y = (x)/(x^2+9)

- Jan 17th 2011, 08:38 PMonanycHow do I find the relative extrema and points of inflection of this function?
y = (x)/(x^2+9)

- Jan 17th 2011, 08:57 PMProve It
Local Maxima: The points where $\displaystyle \displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \displaystyle \frac{d^2y}{dx^2} < 0$.

Local Minima: The points where $\displaystyle \displaystyle \frac{dy}{dx} = 0$ and $\displaystyle \displaystyle \frac{d^2y}{dx^2} > 0$.

Inflection Points may or may not appear where $\displaystyle \displaystyle \frac{d^2y}{dx^2} = 0$. It helps to evaluate these points and compare them to the graph of the function. - Jan 18th 2011, 06:35 AMHallsofIvy
"Inflections Points" are defined as points where the first derivative changes sign. Where that happens the second derivative must be 0 but the second derivative may be 0 where the first derivative does not change sign (example, $\displaystyle f(x)= x^3$ at x= 0). Determine where the second derivative is 0, then look at the first derivative on either side of those points.

(Surely you knew this basic information, onanyc, so what is your difficulty with this particular problem?)