The function F(x)=e^(-3x^2)
has 2 inflection points. Find the x-coordinate of the inflection point when x0, correct to 2 decimal places.

Question

The function F(x)=e^(-3x^2)
has 2 inflection points. Find the x-coordinate of the inflection point when x>0, correct to 2 decimal places.

anonymous · Answer

King, this is the same problem TheFigure was working on.... he just couldn't figure it out. (pun not intended lol)

anonymous · Answer

@amistre64

kinggeorge · Answer

The method he was using is definitely what you want to use. He probably just took one of the derivatives wrong.

anonymous · Answer

could u go see where he went wrong

anonymous · Answer

http://openstudy.com/study#/updates/4f6b71bce4b014cf77c83a61

kinggeorge · Answer

Can you check if a certain answer is correct, and if so I'll tell you how to do it?

kinggeorge · Answer

Try$${1 \over \sqrt{6} }\approx 0.41$$

anonymous · Answer

thats correct!!

kinggeorge · Answer

After looking through Figure's work, I'm pretty sure he took the first derivative wrong. When he wrote$$f'(x)= -6x^2e^{-3x^2}$$It should have been$$f'(x)=-6xe^{-3x^2}$$Since the derivative of $-3x^2$ is $-6x$.

Now, if we use this new $f'$, we have that $$f''(x)=6e^{-3x^2}(6x^2-1)$$Thus, $6x^2-1=0$ which means that $x=\pm1/\sqrt{6}$ But since we're interested in positive x-values, $x=1/\sqrt{6}$

anonymous · Answer

ok thanks for your time man. much appreciated.

kinggeorge · Answer

You're welcome. Once again, sorry I wasn't able to answer this sooner.

anonymous · Answer

at least you helped. thats all that matters.