Let #f(x) = (x^2+1) / (x^2-2)#.

The domain of #f# is #(-oo,-sqrt2) uu (-sqrt2, sqrt2) uu (sqrt2,oo)#.

To find critical numbers, we find #f'(x) = (-6x)/(x^2-2)^2# which is #0# at #x=0# and is never undefined on the domain of #f#.

**The only critical number is #0#.**

To investigate concavity and inflection points, we find #f''(x) = (6(3x^2+2))/(x^2-2)^3# which is never #0# and is undefined at #-sqrt2# and at #sqrt2#

**Sign of #f''#**

#{: (bb "Interval", bb"Sign of "f'',bb" Concavity"),
((-oo,-sqrt2)," "" " +" ", " "" Up"),
((-sqrt2,sqrt2), " "" " -, " " " Dn"),
((sqrt2 ,oo), " "" " +, " "" Up")
:}#

The graph of #f# is concave up on the intervals #(-oo,-sqrt2)# and #(sqrt2,oo)#. The graph is concave down on #(-sqrt2,sqrt2)#.

Although the concavioty changes at #x=+-sqrt2#, there is no point on the graph at #x=+-sqrt2#, so there are no inflection points.