## anonymous 5 years ago locate the three inflexion points of the curve y=xe^(-x^2). please show explanation :)

1. anonymous

$y=x*e^{-x^2}$$y'=e^{-x^2}(1-2x^2)$$y''=2e^{-x^2}x(2x^2-3)$ $y''=0 \rightarrow x=\pm \sqrt{3/2}$

2. anonymous

and x = 0

3. anonymous

When i diffrentiated y=x∗e−x2 i got y′=e−x2(1+2x2) can you show me how u got the first derivative

4. anonymous

use the product rule on x and e^(-x^2) and then the chain rule on e^(-x^2)

5. anonymous

i think you misused a sign somewhere for the quantity (1 + 2x^2)

6. anonymous

i dont see where i went wrong heres my working out y'= uv'+vu' where u= x u'= 1 v=e^(-x^2) v'=2xe^(-x^2) then y'= x(2xe^(-x^2))+e^(-x^2)(1) = e^(-x^2)(2x^2+1) can u tell me where my mistake is

7. anonymous

e should be -e

8. anonymous

v' is -2x e^(-x^2)

9. anonymous

oh yes i see that mistake thanks heaps