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

Mar 31, 2016

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

#### Explanation:

Obtain $f ' \left(x\right)$ by differentiating the original function (product rule):
$f ' \left(x\right) = {e}^{x} \sin \left(x\right) + {e}^{x} \cos \left(x\right) + \sin \left(x\right)$

Obtain $f ' ' \left(x\right)$ by differentiating again:
$f ' ' \left(x\right) = {e}^{x} \sin \left(x\right) + {e}^{x} \cos \left(x\right) + {e}^{x} \cos \left(x\right) - {e}^{x} \sin \left(x\right) + \cos \left(x\right) = 2 {e}^{x} \cos \left(x\right) + \cos \left(x\right)$
$= \left(2 {e}^{x} + 1\right) \cos \left(x\right)$

Evaluate $f ' ' \left(\frac{\pi}{3}\right)$:
$f ' ' \left(\frac{\pi}{3}\right) = \left(2 {e}^{\frac{\pi}{3}} + 1\right) \cos \left(\frac{\pi}{3}\right)$
$= \frac{1}{2} \left(2 {e}^{\frac{\pi}{3}} + 1\right)$
Clearly $f ' ' \left(\frac{\pi}{3}\right) > 0$, so the function will be concave up at $x = \frac{\pi}{3}$

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 $x = \frac{\pi}{3} \approx 1$, 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]}