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

1 Answer
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(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.)