In order to find the concavity of a function at some point #x#, you need to take the second derivative of the function. Then, you need to find the sign of the second derivative:
- If #f''(x)>0#, the function is convex.
- If #f''(x)<0#, the function is concave.
For #f(x)=tanx#:
#f'(x)=sec^2x#
#f''(x)=2secxtanx(secx)=2sec^2xtanx#
Now we have to evaluate this at #x=pi/3#:
#f''(pi/3)=2sec^2(pi/3)tan(pi/3)#
Before we go about this, remember that we're only concerned about the sign of the second derivative, not the actual value. We don't actually need to compute #2sec^2(pi/3)tan(pi/3)# - we just have to find out if it's negative or positive.
Take a close look at this expression. #pi/3# is an angle in the first quadrant (it's in between #0# and #pi/2#), and all the trig functions are positive in the first quadrant. That means all three of the pieces we have here are positive:
#underbrace(2)underbrace(sec^2(pi/3))underbracetan(pi/3)#
#+##color(white)(XX)+##color(white)(XXX)+#
And when you multiply positive numbers, you get a positive result. Therefore, we can conclude that #f''(pi/3)>0#, so the function is convex at that point. You can confirm this by looking at the function's graph.
