What are the points of inflection of #f(x)=4 / (2x^2- 7x - 4#?

1 Answer
Jun 21, 2018

#f'(x)=(-16x+28)/(2x^2-7x-4)^2#
#f''(x)=(8(12x^2-47x+57))/(2x^2-7x-4)^3#
#f''(x)=0# no values, #f''(x)# DNE at #x=4 and x=(-1/2)#

Explanation:

The numerator of the second derivative has no real roots, the only places where the second derivative is 0 or fails to exist come from the denominator. The second derivative fails to exist at at #x=4 and x=(-1/2)#
These are the only points that could be points of inflection. You can only define them as points of inflection if curvature changes sign on either side of each value.