Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 12

h(x)=(x^2-1)/(2x-1)
differentiating by applying the quotient rule,
h'(x)=((2x-1)(2x)-(x^2-1)(2))/(2x-1)^2
h'(x)=(4x^2-2x-2x^2+2)/(2x-1)^2
h'(x)=(2x^2-2x+2)/(2x-1)^2
differentiating again,
h''(x)=((2x-1)^2(4x-2)-(2x^2-2x+2)(2)(2x-1)(2))/(2x-1)^4
h''(x)=(2(2x-1)(2x-1)^2-4(2x-1)(2x^2-2x+2))/(2x-1)^4
h''(x)=(2(2x-1)(4x^2+1-4x-4x^2+4x-4))/(2x-1)^4
h''(x)=-6/(2x-1)^3
There are no points at which h''(x)=0 , but at x=1/2, the function h is not continuous, so test for continuity in the interval (-ooh''(1)=-6
h''(0)=6
Since h''>0 therefore in the interval (- ) , it is concave upward
h''<0 therefore in the interval (1/2