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

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Chapter 3, 3.4 - Problem 9 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=(x^2+1)/(x^2-1)`

differentiating by applying quotient rule,

`f'(x)=((x^2-1)(2x)-(x^2+1)(2x))/(x^2-1)^2`

`f'(x)=(2x^3-2x-2x^3-2x)/(x^2-1)^2`

`f'(x)=(-4x)/(x^2-1)^2`

differentiating again,

`f''(x)=((x^2-1)^2(-4)-(-4x)(2)(x^2-1)(2x))/(x^2-1)^4`

`f''(x)=(-4(x^2-1)(x^2-1-4x^2))/(x^2-1)^4`

`f''(x)=(4(3x^2+1))/(x^2-1)^3`

In order to determine the concavity , determine when f''(x)=0 , however there are no points at which f''(x)=0 but at x=`+-` 1 the function is not continuous.

So test for concavity in the intervals (-`oo` ,-1) , (-1,1) , (1,`oo` ) by plugging in the test values in f''(x).

`f''(0)=(4(3*0^2+1))/(0^2-1)^3=-4`

`f''(-2)=(4(3*(-2)^2+1))/((-2)^2-1)^3=52/27`

`f''(2)=(4(3*2^2+1))/(2^2-1)^3=52/27`

Since f''(2) and f''(-2) are positive , so the function is concave upward in the interval (-`oo` ,-1) and (1,`oo` )

f''(0) is negative , so the function is concave downward in the interval (-1,1). 

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