Points of a curve!

[Synergy]

Hey everyone,

I need to find the points on a curve which is represented by y = x / x+1, where the tangent is parallel to x - y = 2!

I believe I would use the derivative rules, and most probably quotient rule to find derivative of first equation 1 But then after that I can not seem to go on! Any help please!

[opus]

Haven't done that for a long time, but:

You have a function f(x)=x/(x+1)
and you want all the points on that function that have a tangent that is parallel to the line: x - y = 2 which is that same as y=x-2, so this line has a gradient of 1!
So you need to find the points on the function that have a gradient of 1.
To find the gradient of a point on your function, you need to diferentiate it.

f(x)=x/(x+1)
or
f(x)=g(x)/h(x)=x/(x+1)

f'(x)=g'(x)*h(x)-g(x)*h'(x)/h(x)*h(x)
f'(x)=1*(x+1)-x*1/(x+1)^2=1/(x+1)^2

And then solve for 1

1=1/(x+1)^2

and that's only true for x=0


I'm crossing my fingers on that one!

[krtxmrtz]

...
...
...
And then solve for 1

1=1/(x+1)^2

and that's only true for x=0

...and for x = -2

[opus]

I knew it, missed that one again