Proving x1 + x2 = -2 for Parallel Tangents to f(x) = 1/(x+1)

[rshalloo]

Homework Statement

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

If tangents to curve at x=x₁ and x=x₂ are parrallel and if x₁ is not equal to x₂
show that x₁ + x₂ = -2

The Attempt at a Solution

Well i found my equations for the asymptotes
Horizontal: x=0
Vertical: x= -1

and then i would say that if they were parrallel that the slopes are equal and therefore at the point x₁ and the point x₂ the slopes are equal

f¹(x) = -1/(x+1)²

If i sub in x₁ and x₂, they will be equal and the question says they arent?

Can anyone help me please?

 

[Mark44]

Homework Statement

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

If tangents to curve at x=x₁ and x=x₂ are parrallel and if x₁ is not equal to x₂
show that x₁ + x₂ = -2

The Attempt at a Solution

Well i found my equations for the asymptotes
Horizontal: x=0
Vertical: x= -1

and then i would say that if they were parrallel that the slopes are equal and therefore at the point x₁ and the point x₂ the slopes are equal

f¹(x) = -1/(x+1)²

If i sub in x₁ and x₂, they will be equal and the question says they arent?

Can anyone help me please?

Your title is misleading - this problem doesn't have anything to do with asymptotes of either kind.

You are give that the tangent lines are parallel at x = x₁ and x = x₂, so f'(x₁) = f'(x₂). This means that -1/(x₁ + 1)² = -1/(x₂ + 1)². Solve that equation, keeping in mind that x₁ $\neq$ x₂, and that if a² = b² ==> a = b or a = -b.

 

[rshalloo]

Your title is misleading - this problem doesn't have anything to do with asymptotes of either kind.
QUOTE]

Sorry about that it was just a part of a question with asymptotes and i asumed it was something to do with asymptotes :S thanks for the help