Is #f(x) =sinx-x^2# concave or convex at #x=pi/3#?

1 Answer
Jul 13, 2016

Since the function is twice differentiable, the answer depends on whether the second derivative #f''(pi/3)# is positive or negative

Explanation:

Let's calculate the #f''(x)#. First #f'(x) = cosx-2x#, then #f''(x)=-sinx-2#. So, #f''(pi/3)=-sin(pi/3)-2#.

But #sin(pi/3)=sqrt3/2#, and then #f''(pi/3)=-sin(pi/3)-2=-sqrt3/2-2#, which is negative. Thus, since #f''(pi/3)<0#, the function is concave downward (convex)