Question
The point P(2,-1) lies on the curve y=1/(1-x)
If Q is the point (x, 1/(1-x) find slope of secant line.
these are the points
1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.01 2.001
when i calculated it i got a negative number
1/(1-1.5)=-2 but the answer is 2. also would the points be (1.5, 2) 0r (2,2)?
If Q is the point (x, 1/(1-x) find slope of secant line.
these are the points
1.5 1.9 1.99 1.999 2.5 2.1 2.01 2.01 2.001
when i calculated it i got a negative number
1/(1-1.5)=-2 but the answer is 2. also would the points be (1.5, 2) 0r (2,2)?
Answers
Damon
first let's look at the derivative (slope at a point)
y = 1/(1-x)
dy/dx = slope = [-1(-1)]/(1-x)^2
bottom always +
so the slope is always positive
now secant from (2,-1) to
( x , 1/(1-x) )
secant slope P to Q =
(1/(1-x) +1) / (x-2)
(1 + 1 - x )/ [(1-x)(x-2)]
= (x-2) / (x^2 - 3x + 2)
if x = 1.5 then
= (-.5 /(2.25 - 4.5 + 2
= -.5 / (-.25)
= + 2 sure enough
y = 1/(1-x)
dy/dx = slope = [-1(-1)]/(1-x)^2
bottom always +
so the slope is always positive
now secant from (2,-1) to
( x , 1/(1-x) )
secant slope P to Q =
(1/(1-x) +1) / (x-2)
(1 + 1 - x )/ [(1-x)(x-2)]
= (x-2) / (x^2 - 3x + 2)
if x = 1.5 then
= (-.5 /(2.25 - 4.5 + 2
= -.5 / (-.25)
= + 2 sure enough
feather
it says to round to 6 decimal numbers
so it would for example the first point
(1.5, 2)
then it would be
(1.9,1.111111) right?
so it would for example the first point
(1.5, 2)
then it would be
(1.9,1.111111) right?