`y = 2x - tan(x) , (-pi/2, pi/2)` Determine the open intervals on whcih the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 13 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`y=2x-tanx`

differentiating,

`y'=2-sec^2(x)`

differentiating again,

`y''=-2sec(x)sec(x)tan(x)`

`y''=-2sec^2(x)tan(x)`

`y''=-2(1/(cos^2(x)))(sin(x)/cos(x))`

`y''=-2sin(x)/(cos^2(x))`

In order to determine the concavity , determine the value of x when y''=0

`(-2sinx)/(cos^2(x))=0`

sinx=0 , so x=0 , pi, 2pi ,.....

Now let us test for concavity in the intervals (-pi/2,0) and (0,pi/2)

y''(-pi/4)=2*(-pi/4)- tan(-pi/4)= -pi/2+1 ( negative)

y''(pi/4)=2*(pi/4)-tan(pi/4)=pi/2-1 (positive)

So, the graph is concave upward in the interval (0,pi/2) and

concave downward in the interval (-pi/2,0)

 

 

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