WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
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I know we did a lot of example problems in the last lesson, when we discussed how to solve limits analytically.
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But I thought it would be nice to actually do a little bit more.
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That is what we are going to do in this lesson.
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Let us get started.
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Evaluate the following limit by showing each application of a limit law.
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Now that I actually say this problem, I do not want to go through and show the evaluation of each limit law.
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Basically, the idea is this.
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The limit of a quotient is the limit of this/ the limit of this.
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These are sums, it is going to be the limit of this - the limit of that + the limit of this/ the limit of this - the limit of that.
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And then, each of those².
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You know what, let us just do what we normally do.
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Plugging in and see what we get, that is the quick way of doing it.
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Just plug it in, plug in 2 into here and see what you get.
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That is what we want.
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2³ is 8, 4 × 8 is 32.
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2² is 4, this is -12 + 6, 24 – 2³ is 8, all this².
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32 – 12, 26/ 16², that is your answer, you are done.
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You have a nice finite number, that is your limit.
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Evaluate the following limit, if it exists.
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The limit is x approaches 4 of x² - 6 - 12 divided by x – 4.
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If we plug in 4 here, we get a 0 in the denominator.
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There is not much we can do with that.
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Let us see if we can actually manipulate this.
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And whenever you see something like this, I think factoring is probably the best approach.
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This is going to be, this x² – x - 12/ x – 4.
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We can factor that into x – 4, x + 3/ x – 4.
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Those cancel, we are just left with x + 3.
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We take the limit again.
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The limit as x approaches 4 of x + 3, you plug it in again like always.
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We get a finite number 7, we can stop, we are done.
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Notice the limit exists, the limit is 7.
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The limit as x approaches 4.
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But go back to the original function, 4 is not in the domain.
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The function is not even defined.
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The limit as x approaches 4 is defined, it is 7.
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f(4) is not defined.
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That is all that is going on here.
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The limit exists but the value of the function there does not exist.
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Those are two independent things.
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Evaluate the following limit if it exists.
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5x – 45/ the absolute of x – 9.
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Here we go, the absolute value sign again.
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Great, that is all we need.
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Let me see, where shall I write this.
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I will write it over here.
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The absolute value of x – 9 is equal to x – 9, whenever x - 9 is greater than 0.
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It is equal to -x – 9, whenever x - 9 is less than 0.
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It is the whole thing in there, whatever the whole thing is.
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There we go.
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Let us take a look here, what is it that we are actually going to be doing?
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Let us go ahead and solve for x - 9 greater than 0, which is the same as x is greater than 9,
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which is the same as x approaching 9 from above.
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The limit as x approaches 9 of 5x - 45/ the absolute value of x – 9.
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Since this is 4x approaching 9 from above which is x is bigger than 9, which is x - 9 is bigger than 0, it is this one.
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That is going to equal the limit,
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I like to do it over here, not a problem.
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= the limit as x approaches 9 from above of 5x - 45/ x - 9 = 5 × x - 9/ x - 9 = 5.
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There you go.
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Now for x - 9 less than 0 which is equivalent to x is less than 9, which is equivalent to x approaching 9 from below.
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Now we have the limit as x approaches 9 from,
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I think I got another page that I can do this.
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I do, let us go ahead and do that.
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Let us try this again.
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For x - 9 less than 0 which is equivalent to x less than 9,
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which is equivalent to x approaching 9 from below, numbers less than 9.
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The limit as x approaches 9 from below of 5x - 45/ the absolute value of x – 9
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is equal to the limit as x approach, this was the original.
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Now 9, 5x – 45/ -x – 9 = the limit as x approaches 9 from below of 5 × x - 9/ -x - 9 = -5.
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The left hand limit, -5 does not equal the right hand limit 5.
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The limit does not exist.
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5 does not equal -5, this limit, it does not exist.
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What you are looking at is the following.
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This is 9, this is the Cartesian coordinate system.
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When x is less than 9, you are down here at -5.
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When it is bigger than 9, you are up here at 5.
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That is what is going on here.
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Clearly, 9 is not in that domain.
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It is open circle, open circle, it is not defined at 9 because then you would have a 0 in the denominator.
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But the limits exist, but the left hand limit and the right hand limit individually exist, but they are not the same.
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The limit itself does not exist.
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This is a discontinuous function, a huge discontinuity, a difference of 10.
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Evaluate the following limit, if it exists.
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When you plug in h = 0, in this case, 0 in the denominator, 0 in the denom.
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When we plug in, means we have to manipulate.
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Whenever you see radicals, you are probably going to be rationalizing something.
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Let us take x + h - √x/ h and let us multiply by the conjugate of the numerator.
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It is going to be x + h, under the radical, + √x/ x + h + √x.
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And that is going to equal x + h - x/ h × √x + h + √x.
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That and that cancels, we are left with h/ h × √x + h + √x.
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h and h cancels, we are left with 1/ √x + h + √x.
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We have gone as far as we can go, as far as simplification is concerned.
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We take the limit again.
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We are going to take the limit again.
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As h goes to 0 of this, that goes to 0.
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What you are left with is 1/ √x + √x 1/ 2 √x.
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That is our limit.
