[tex]\bf f(x)=\sqrt{x^2+4x+5}\implies f(x)=(x^2+4x+5)^{\frac{1}{2}} \\\\\\ \cfrac{df}{dx}=\stackrel{\textit{using the chain rule}}{\cfrac{1}{2}(x^2+4x+5)^{-\frac{1}{2}}(2x+4)}\implies \cfrac{df}{dx}=\cfrac{2x+4}{2\sqrt{x^2+4x+5}} \\\\\\ \cfrac{df}{dx}=\cfrac{2(x+2)}{2\sqrt{x^2+4x+5}}\implies \cfrac{df}{dx}=\cfrac{x+2}{\sqrt{x^2+4x+5}} \\\\\\ \stackrel{\textit{now, setting the derivative to 0}}{0=\cfrac{x+2}{\sqrt{x^2+4x+5}}}\implies 0=x+2\implies \boxed{-2=x}[/tex]
recall that a tangent line that's horizontal has a slope of 0 and the value of "x" is simply the critical points.
one may note that we can get two more critical points from zeroing out the denominator, but those critical points are usually asymptotic and the equation is not differentiable at those points.
