# How do you find the inflection points for the function f(x)=x/(x^2+9)?

Aug 18, 2014

Inflection points on $y = f \left(x\right)$ are places on the curve where the concavity (measured by $f ' ' \left(x\right)$) 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:

$f \left(x\right) = \frac{x}{{x}^{2} + 9}$ now use the quotient rule (derivs in [ ])

$f ' \left(x\right) = \frac{\left({x}^{2} + 9\right) \left[1\right] - \left(x\right) \left[2 x\right]}{{\left({x}^{2} + 9\right)}^{2}} = \frac{- {x}^{2} + 9}{{x}^{2} + 9} ^ 2$; do again:

$f ' ' \left(x\right) = \frac{{\left({x}^{2} + 9\right)}^{2} \cdot \left[- 2 x\right] - \left(- {x}^{2} + 9\right) \left[2 \left({x}^{2} + 9\right) \left(2 x\right)\right]}{{\left({x}^{2} + 9\right)}^{4}}$
$= \frac{\left({x}^{2} + 9\right) \cdot \left[- 2 x\right] - \left(- {x}^{2} + 9\right) \left[2 \left(2 x\right)\right]}{{\left({x}^{2} + 9\right)}^{3}} = \frac{- 54 x + 2 {x}^{3}}{{x}^{2} + 9} ^ 3$

If we set this equal to zero, and solve:
$- 54 x + 2 {x}^{3} = 0 \implies 2 x \left({x}^{2} - 27\right) = 0$; we get three answers:
$x = 0 \mathmr{and} x = \pm \sqrt{27} = \pm 3 \sqrt{3}$. The sign of $f ' ' \left(x\right)$ does change across $x = 0$, so the inflection points are at $x = 0 , \pm \sqrt{3}$.

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!/