We have the expression:
[tex]f(x)=\frac{-5}{x^2+1}[/tex]
In orde to determine it's relative maximum and minimum, we operate as follows:
[tex]f^{\prime}(x)=\frac{10x}{(x^2+1)^2}[/tex]
When we equal x to 0, we will have the point (0, -5).
And replacing values near 0, we will have that before 0 decreases and after 0 increases, from this, we have that the point (0, -5) is a relative minimum.
