find the equations of all three tangent to the curve x+y^3-y=1 at x=1.

Question

a) dy/dx

b) ordered pairs when x=1

c) slope of tangent lines

d) equation of tangent lines

(be sure to show all work)

Answer 1

Graphing this (to get your head wrapped around it) gives you this:

At x = 1 we can determine the three points by plugging in x = 1 and solving:

x + y³ - y = 1

1 + y³ - y = 1

y³ - y = 0

y(y² - 1) = 0

y(y + 1)(y - 1) = 0

y = 0

y = -1

y = 1

So the points are (1, 1), (1, 0), and (1, -1)

The derivative can be found using implicit differentiation:

Apply the derivative operator:

d (x + y³ - y) = d (1)

dx dx

1 + 3y²(dy/dx) - dy/dx = 0

3y²(dy/dx) - dy/dx = -1

dy/dx(3y² - 1) = -1

dy/dx = -1/(3y² - 1)

At y = 1: dy/dx = -1/2

At y = 0: dy/dx = 1

At y = -1: dy/dx = -1/2

Using the point-slope form of a line:

Line 1 [at a slope of -1/2 and point (1, 1)], the equation of the line is y - 1 = -1/2(x - 1) or y = -1/2x + 3/2

Line 2 [at a slope of 1 and point (1, 0)], the equation of the line is y - 0 = 1(x - 1) or y = x - 1

Line 3 [at a slope of -1/2 and point (1, -1)], the equation of the line is y - -1 = -1/2(x - 1) or y = -1/2x - 1/2