# Stationary Points

• Dec 2nd 2008, 04:06 AM
Jen1603
Stationary Points
Locate 1) the position and nature of the stationary points and 2) the points of inflection of the following curve

y = x^4 - 6x^2 +15
• Dec 2nd 2008, 04:12 AM
Leodinas
Find $\displaystyle \frac{dy}{dx}$ of the $\displaystyle y$, then put that expression equal to zero, which will give you the x co-ordinates of the points at which the gradient is 0 i.e. stationary points etc.
To classify the inflexion point, find the gradient at the points either side of the place where the gradient is equal, if both results provide gradients with the same sign i.e. +ve or -ve, then it is inflection.

$\displaystyle \frac{dy}{dx} = 4x^3 - 12x$
• Dec 2nd 2008, 04:19 AM
Jen1603
i got that far , thanks.

i just got stuck after that , when i simplified i was ended up with x + / - roots and i wasnt sure if both were roots, or only one was ...etc
• Dec 2nd 2008, 04:23 AM
Leodinas
Could you post your working so that I can see what you mean?
• Dec 3rd 2008, 12:18 PM
Jen1603
i got dy/dx = 4x^3 - 12x

which i simplified to 4x ( x^2 - 3) = 0

so then x values of stationary points are x= o and x = + or - root 3

and this is where i get confused