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

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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?
 
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rshalloo said:

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.
 


Mark44 said:
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
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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