help with finding critical points

[delgeezee]

I am trying to see how I can find the critical points of
\(\displaystyle f(x)=-e^{\frac{-x^2}{2}\)

The derivative should be
\(\displaystyle f'=xe^{\frac{-x^2}{2}\)

According the the rules of finding critical points, I should set the derivative =0


Mathlab says I should get critical points at x=-1 and x=1,
Unfortunately, I do not see how this can be. Can someone explain this to me. It looks to me that I should get x=0, unless my derivative is wrong. Also I read the \(\displaystyle e^x !=0\) so I don't know.


\(\displaystyle xe^{\frac{-x^2}{2}=0 \)??

 

[renegade05]

I am trying to see how I can find the critical points of
\(\displaystyle f(x)=-e^{\frac{-x^2}{2}\)

The derivative should be
\(\displaystyle f'=xe^{\frac{-x^2}{2}\)

According the the rules of finding critical points, I should set the derivative =0


Mathlab says I should get critical points at x=-1 and x=1,
Unfortunately, I do not see how this can be. Can someone explain this to me. It looks to me that I should get x=0, unless my derivative is wrong. Also I read the \(\displaystyle e^x !=0\) so I don't know.


\(\displaystyle xe^{\frac{-x^2}{2}}=0 \)??

I think you are getting inflection points and absolute max/min confused. These are the critical points.

You are right to set the derivative to zero and solve for x to find absolute/local max/min.

\(\displaystyle xe^{\frac{-x^2}{2}}=0 \)

It may be easier to rewrite this as:

\(\displaystyle \frac{x}{\sqrt{e^{x^2}}}=0 \)

Now will \(\displaystyle \sqrt{e^{x^2}}\) ever equal zero? Try graphing it and seeing if it has zeroes. Or plug in values for X.

So you can conclude there is only one critical point when it comes to max/min, which is?

For inflection points you must now take the second derivative of f(x) and set that to zero.

You should review your notes to see what the difference is between the two and get some intuition to why we set the derivative and second derivative to zero to attain max/min and inflection points, respectively.