# How do you find the inflection points of the graph of the function: (1-x^2)/x^3?

Oct 20, 2015

An inflection point is a point on the graph at which concavity changes. So, it is a point on the graph at which the sign of the second derivative changes.

#### Explanation:

For $f \left(x\right) = \frac{1 - {x}^{2}}{x} ^ 3$, the second derivative is

$f ' ' \left(x\right) = \frac{- 2 \left({x}^{2} - 6\right)}{x} ^ 5$

Key numbers for $f ' ' \left(x\right)$ are $\pm \sqrt{6}$ and $0$.

These key numbers cut the number line into 4 intervals.

We look at the sign of $f ' '$ on each interval to determine whether $f$ is concave up or concave down on the interval

{: (bb "Interval", bb"Sign of "f'',bb" Concavity"), ((-oo,-sqrt6)," " +" ", " ""Up"), ((-sqrt6,0)," " -" ", " ""Down"), ((0,sqrt6), " " + " ", " ""Up"), ((sqrt6,oo), " " -, " ""Down") :}

The concavity changes at $x = - \sqrt{6} , 0 , \text{and} \sqrt{6}$.

$0$ is not in the domain of $f$, so there is no point on the graph at $x = 0$.

The inflection points are $\left(- \sqrt{6} , f \left(- \sqrt{6}\right)\right)$ and $\left(\sqrt{6} , f \left(\sqrt{6}\right)\right)$.

The arithmetic is left as an exercise. (But note that $f$ is odd, so we only need to do one arithmetic.)