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The point $ P(2, -1) $ lies on the curve $ y = 1/(1-x) $.

(a) If $ Q $ is the point $ (x, 1/(1-x)) $, use your calculator to find the slope of the secant line $ PQ $ (correct to six decimal places) for the following values of $ x $: (i) $ 1.5 $ (ii) $ 1.9 $ (iii) $ 1.99 $ (iv) $ 1.999 $ (v) $ 2.5 $ (vi) $ 2.1 $ (vii) $ 2.01 $ (viii) $ 2.001 $

(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at $ P(2, -1) $.

(c) Using the slope from part (b), find an equation of the tangent line to the curve at $ P(2, -1) $.

A.(a) $y=\frac{1}{1-x^{2}}, P(2,-1)$$$\begin{array}{|r|l|l|l|}\hline & x & Q(x, 1 /(1-x)) & m_{P Q} \\\hline(i) & 1.5 & (1.5,-2) & 2 \\(i i) & 1.9 & (1.9,-1.111111) & 1.111111 \\(\text { ii }) & 1.99 & (1.99,-1.010101) & 1.010101 \\(i v) & 1.999 & (1.999,-1.001001) & 1.001001 \\(v) & 2.5 & (2.5,-0.666667) & 0.666667 \\(v i) & 2.1 & (2.1,-0.909091) & 0.509091 \\(\text { vin }) & 2.01 & (2.01,-0.900099) & 0.990099 \\( \text { viii) } & 2.001 & (2.001,-0.999001) & 0.999001 \\\hline\end{array}$$B.1C.$y=x-3$

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the point two negative one called P lies on the curve. Michaels 1/106. We know that that is true because if we put x equals two in this equation here we get Y equals -1. So the point to negative one less than the curve. Why? It was 1/1 -6 in part A. If we take any point of that care Q. That is with coordinates X. and one over one minute six. We want to calculate the slope of the second line peak. You for the following values of X 1.519199. 2.5 to bring 1 2.01 2.001. That is we're taking values closer and close to two from the left and from the right. Okay. In part cheerier use we're gonna use the results in part A. And with that we're going to get the value of the slope of the tangent line to the cure at B 2 -1. You know these slow will be the values that are approaching those slopes of the second lines you perceive? We use the slow guest in barbie To find an equation to turn to the lines of the cure at -2 -1. So we're going to calculate them for a all the values. So first we know it. You're following. All right, mm hmm. This lobe of this second nine through the point p 82 and 81 and Q Given by X 1/1 -X. Supposing X is not zero. Sorry X is not one which is the value that notifies dominator and the value is undefined. So if we write that slope of that second line we get Mhm The following it's called the globe M. And we know the equation will be won over one minus X. Which is the Y coordinate of the point Q minus. Do I coordinate of the point P. That is negative one. My understand the one and in the denominator yet the first coordinative Q. That is X minus the first coordinator of P. That is to And this give us one over 1 -6 plus one over eggs minus two. And in fact this is equal to one minus eggs as denominator in the numerator and um one minute six denominator And where do we get one plus one minus x. Which is two minus six. The denominator of the big fraction we get x minus two and then this is equal to two minus x over the product of one minus x. Times explain this to. We noticed that in the numerator we can get I can factor negative one out and then this negative parenthesis X -2 over. Okay 1 -6 times X -2. And because X is if we take X different front to for example and in this case we verify we are not taking the value to we are close enough to to where we are never taking the value to. So we can say that this equal for X. Different from two. That we can simplify the factor X -2. And we get 91 over one minus six. Or in fact one over X -1. So the slope of the second line, mhm BQ is m equals one over x minus or so. This is our expression here. And now we can calculate all these slopes for x equal 1.5 which is the first value here we get then the slope one over 1.5 minus one is 1/0 10.5. Remember 0.51 house. So I want to now far too. We take x equal 19. And for 1.9 we get em equal 1/1 9 minus one which is 1/0 10.9. And here we use the calculator and we find that this is approximately equal to and we are going to use six testicle figures. So at one one in all the vegetables. So this is uh the slope of the second line peak you at x equals 1.9. Now, Part three. We take 1.99. Yeah. Yeah. Okay. And so the slope of the 10. 2nd line is one over 199 -1 which is one over 0.99. And is this approximately equal to on 21? Yeah 0101. Then our four x is 1.999. And then we get em equal 1/1 0.999 minus one which is 1/0 10.999. And that's approximately equal to 1.001 your little one. No Part five. We use x equal We did this already. We take 2.5. and so m is one over 2.5 -1, sequel to one over 1.5. And this is approximately equal to some 4.6666 67. Mhm. Then for six x is taken equal, 2.1 Mm is one over 2.11 is one is 1/1 10.1. That's approximately equal to 09 zero 90 90, not zero because the following pressing over the announcer, we won here and for seven is equal to 2.01. and so um is one over 2.01 -1, which is one over 1.01 and that's approximately two. It's your point 99 zero 99 mm Yes. Sarah, So this is seven now eight exist. 2.001. So um is 1/2 0.1 minus one, which is one over 1.001. That's approximately equal to to your point three times nine 001. Okay, yeah. Okay, so these are approximations. So we see that yes, first approximations of X closer to the value to hear. But from the left is all this result up to here and for values of X Closer to two. But from left, the value of the slope of the second line peak, you are getting Closer and closer to one with values greater than one. Now the other part here is for ex closer to two. But from the right and for those values of X closer to two from the right again develops of the slope of the second line peak. You are closer to one again but now from values less than But in both cases we see that the slope of the second line peak, you are getting closer and closer to one so far. B we can say that mhm. The slope of the tangent line which is in fact the limits of this or the values that are approaching these slopes of second lines. When we are close to be acceptable to slope of the tangent line to the cure at P 2 -2. Because here we are, we have the first question equal to which is the value. We are approaching for the variable X. And when we approached the X22 from the left from the right, the values of the slope of the second lines Approaches of only one. And so we can say that the slope of the tangent line which is defined as those mm limits Which got to be equal in fact from the left and from the right. And then we say that that slope of the tangent line to the care at that point to an 81 is one. And in party we know then that the line is tangent line has an equation of the tangent. Flying to the curve at B291 is given by Why -2. Using the coordinates of the point equal to slow. Calculated by using the slopes of the second lines. When x is approaching the value to one times X minus two. So we get y plus two, Y Plus one. Sorry, Equal X -2. And if we put all from to the right of the equation we get X -Y -3 equals zero. So this is equation of the tangent line to the curve At P 2 91. Yeah. And we have calculated the slope of the tangent line to the curve at that point. B tune 81. Using the slopes of the second lines to the cure fixing the point P two negative one and moving the point Q. By changing its first coordinate X. And putting it closer and closer to two from the writing from the left. And in both cases we have seen that the slopes of that of those second lines approaches the value one. So the slope of the tangent line will be one. And with that we have calculated the equation of the tangent line at the point two negative one.

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