WEBVTT
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given F of X which is equal to sign of
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tent by over X which passes through the point P
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Which is coordinates 10. We need to find the
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following. 1st the slope of the second lines through
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the X values. And then we need to explain
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why these slopes of the second lines in part A
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are not close to the slope of the tangent line
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at P. And lastly we need to choose appropriate
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sequence lines to estimate the slope at P. For
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the first one to find the slope of the 2nd
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line. We used the formula for the slope of
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the line which is just the change and why over
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the change in X, which in this case will
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be F of x minus ever won over. That
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will be x minus one. Since f of one
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is zero. Then we have F of X.
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All over X-1 is the slope of the second
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line. So we will use this formula to find
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me slope of the second lines. Through that these
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given X values. Now for us to see a
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pattern, we will do a table and in the
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stable we will write our X values in order So
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for extra 0.5 we have signs of 10 pie Over
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0.5 that's divided by 0.5-1. We have Value
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equal to zero And if it's 0.6 we have Sign
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of 10 by over 0.6 Divided by 0.6-1.
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We get-2.1651 At X equals 0.7. We have
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signed up 10 pi over 0.7 over 0.7-1.
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That gives us-2.6061 at X 0.8 We have signs
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of 0.8 Rather 10 pi over 0.8 Over 0.8-1
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. This gives us-5. And if it's your
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.9 we have sign of 10 by Over 0.9 over
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0.9-1. This gives us 3.4-0 20. If
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access 1.1 We have signs of 10 pi over 1.1
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divided by 1.1-1, we have negative two point
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8173. and if X is 1.2 we have signs
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of 10 pi over 1.2, All over 1.2-1
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, that gives us 4.3301. Yes Access 1.3 We
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have signs of 10 pi over 1.3 Over 1.3-1
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. This gives us-2.7433. and if it's 1.4
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we have a sign of Then by over 1.4 divided
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by 1.4-1 we have-1.0847. And if it's
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1.5 we have sign of 10 by Over 1.5 over
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1.5-1. This is just one point 7321.
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And lastly if it's too we have signs of 10
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by Over to over 2-1, we have Slope
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equal to zero Know that as we move closer to
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x equals one. The value of the slope is
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not really approaching a value close to each other.
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So we're not really seeing a pattern here. So
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to estimate the slope of the tangent line at P
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. Let's look at the graph and from there,
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choose the correct points to form the second lines,
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the lowest a graph of the function. And here
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we see that the function is changing rapidly from increasing
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to decreasing. And so this is the reason why
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we're not seeing a pattern for the slope of the
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tangent line. And so for part C to choose
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the correct points, we will use the X values
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just right here. That will be X values in
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between 0.95 and 1.05. Let's say X is 0.9
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brian 0.999. And then for X values right of
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one, let's say we have one point 0001 And
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then 1.001. And then from here we find these
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slopes of the second lines. We do the same
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process as what they did in party. We make
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a table and then find the slope there below is
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the table of values now, yes, X 0.99
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We have signs of 10 by over 0.99 over 0.99
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-1. We get negative 31 point 2033. And
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if it's 0.999, we have Sign of 10 by
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over 3.999 Over 0.999-1. This gives us-31.4422
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. Now, if X is 1.0001, we have
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sign of 10 by over 1.0001 Over 1.00 1-1
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. That gives us-31.41 27 At X equals 1.001
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. We have sign of 10 pies over 1.001 Over
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1.001-1. We get-31.3794. Therefore the slope
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of the tangent line at p is approximately-31.4.
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This is because as we Move closer to one,
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The value of the slope is approaching negative 31.4