`h(x) = (x^2 - 1)/(2x - 1)` Determine the open intervals on which the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 12 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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differentiating by applying the quotient rule,




differentiating again,





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 (-`oo<x<1/2` ) and            (1/2`<x<oo` ).



Since h''`>0` therefore in the interval (-`` ) , it is concave upward

h''`<0`  therefore in the interval (1/2`<x<oo`  ), it is concave downward. 

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