Check a cancellation exception before claiming a strict parameter range

$y'=[(4x+k)(x+1)-(2x^2+kx-1)]/(x+1)^2=[4x^2+(k+4)x+k-2x^2-kx+1]/(x+1)^2=(2x^2+4x+1+k)/(x+1)^2=[2(x+1)^2-1+k]/(x+1)^2$. The source prints 1−kin one intermediate numerator; its subsequent completed-square form and discriminant use the correct 1+k.

The source sets $2x^2+4x+1+k=0$ and excludes real roots by requiring $4^2-4(2)(1+k)<0$. Thus 1+k>2 and k>1. This correctly gives the range with no real numerator roots.

At k=1, the numerator factors as(2 x−1)(x+1), so G is y=2 x−1 with x=−1 removed. Its derivative is 2 everywhere in its domain, hence it also has no stationary point. For k<1 there are two admissible derivative zeros. Thus the literal no-stationary-point condition is k≥1; the printed target k>1 is valid only if the cancellation case is excluded. The prompt does not explicitly exclude it.

The source writes $y=2x+1+r/(x+1)=[2x^2+3x+1+r]/(x+1)$, calling the remainder r by the auxiliary letter a. Compare with the original numerator: k=3 and 1+r=−1, so r=−2. Hence $y=2x+1-2/(x+1)$.

Asymptotes are x=−1 and y=2 x+1. Intercepts are $((-3\pm\sqrt{17})/4,0)$, approximately (−1.78,0),(0.281,0), and (0,−1). The derivative is $2+2/(x+1)^2>0$, so no stationary points occur.