## anonymous 4 years ago The graph of the function y=e^(-x^2/2) isn't the same as the graph of the function y=e^(-x) is it because that the first expression wants a positive value of all x? And does anyone know d/dx (x^2-1)e^(-x^2/2)?

1. anonymous

Both $f(x)=e^{(-x^2/2)}$$y=e^{-x}$may have negative value for x. Following links provide the graph of each function. f(x) --> http://www.wolframalpha.com/input/?i=y%3De^%28-x^2%2F2%29 g(x) --> http://www.wolframalpha.com/input/?i=y%3De^%28-x%29 Note something: 1) ^2 cause concave up. 2) - sign cause mirror image of the + graphs. Hence g(x) is a mirror image of e^x , whereas f(x) has a concave, but it is down due to the negative sign.

2. anonymous

On the second question First of all, multiply/rearrange the function. $f(x)=(x^2-1)(e^{-x^2/2})=x^2 e^{-x^2/2}-e^{-x^2/2}$Next, we use chain rule.$f'(x)=2xe^{-x^2/2}+x^2(-2x/2)e^{-x^2/2}+(-)(-2x/2)e^{-x^2/2}$$f'(x)=2xe^{-x^2/2}-x^3e^{-x^2/2}+xe^{-x^2/2}$$f'(x)=e^{-x^2/2}(2x-x^3+x)$$f'(x)=e^{-x^2/2}(3x-x^3)$

3. anonymous

P.S. g(x) I mentioned on the first response is y=e^-x

4. anonymous

This is what I meant and I graphed and expected to see the same graph...

5. anonymous

$\frac{1}{\sqrt{e^{x^2}}} \neq \frac{1}{e^x}$ I think you mistaken $\frac{1}{\sqrt{e^{x^2}}} \text{as} \frac{1}{\sqrt{e^{2x}}}$See? say for example e=2.7 and let x=2. If you subtitle those values to what you meant before, the left hand side (LHS) will be diferrent to right hand side (RHS).

6. anonymous

of course haha it's been a long day thanks! I was just starting to freak out...

7. anonymous

LOL. You just need to take a fresh air. Observe that the ^2 here is for x only; not (e^x) altogether. $e^{(x^2)} \neq (e^x)^2$

8. anonymous

Good idea! Going for a run and pump some more blood into my head. Thanks!

9. anonymous

student 92.... i do not think e(x^2)≠(e^x)^2 They are both equal....

10. anonymous

let just say $$x=1$$. $e(x^2)=e(1^2)=e$$(e^x)^2=(e^1)^2=e^2$Are they still the same?