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

h''(1)=-6

h''(0)=6

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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