Let: #y = tanx + 2x# on #(-pi/2, pi/2)#
#y' = sec^2x +2# (For finding #y''#, remember that #sec^2x = (secx)^2# so we'll use the chain rule)
#y'' = 2secx secx tanx # (The derivative of #2# is #0#.)
So #y'' = 2sec^2x tanx#.
#y''# is never undefined on #(-pi/2, pi/2)#, so its only chance to change sign is at its zero(s)..
#abs(sec x) >= 1#, so #sec^2x > 1# So the only zero of #y''# is where #tanx = 0#, which is at #x = 0#
For #x# in #(-pi/2, 0)#, #tanx # is negative, so #y''# is negative and the graph of the function is concave down.
For #x# in #(0, pi/2)#, #tanx # is positive, so #y''# is positive and the graph of the function is concave up.
The function is concave down on #(-pi/2, 0)#, and it is
concave up on #(0, pi/2)#.
The point #(0,0)# is an inflection point.
