`g(x) = (x^2 + 4)/(4 - x^2)` Determine the open intervals on whcih the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 11 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`g(x)=(x^2+4)/(4-x^2)`

differentiating by applying quotient rule,

`g'(x)=((4-x^2)(2x)-(x^2+4)(-2x))/(4-x^2)^2`

`g'(x)=(8x-2x^3+2x^3+8x)/(4-x^2)^2`

`g'(x)=(16x)/(4-x^2)^2`

differentiating again by applying quotient rule,

`g''(x)=16((4-x^2)^2-x(2)(4-x^2)(-2x))/(4-x^2)^4`

`g''(x)=(16(4-x^2)(4-x^2+4x^2))/(4-x^2)^4`

`g''(x)=(16(3x^2+4))/(4-x^2)^3`

There are no points at g''(x)=0,but at x=2 and x=-2 the function is not continuous.

So test for concavity in the intervals (-`oo` ,-2) , (-2,2) and (2,`oo` )

g''(-3)=-496/125

g''(0)=1

g''(3)=-496/125

Since g''(-3) and g''(3) are less than 0 , so the graph is concave downward in the intervals (-`oo` ,-2) and(2,`oo` )

and g''(0) is greater than 0 , so the graph is concave upward in the interval (-2,2).

 

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