Inflection points on #y=f(x)# are places on the curve where the concavity (measured by #f''(x)#) changes from positive to negative, or vice versa. We can see where the second derivative is zero to look for possibilities. Let's take the first two derivatives:

#f(x) = x/(x^2+9)# now use the quotient rule (derivs in [ ])

#f'(x) = ((x^2+9)[1]-(x)[2x])/((x^2+9)^2)=(-x^2+9)/(x^2+9)^2#; do again:

#f''(x) = ((x^2+9)^2*[-2x]-(-x^2+9)[2(x^2+9)(2x)])/((x^2+9)^4)#

# = ((x^2+9)*[-2x]-(-x^2+9)[2(2x)])/((x^2+9)^3)=(-54x+2x^3)/(x^2+9)^3#

If we set this equal to zero, and solve:

#-54x+2x^3=0 => 2x(x^2-27)=0#; we get three answers:

#x = 0 and x = +-sqrt(27) = +-3sqrt(3)#. The sign of #f''(x)# does change across #x = 0#, so the inflection points are at #x=0, +-sqrt(3)#.

Plug these x's into the original function to get the y-coordinates.

Going further: Now sketch the graph. You might want to find the critical points too. I'll leave that to you so your brain gets some more exercise!

\dansmath to the rescue!/