Finding Extrema and Points of Inflection

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kinerry

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Jul 7, 2006
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I'm assuming I have to use some sort of derivative test to find each.

Here's the problem in a nutshell:

Need to find the extrema and points of inflection of the following problem;

n(x) = (1/(sqrt2pi))e^(-x^2/2)

Also, what is the function (the graph of this function) known as?

As usual, any help is appreciated.
 
\(\displaystyle \L n(x) = \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}\)
is the normal probability density function.

why don't you start out by finding \(\displaystyle n'(x)\) and \(\displaystyle n''(x)\), then showing us what you think needs to be done to find extrema and points of inflection.

small hint ... you'll get one extrema value and two inflection points.
 
n'(x) = -(xe^((-1/2x)^2))/(sqrt(2)*sqrt(pi))
n''(x) = x^2e^((-1/2x)^2)/(sqrt(2)*sqrt(pi))-(e^((-1/2x)^2)/(sqrt(2)*sqrt(pi))

I can only hope I typed that out correctly lol
 
let's make life easy by naming that "ugly" constant, \(\displaystyle \L \frac{1}{\sqrt{2 \pi}\), k.

\(\displaystyle \L n(x) = ke^{\frac{-x^2}{2}}\)

\(\displaystyle \L n'(x) = -kxe^{\frac{-x^2}{2}}\)

\(\displaystyle \L n''(x) = k(x^2 - 1)e^{\frac{-x^2}{2}}\)

if we set the first derivative equal to zero, we get a critical value at x = 0.
note also that n''(0) < 0, indicating that n(x) is concave down at x = 0.
so ... what does that tell you about the extremum located at x = 0?

if we set the second derivative equal to zero, we get two critical values ... one at x = -1 and the other at x = 1. It is easy to confirm that n''(x) changes sign at each of these x-values, indicating what?
 
You seem to be guessing.

Relative maxima come from the first derivative.

Points of Inflection come from the second derivative.
 
That's what I get for second guessing myself.
I assumed they were inflection points, but when I graphed it, I couldn't quite pin-point them just by looking.

Thank you for the help.
 
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