Question #aa795

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Answer 1 · 2016-09-17T16:13:09

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

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

$y'=(s^x)'sin x+e^x(sin x)'$

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

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

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

$=e^x(sin x + cos x + cos x - sin x)$

$=2e^x cos x=0, when cos x =0 to x=+-pi/2$

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

$=2e^x(cos x-sin x)$

At $x=+-pi/2$ , y'''is not 0

Thus, $x = +-pi/2$ are points of inflection. But these ar outside $(-pi/2, pi/2)$ .

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

Here, y' is not 0 at $x = +-pi/2$ .