Question

Is `f(x) =e^xsinx-cosx` concave or convex at `x=pi/3`?

Answer 1

Concave up (my naming convention is different to yours, will explain further in explanation)

Explanation:

Obtain `f'(x)` by differentiating the original function (product rule):
`f'(x)=e^xsin(x)+e^xcos(x)+sin(x)`

Obtain `f''(x)` by differentiating again:
`f''(x)=e^xsin(x)+e^xcos(x)+e^xcos(x)-e^xsin(x)+cos(x)=2e^xcos(x)+cos(x)`
`=(2e^x+1)cos(x)`

Evaluate `f''(pi/3)`:
`f''(pi/3)=(2e^(pi/3)+1)cos(pi/3)`
`=1/2(2e^(pi/3)+1)`
Clearly `f''(pi/3)>0`, so the function will be concave up at `x=pi/3`

Note that the naming convention I learnt was concave up and concave down, so I am not sure which is concave or convex in your terms.

Graphing the original fucntion is a good way to check your answer, at `x=pi/3~~1`, we see that the rate of change of gradient is indeed positive and therefore concave up.
graph{e^x*sinx-cosx [-3, 3, -5, 5]}