Recent content by pip_beard

Then sub in -1 and i get dy/dx=0.. but how do i find the second stat point?>?

Homework Statement A curve is defined for x<-2 by the equation y=((x^2)+3)sqr(x+2) a) Show that dy/dx=0 when x=-1 and find the x-coordinate of the other stationary point. b) Find the value of d^2y/dx^2 when x=-1 hence determine whether the turning point is max or min. Homework...

ok. i'll have a play and post back if i can't do it.. Thanks for the inspiration!

ohhh... did sin^-1... so ... x principle = 0.479425 so pi - 0.4879425 = 2.66216

principle value of -0.523 (but out of range) the pi-(-0.523) = 3.6651 therfore value = 3.6651 and 6.2832?

-2sinx=1 sinx=-1/2

how to i rearrange this to work it out?

so do i differentiate it and make it = 0? so i get -2cosx=0

Sorry. the qu reads 'the curve y=1-2sinx has domain 0<x<pi. find the gradients of the curve at the points where the curve crosses the x-axis

because stationary points lie on the x axis??

so therefore: dy/dx=cosx+sinx. stationary points when diff = 0 so cosx+sinx=0 where do i go from here??

Homework Statement gradient of the stationary points of y=1-2sinx domain 0<x<2piHomework Equations The Attempt at a Solution dy/dx = -2cosx -2cosx=0...?

ive got the answers as.. (-pi/4,-rt2) and (3pi/4, rt2) Are the answers wrong? I don't know how they got these?? because surly the x co-ordinate is 0?

Homework Statement Solve the stationary points of y=-sinx+cosx for domain -pi<x<pi Homework Equations The Attempt at a Solution Differentiate: d/dx=cosx+sinx But how do i solve?