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!/
