finding extremas of natural log functions

[roxstar1]

find the extremas and the points of inflections of the following:

g(x) = 1/((2pi)^1/2) e^((-(x-2)^2)/2)

I have found g'(x) to = (e^(-(x-2)^2)/2) but I can't figure out how to find the

critical numbers using this information. Any suggestions would be much appreciated.

Ps: sorry for the ugly equations, I'm not sure how to copy the equation (using equation editor) onto this space

 

[soroban]

Hello, roxstar1!

Find the extremas and the points of inflections of the following:

\(\displaystyle \;\;g(x)\;=\;\:\frac{1}{\sqrt{2\p}}\cdot e^{-\frac{1}{2}(x-2)^2}\)

I have found: \(\displaystyle \,g'(x)\;=\;e^{-\frac{1}{2}(x-2)^2}\;\) . . . This is incorrect

The derivative of \(\displaystyle e^u\) is: \(\displaystyle e^u\cdot u'\)
\(\displaystyle \;\;\)the function itself ... time the derivative of the exponent.

Since -\(\displaystyle \frac{1}{2}(x\,-\,2)^2\:=\:-\frac{1}{2}(x^2\,-\,4x\,+\,4)\:=\:-\frac{1}{2}x^2\,+\,2x\,-\,2\),

\(\displaystyle \;\;\)we have: \(\displaystyle \,g(x)\;=\;\frac{1}{\sqrt{2\pi}}\cdot e^{(-\frac{1}{2}x^2+2x-2)}\)


Then: \(\displaystyle \:g'(x)\;=\;\frac{1}{\sqrt{2\pi}}\cdot e^{(-\frac{1}{2}x^2+2x-2)}\cdot(-x\,+\,2)\)


Can you finish it now?

 

[Gene]

z=1/(2pi)^.5 to save typing
You erred on g'(x)
y = z*e^(-(x-2)²/2)
ln(y)=ln(z)+(-(x-2)²/2)
dy/y=-(x-2)dx
dy/dx=-(x-2)*y =
-(x-2)*z*e^(-(x-2)²/2))
looks better.
-(x-2)*z*e^(-(x-2)²/2)) = 0
has only one extrema which you should see easily.
Would you care to try for g''(x) for the two inflexion points?

PS. Soroban took the shortcut and beat me, as usual. I'm glad we agree.