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The point of P(2,-1) lies on the curve y=1/(1-x). If Q is the point (x, 1/(1-x)), use your calculator to find the slope of the secant line PQ for the following values of x:

i) 1.5

ii) 1.9

iii) 1.99

iv) 1.999

v) 2.5

 

For parts i and ii of the question, I keep on getting the wrong answer... Could someone please help me out with this? Help with at least just "i" least would be appreciated. 

 Aug 21, 2014

Best Answer 

 #1
avatar+128061 
+10

I'll do the first one, the others are similar....(only the values have changed to protect the innocent...)

For, 1.5, we have the point on the curve

f(1.5) = 1 /(1 - 1.5) = 1 / (-.5) = -2

So we have two points  (2, -1) and (1.5 , -2)   and the slope of the secant line is  [-2 - (-1)] / [1.5 - 2] =         -1 / -.5    = 2

Here's the graph, here......note that the secant line drawn with these two enpoints would have a positive slope.

The point of this exercise is that is to show that the closer we get to x = 2, the more the secant line models the tangent line to the function at that point (i.e., the "derivative" at that point)

Hope this helps...

 

 Aug 21, 2014
 #1
avatar+128061 
+10
Best Answer

I'll do the first one, the others are similar....(only the values have changed to protect the innocent...)

For, 1.5, we have the point on the curve

f(1.5) = 1 /(1 - 1.5) = 1 / (-.5) = -2

So we have two points  (2, -1) and (1.5 , -2)   and the slope of the secant line is  [-2 - (-1)] / [1.5 - 2] =         -1 / -.5    = 2

Here's the graph, here......note that the secant line drawn with these two enpoints would have a positive slope.

The point of this exercise is that is to show that the closer we get to x = 2, the more the secant line models the tangent line to the function at that point (i.e., the "derivative" at that point)

Hope this helps...

 

CPhill Aug 21, 2014
 #2
avatar+564 
0

Thanks so much CPhill!

 Aug 22, 2014

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