Question

Find all values of x at which the tangent line to the given curve satisfies the stated property y=xsquared +1 over x+1; parallel to the line y = x the answer is supposed to be none

Answer 1

$y=\dfrac{x^2+1}{x+1}\implies y'=\dfrac{2x(x+1)-(x^2+1)}{(x+1)^2}=\dfrac{x^2+2x-1}{(x+1)^2}$

Any line parallel to $y=x$ will have the same slope of 1, so you're looking for all $x$ such that $y'$ above also evaluates to 1.

$\dfrac{x^2+2x-1}{(x+1)^2}=1\implies x^2+2x-1=(x+1)^2$

This assumes $x\neq-1$, which is of course the case because $x=-1$ lies outside the function's domain.

$x^2+2x-1=x^2+2x+1\implies -1=1$

which is not true. This means no tangent line to $y=\dfrac{x^2+1}{x+1}$ will ever be parallel to $y=x$.