Question

locate the three inflexion points of the curve y=xe^(-x^2).

please show explanation :)

Answer 1

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

Answer 2

and x = 0

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

e should be -e

Answer 8

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

Answer 9

oh yes i see that mistake thanks heaps