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That is our critical value.
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We go ahead and evaluate the critical values.
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We have f(π/4), we put it back into the original.
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That is going to equal, we put π/4 back into the original.
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2 × sin(π/4) + 2 × cos(π/4).
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We end up with the answer, it is going to be 4/√2, when you work it out.
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That is equal to 2.83, that is one possible value.
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We evaluated at 0 and π/3.
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When we do f(0), when we put 0 in for here, we are going to end up with 2.
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When we put f(π/3), we end up with √3 + 1 which = 2.732.
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Of these numbers, the biggest number is 2.83.
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Our absolute max = 2.83 and it happens at x = π/4.
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Our minimum value is 2.
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Our absolute min is equal to 2 and it happens at x = 2.
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Let us take a look at what this looks like.
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π/4, it looks like we are going from 0 to π/3.
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I’m sorry, it did not happen at 2, it happened at 0, I apologize.
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Here is your absolute min at 0 and at π/4, just right about there, we achieve our absolute max.
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There you go, that is about it.
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Our domain was 0 to π/3.
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π/3 puts us right about there.
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The graph that we are looking at is just about like that.
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There you go, you see that this is the absolute max because if you go to the right of it, it drops a little.
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You go to the left of it, it drops a little.
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This is the highest point on the domain.
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Find the absolute max and min of ln of 2x/x on the interval 1,3, nice and simple.
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Once again, we form f’(x).
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F’(x) this is a quotient, it is going to be this × the derivative of that - that × the derivative of this/ this².
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This × the derivative of that, x × the derivative of ln 2x is 1/2x × 2 – the ln of 2x × the derivative of x which is × 1/x².
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This is equal to, the 2 cancel, the x’s cancel, w get – ln of 2x/x².
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We set that equal to 0 and that is equal to 0, when the numerator is equal to 0.
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We have 1 – ln of 2x is equal to 0.
00:25:49.800 --> 00:25:57.200
We have ln of 2x is equal to 1, we exponentiate both sides, e to this this or e to this power.
00:25:57.200 --> 00:26:10.400
I'm left with 2x = e, that means x = e/2 which actually equals 1.359.
00:26:10.400 --> 00:26:14.000
This is our one of our critical points.
00:26:14.000 --> 00:26:17.100
Now we are going to evaluate.
00:26:17.100 --> 00:26:24.900
We are going to take f(1.359), we are going to evaluate at the critical points.
00:26:24.900 --> 00:26:32.900
When we do that, it is going to be, we put the 1.359 at the original equation.
00:26:32.900 --> 00:26:42.500
It is going to equal ln(2) × 1.359 divided by 1.359.
00:26:42.500 --> 00:26:48.300
We get the value 0.736.
00:26:48.300 --> 00:26:51.000
That is the only critical value.
00:26:51.000 --> 00:26:54.900
We evaluate at 1 and evaluate the function at 3.
00:26:54.900 --> 00:27:08.500
F(1) is equal to ln(2) × 1/1 = ln(2) and that = 0.693.
00:27:08.500 --> 00:27:24.100
F(3) that is going to equal the nat-log of 2 × 3/3 which is going to be ln(6)/3, that is going to give us 0.597.
00:27:24.100 --> 00:27:30.700
The largest value here is this one, this is our absolute max.
00:27:30.700 --> 00:27:43.500
The smallest value is this one, and this is going to be our absolute min.
00:27:43.500 --> 00:27:46.600
Is that correct? Yes.
00:27:46.600 --> 00:27:56.600
It is still the numbers that I got.
00:27:56.600 --> 00:28:01.300
Let us take a look at what this looks like.
00:28:01.300 --> 00:28:09.600
We are moving from 1 to 3, that is our domain.
00:28:09.600 --> 00:28:14.300
The part of the graph that we are concerned about is this part right here.
00:28:14.300 --> 00:28:22.100
Sure enough, as you can see, at about 1.359 that is where it hits its max.
00:28:22.100 --> 00:28:24.300
Here, notice that it goes this way.
00:28:24.300 --> 00:28:27.600
It drops a little bit, if you look really carefully.
00:28:27.600 --> 00:28:32.600
Here is your absolute max on this domain.
00:28:32.600 --> 00:28:35.700
Here is your absolute min.
00:28:35.700 --> 00:28:44.000
That is all, if you have a function to find on a closed interval, find the critical values, evaluate f(x) at those critical values.
00:28:44.000 --> 00:28:46.700
Evaluate f(x) at the two endpoints.
00:28:46.700 --> 00:28:53.600
The largest number is your absolute max, the smallest number is your absolute min on that domain.
00:28:53.600 --> 00:28:55.900
Thank you so much for joining us here at www.educator.com.
00:28:55.900 --> 00:40:44.000
We will see you next time, bye.