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Find all the maxima and inflection points of the following function:

$$f(x)=\frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}}$$

Answer is local Max is at $(0,\frac{1}{\sqrt{2\pi}})$

inflection points $(1,\frac{1}{\sqrt{2\pi{e}}})$ & $(1,\frac{-1}{\sqrt{2\pi{e}}})$

MCO
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  • What makes you think they are wrong? – Henry Mar 12 '22 at 20:42
  • @Henry I'm not sure if the e is part of the answer or if I should omit it – MCO Mar 12 '22 at 21:26
  • Since $e^{-1/2}=\frac1{\sqrt e}$ you can't just make the $e$ disappear; I don't think that's the error. On a piece of scrap paper, try writing these out more explicitly. So the maximum is at $(x, f(x)) = \left(0,\frac{1}{\sqrt{2\pi}}\right),$ or even better write it as $x=0,$ $f(x)=\frac{1}{\sqrt{2\pi}}.$ Now one inflection point written out like this. Then the other inflection point. You should now have how many different $x$ values? And $f(x)$ is what at each $x$? – David K Mar 12 '22 at 22:36
  • $(1,\frac{-1}{\sqrt{2\pi{e}}})$ is wrong, in a sense doubly wrong – Henry Mar 12 '22 at 23:17

1 Answers1

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If

$$f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},$$

then

$$f'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}},$$

and

$$f''(x)=\frac{x^2-1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.$$

Now as $e^{-\frac{x^2}{2}}>0$ for all $x\in\mathbb{R}$, we get that

$$f'(x)=-\frac{x}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}=0 \iff x=0$$

and

$$f''(x)=\frac{x^2-1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}=0 \iff x^2-1=0 \iff x=\pm 1.$$

It is easy to check that these give a global maximum and two inflection points respectively. Furthermore

$$f(0)=\frac{1}{\sqrt{2\pi}}$$

and

$$f(\pm 1)=\frac{1}{\sqrt{2\pi}}e^{-\frac{(\pm 1)^2}{2}}=\frac{1}{\sqrt{2\pi e}}.$$

Thus the global minimum is at

$$\left(0,\frac{1}{\sqrt{2\pi}}\right)$$

and the inflection points are at

$$\left(-1,\frac{1}{\sqrt{2\pi e}}\right)\text{ and }\left(1,\frac{1}{\sqrt{2\pi e}}\right).$$

Lorago
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