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darn forgot the points (3,4)
hmmm i forgot theres also a part b to the question find the x co-ordinate of the point at which the tangent is parallel to the tangent at the point.
sweet thx so much
so x=2 is min and x=0 is max?
yep i understand why they a max and min but i dont get what to do with the -6 and 6
ops i mean min i get x=2 y=8.77 x 10^-24
x = 0 and x = 2?
ah ok
so its meant to be 6x-6?
(x+1)2(x-2) = (x+1)(2x-4)
o thought so, how about this?
i expanded this first (x +1) (2x-4) = 2x^2 - 4x + 2x -4
then
(x - 2)(x-2) = x^2 + 4
so 2x^2 - 2x - 4 + x^2 + 4 = 3x^2 -2x
f'(X) = 6x -2
something like this?
d(x+1)(x-2)^2/dx = (x+1)2(x-2) + (x-2)^2
Hi all
i am stuck on this question
Find the x co-ordinates of the points on the curve y= (x+1)(x-2)^2 at which the gradient is zero. test whether the points are local maxima or minima.