Question

find the stationary points on the curve y=x^3e^x and determine their nature.

Answer 1

There aren't any.

The derivative of y=e^x is y'=e^x, which is never equal to zero.

Answer 2

then hw do i know if there is a point of inflextion ?

Answer 3

sorrry there is an error the question states y=x^3e^x

Answer 4

Oh, that makes a pretty big difference :P

Differentiate it to find y'. Use the product rule, then set y' = 0 and find x.

To find inflection points, find y'' and set y''=0 and solve for x.

Answer 5

yeah i did that then got rlly confused i need working out:/

Answer 6

Using the product rule...
\[\frac{dy}{dx}=3x^2*e^x+x^3*e^x=e^x(3x^2+x^3)=x^2e^x(3+x)\]

Answer 7

then for stationary points dy/dx = 0
\[0=x^2e^x(3+x)\]
e^x is never 0
x^2 is 0 at x = 0
(3+x) is = at x = -3

so you have stationary points at x = 0, -3

Answer 8

then for the nature, you could differentiate again, but I think it's easier just to use a table of values

Answer 9

therefore at x = -3, it is a relative minimum
and at x = 0, it is a horizontal inflexion

Answer 10

how do you find the point of inflexion using the second derivative ?

Answer 11

The same way you found the stationary points - set f'' = 0 and solve for x.

Answer 12

yeah but i get confused when i differentiate it

Answer 13

\[\large y' =x^2e^x(3+x)\] maybe foil/expand this first, then differentiate.

Answer 14

You'll need the product rule again: http://www.youtube.com/watch?v=h78GdGiRmpM