Another problem (Derivatives)
Find all values of x at which the tangent line to the graph curve is parallel to the line y = x
y = (X^2 + 1)/(X+1)
Y' = (2x(x + 1) - (X^2 + 1)) / (X^2 + 2X +1) (Quotient rule)
if y = x, the slope is 1, so Y' = 1
1 = (2x^2 + 2x - X^2 - 1) / (X^2 + 2X + 1)
X^2 + 2X + 1 = X^2 + 2X - 1
2 = 0
...? What'd I do wrong? (Crying)
Re: Another problem (Derivatives)
First of all you should factor your function since X^2 - 1 + 2 = (x+1)(x-1) + 2 which implies y = (x-1) + 2/(x+1) as long as x != -1.
dy/dx = 1 - 2/(x+1)^2 = 1 which implies
1 = 1 - 2/(x+1)^2 which implies
-2/(x+1)^2 = 0 which only happens when x tends to infinity (positive or negative) which implies that there is no solution (which confirms your answer).
Notice what happens when x -> infinity: the 2/(x+1) cancels out and you get y = x - 1 which is indeed parallel to y = x.