How do you determine whether the function #f(x)=(2x-3) / (x^2)# is concave up or concave down and its intervals?

1 Answer
Jul 28, 2015

#f(x)=(2x-3)/(x^2)# is concave up when #x > 9/2# and concave down when #x < 9/2# (and #x!=0#).

Explanation:

By the Quotient Rule, for #x!=0#, the first derivative is #f'(x)=(x^2*2-(2x-3)*2x)/(x^4)=(-2x^2+6x)/(x^4)=(-2x+6)/(x^3)#

and the second derivative is
#f''(x)=(x^3*(-2)-(-2x+6)*3x^2)/(x^6)=(4x^3-18x^2)/(x^6)=(4x-18)/(x^4)#

Since #x^4\geq 0# for all #x#, the sign of #f''(x)# is the same as the sign of its numerator #4x-18#. This expression is positive when #4x>18\Leftrightarrow x > 9/2# and negative when #4x < 18 \Leftrightarrow x < 9/2#.

The value #x=9/2# is the first coordinate of the unique inflection point of the graph of #f#. The second coordinate is #f(9/2) = (9-3)/((9/2)^2)=6/(81/4)=24/81=8/27#