Answer
No values of $x$.
Work Step by Step
We have: $y=\dfrac{x}{x+1}$
We differentiate both sides with respect to $x$.
$y^{\prime}=\dfrac{d}{dx} [\dfrac{x}{x+1}] \\=\dfrac{1}{(x+1)^2}$
The tangent line to the graph will be horizontal when $ \dfrac{dy}{dx} =0$
Therefore, $\dfrac{1}{(x+1)^2} \implies 1=0$
So, we find no values of $x$.
