If f(x) = ( x^3 )/( x^2 - 25 ), what are the points of inflection, concavity and critical points?

1 Answer
Nov 30, 2016

Critical numbers: 0 and +-sqrt75

Explanation:

Concave down on (-oo,-5) and on (0,5)

Concave up on (-5,0) and on (5,oo)

Inflection point (0,0)

f'(x) = (x^2(x^2-75))/(x^2-25)^2

f'(x) DNE at x=+-5 but those are not in the domain, so they are not critical.

f'(x) = 0 at x=0 and at x + +-sqrt75 which are in the domain, so they are all critical numbers.

f''(x) = (50x(x^2+75))/(x^2-25)^3 could change sign at 0 and at +-5

On (-oo,-5), f''(x) < 0, so f is concave down.
On (-5,0), f''(x) > 0, so f is concave up.
On (0,5), f''(x) < 0, so f is concave down.
On (5,oo), f''(x) < 0, so f is concave up.

The concavity changes at x=-5, 0 and 5. An inflection point is a point of the graph where concavity changes. Since -5 and 5 are not in the domain of f, there are no IPs there, but f(0) = 0, so (0,0) is an Infle pt.