Calculus problem?

Question

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). 

Answer 1

(a).

The points are P = (2, - 1) and Q = [x, 1/(1 - x)] and the curve : y = 1/(1 - x).

Calculate the slope of the secant line PQ by using formula : m = (y₂ - y₁) / (x₂ - x₁), where (x₁, y₁) and (x₂, y₂) are two points.

To find the slope, substitute (x₁, y₁) = (2, - 1) and (x₂, y₂) = [x, 1/(1 - x)] in the slope formula.

 

(b).

From the above conclusion (part a), the x-values approaches to 2 then the slope approaches to 1.

The x-coordinate of the point P(2, - 1) is 2, so the value of the slope of the tangent line to the curve at P(2,-1) is 1.

Answer 2

(c).

From part (b), the slope is m = 1 and the point P(2, - 1).

Find the tangent line by using Point-slope form of line equation : y - y₁ = m(x - x₁),, where (x₁, y₁) = point and m = slope.

Substitute (x₁, y₁) = (2, - 1) and m = 1 in the Point-slope form of line equation.

y - (-1) = (1)(x - 2)

y + 1 = x - 2

y = x - 3.

The tangent line equation is y = x - 3.