Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 13

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)