Question

How do you find the intervals which are concave up and concave down for `f(x) = x/x^2 - 5`?

Answer 1

Assuming that this should be `f(x) = x/(x^2 - 5)`, see below.

Explanation:

To determine concavity, investigate the sign of the second derivative.

`f''(x) = (2x(x^2+15))/(x^2-5)^3`

`f''` is zero at `0` and undefined at `+-sqrt5`, so we investigate the sign in four intervals:

#{: (bb "Interval", bb"Sign of "f'',bb" Concavity"), ((-oo,-sqrt5)," " -" ", " "" Down"), ((-sqrt5,0), " " +, " " " Up"), ((0,sqrt5), " " -, " " " Down"), ((sqrt5,oo), " " +, " "" Up") :}#

The only inflection point is `(0,0)`.

(`f` is undefined at `+-sqrt5`, so there is no point on the graph at `x = +-sqrt5`.)