`f(x) = (2x^2)/(3x^2 + 1)` Determine the open intervals on whcih the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 8 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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

differentiating by applying quotient rule,

`f'(x)=((3x^2+1)(4x)-(2x^2)(6x))/(3x^2+1)^2`

`f'(x)=(12x^3+4x-12x^3)/(3x^2+1)^2`

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

differentiating again,

`f''(x)=((3x^2+1)^2(4)-4x(2)(3x^2+1)(6x))/(3x^2+1)^4`

`f''(x)=(4(3x^2+1)(3x^2+1-12x^2))/(3x^2+1)^4`

`f''(x)=(4(1-9x^2))/(3x^2+1)^3`

In order to determine the concavity , first determine when f''(x)=0,

`4(1-9x^2)/(3x^2+1)^3=0`

`1-9x^2=0`

`x=+-1/3`

Now let us consider the intervals (-`oo` ,-1/3) , (-1/3,1/3) and (1/3,`oo` ) and determine the signs of f''(x) by plugging in the test values in f''(x).

`f''(0)=(4(1-9*0^2))/(3*0^2+1)^3=4`

`f''(-1)=(4(1-9*(-1)^2))/(3*(-1)^2+1)^3=-1/2`

`f''(1)=(4(1-9*(1)^2))/(3*1^2+1)^3=-1/2`

So the graph is concave downward in the interval (-`oo` ,-1/3) and (1/3,`oo` )

graph is concave upward in the interval (-1/3,1/3).

 

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