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Notice in this case, our limit is a function, which is fine, because we did not specify what came as a function.
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Therefore, it is going to be a function.
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For different values of x, you are going to get a number.
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If x is 4, the √4 is 2.
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You are going to end up with ¼.
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You still get a finite number.
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You just get different numbers for different values of x.
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Let us see what we get, evaluate the following limit if it exists.
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X³ – 8/ x² – 4.
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If you plug in 2, you are going to end up with a 0 in the denominator.
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This is going to require a little bit of something.
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Let us do some factoring, x³ - 8/ x² – 4.
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This is going to equal x - 2 × x² + 2x + 4.
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Hopefully, you remember how to factor a difference of cubes.
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If not, just Google it.
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/ x - 2 × x + 2.
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We notice that goes away.
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Now we can take the limit of this function, since this is now the equivalent of the original function.
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The limit as x approaches 2 of x² + 2x + 4/ x + 2 is equal to 4 + 4 + 4/4 is 12/4 = 3.
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Again, notice the limit of the function exists but 2 is not in the domain.
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The limit of the function as x approaches 2 exists.
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It is equal to 3.
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But the value of f at 2 does not exists, because 2 is not in the domain.
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We get to go back to the original function.
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2 is not in the domain, it does not work.
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Just because it cancels here, and you are left with this, it does not mean that 2 is in the domain.
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You have to go back to the original function.
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This is going to be some graph with a hole in it, at x = 2.
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That is what it going on here.
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Anytime you have a cancellation like this, there is a hole in the graph.
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3x – 13 is less than or equal to f(x), less than or equal to x² – 9x + 7.
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Find the limit of f(x) as x approaches 2.
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I notice that the function is between two things.
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Let us take the limit of everything.
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If we have this, therefore, the limit as x approaches 2 of 3x - 13 is going to be less than or equal to
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the limit as x approaches 2 of f(x), which is what we want,
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which is going to be less than or equal to the limit as x approaches 2 of x² – 9x + 7.
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If you have a relation, anything you do, if you do it to every single element in this relation, the relation is retained.
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The limit as x approaches 2 of 3x – 13.
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You plug in the 2, 6 – 13, you will get to get -7, less than or equal to the limit as x approaches 2 of f(x).
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Less that or equal to, plug 2 in, you get 4 - 18 + 7 is – 7.
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Therefore, this limit = -7, the squeeze theorem.
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Nice and straightforward application.
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Let f(x) = x², form the quotient f(x) + h - f(x)/ h, then algebraically simplify as possible.
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Now take the limit of the expression that you just got.
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f(x) = x², form the quotient f(x) + h.
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f(x) + h, whenever you see in parentheses, that goes in there.
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This thing is going to be x + h², that is f(x) + h - f(x) which is x²/ h.
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That = x² + 2x h + h² – x²/ h.
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That cancels, I’m going to factor out an h.
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h × 2x + h/ h, that cancels.
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I'm left with 2x + h.
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Now take the limit as h approaches 0.
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That is it, nice and straightforward.
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What you just did, let me change colors here.
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What you just did, what we just did is the limit as h approaches 0 of f(x) + h - f(x)/ h, for a specific f(x).
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In other words, x².
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We mentioned in the first lesson that this was the definition of the derivative, that this was the how.
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Remember, we said the derivative why and how.
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What is it and how do we find it?
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This was the how, for finding the derivative.
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I know I get too ahead of myself here, derivative of f(x).
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Here we have f(x) = x², we just found f’(x) = 2x.
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Remember, we said if you have an f(x), you take the derivative.
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You are going to end up with some other function of x,
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which is going to give you the slope of the tangent line to the curve at any given value of x.
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That is what we did.
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When our functions is x², my slope at any value of x,
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the slope of my tangent line that is touching the curve at 1 point is going to be 2x.
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That is what I just did.
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For different values of x, f’(x) = 2x will give us the slope of the tangent line thru the point x f(x).
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That is it, that is all we have done.
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We said that this is how you find the derivative and that is exactly what we did for a specific function.
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We found the derivative.
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We started with f(x) and we derived f’(x).
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Now no matter where I am, let us say x = 15, if I go to x = 15, the value for the y value was going to be 15² which is 225, I hope.
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At 15, 225, there is a line that touches that point 15, 225 which is tangent to the curve.
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That is the curve at that line.
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2 × 15 is going to be the slope of that tangent line.
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That is what we have done, finding the derivative.
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It involves taking the limit of this thing.
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In case some of you are wondering, we took x², we took the derivative and we ended up with 2x.
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Is it a coincidence that this 2 and this 2, that this is a 2 and this is a 2 is a coincidence, it is not.
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Is it a coincidence, the answer is no.
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It is not a coincidence.
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This is the limit process for finding the derivative.
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We are going to spend some time doing that, so we actually get accustomed to the process.
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But in calculus, there is always a shortcut.
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The shortcut is x² to 2x and we will see what the shortcut is.
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We do not actually have to calculate this whole long process of simplifying some bizarre function.
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And then, having to take the limit of it still.
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We can just take the derivative of it but that is exactly what we have done.
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There you go, a little bit of an introduction to some things that you are going to see in some future lessons.
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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.