QUESTION SOLVED, HELP RECEIVED
So the question is: "At which points do the tangent lines through (1,2) touch y=x/x+1?"
So I've gone through basically everything (fried my brain this one did):
1. Found derivative using quotient rule; "1/(x+1)^2
2. Equalled the slope of the tangent line that passes through (1,2) and (x,x/x+1) to the derivative and solved for x
3. Got the roots x=-1 and x=-2(+or-)sqrt3
4. Rejected x=-1 because of restrictions
Now I believe that the next step is to plug in x into the original y to complete the point. So, err, how do I go about doing that?
Thanks in advance for any and all help.
Ah, well, for curiosity's sake, this is how I did those steps:
[2-(x/x+1)] / (1-x) = 1/(x+1)^2 (Initial Equation)
[2(x+1)-x] / (1-x)(x+1) = 1/(x+1)^2 (Found Common Denominator for Left side)
x+2/(1-x)(x+1) = 1/(x+1)^2 (Simplified left numerator)
(x+2)(x+1)^2 = (1-x)(x+1) (Cross multiplied; I know now that I could have reduced (x+1) in the denominators before cross multiplying which would mean that I wouldn't have ended up with x=-1 as a root...I also could have cross multiplied without simplifying >//<; )
x^3 + 5x^2 + 5x + 1 = 0 (Simplified and moved all terms to left side)
(x+1)(x^2 + 4x + 1) = 0 (Factored using remainder and factor theorem)
I then rejected x=-1 and used the quadratic equation on the trinomial:
= [-4 (+or-) sqrt (16-4(1)(1)] / 2(1)
= [-4 (+or-) sqrt (12)]/ 2
= [-4 (+or-) 2*sqrt (3)]/ 2
= -2 (+or-) sqrt 3
And that's that. I hope that was clear and if you see any mistakes in MY calculations please let me know. ^_^