Question

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

Answer 1

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(x) = (1-x^2)/x^3`, the second derivative is

`f''(x) = (-2(x^2-6))/x^5`

Key numbers for `f''(x)` are `+-sqrt6` 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 = -sqrt6, 0, "and" sqrt6`.

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

The inflection points are `(-sqrt6, f(-sqrt6))` and `(sqrt6, f(sqrt6))`.

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