Hey, y'all. Find equations of the tangent lines to the curve y= (x-1)/(x+1) that are parallel to the line x-2y=1

Answer 1

y = 1/2x - 1/2; slope = 1/2; y' = [(x + 1) - (x - 1)]/(x + 1)² = 2/(x + 1)² = 1/2; x = - 1 ± 2; x = 1 or x = - 3

x = 1, y = 0; equation of tangent line: y = 0 + 1/2(x - 1); y = 1/2x - 1/2.

x = - 3; y = 2; equation of tangent line: y = 2 + 1/2(x + 3); y = 1/2x + 7/2

Answer 2

f(x) = (x-1)/(x+1)

f'(x) = [(x+1) - (x-1)]/(x+1)² = 2/(x+1)²

x-2y=1 can be written as y = (1/2)x - 1/2, so the slope of all lines parallel to x-2y=1 must be 1/2. Hence, we can solve for x in f'(x) using our slope of 1/2:

1/2 = 2/(x+1)²

(1/2)(x+1)² = 2

(x+1)² = 4

x² + 2x + 1 = 4

x² + 2x - 3 = 0

(x+3)(x-1) = 0

x = -3, 1

If x=-3, then f(-3) = ((-3)-1)/((-3)+1) = -4/(-2) = 2 --> (x₁,y₁) = (-3,2)

If x=1, then f(1) = (1-1)/(1+1) = 0/2 = 0 --> (x₂,y₂) = (1,0)

Hence, we have two possible tangent line equations for the original function that are parallel to x+2y=1:

y-y₁ = m(x-x₁)

y-2 = (1/2)(x-(-3))

y-2 = (1/2)x + 3/2

y = (1/2)x + 7/2

y-y₂ = m(x-x₂)

y-0 = (1/2)(x-1)

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

The two equations are thus y = (1/2)x + 7/2 and y = (1/2)x - 1/2. Hope this helped!

Answer 3

Finding y' using the quotient rule yields y' = 2/(x+1)².

The slope of the given line is 1/2.

Determine when y' = 1/2 to find x-coordinates of points of tangency.

1/2 (x+1)² = 2

(x+1)²=4

x+1 = ±2

x₁=1, x₂=-3

y₁=0, y₂=2

The equations of the tangent lines given m = 1/2 and the points of tangency (1, 0) and (-3, 2) are:

y=1/2(x-1)

and

y - 2 = 1/2(x+3)

Of course you can rearrange those equations into other forms (slope-intercept, for example).

Here is a Desmos graph depicting the above:

desmos.com/calculator/dhrqm62qpk