# Question aa795

Sep 17, 2016

$x = \pm \frac{\pi}{2}$. But these are outside the open interval (-pi/2, pi/2)). So, within here, there is no point of inflexion.

#### Explanation:

Let $y = {e}^{x} \sin x$.

$y ' = \left({s}^{x}\right) ' \sin x + {e}^{x} \left(\sin x\right) '$

=e^x(sin x + e^x cos x

$= {e}^{x} \left(\sin x + \cos x\right)$

$y ' ' = \left({e}^{x}\right) ' \left(\sin x + \cos x\right) + {e}^{x} \left(\sin x + \cos x\right) '$

$= {e}^{x} \left(\sin x + \cos x + \cos x - \sin x\right)$

$= 2 {e}^{x} \cos x = 0 , w h e n \cos x = 0 \to x = \pm \frac{\pi}{2}$

y'''=2((e^x)'cos x+e^x(sin x)'

$= 2 {e}^{x} \left(\cos x - \sin x\right)$

At $x = \pm \frac{\pi}{2}$, y'''is not 0

Thus,$x = \pm \frac{\pi}{2}$ are points of inflection. But these ar outside $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$.

Note that y'=0 is not a necessary condition.

Here, y' is not 0 at $x = \pm \frac{\pi}{2}$.