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# Core 3 differentiation stationary points

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1. Find the co-ordinates of the stationary points on the curve y = (x+3)^2 (2-x)^3

Is this correct??
y' = u'v +v'u

y' = (x +3)^2 - 3(2 - x )^2 + (2-x)^3 2(x+3)

0 = (2-x)^2 (x+3) [-5x + 7]

so stationary points are (0, 72) and (7/5, 13068/3125)
2. (Original post by Custardcream000)
Find the co-ordinates of the stationary points on the curve y = (x+3)^2 (2-x)^3

Is this correct??
y' = u'v +v'u

y' = (x +3)^2 - 3(2x-1)^2 + (2-x)^3 2(x+3)

0 = (2-x)^2 (x+3) [-5x + 7]

so stationary points are (0, 72) and (7/5, 13068/3125)
Check the bold bit. That is not right.
3. (Original post by RDKGames)
Check the bold bit. That is not right.
i've typed it out wrong, i've changed the main post
4. (Original post by Custardcream000)
i've typed it out wrong, i've changed the main post
In that case check the factorisation. There should not be a 7.

Also, where did you get (0,72) as a stationary point when x=0 isn't even a solution to the differential??? Plus there are 3 roots so there should be 3 stationary points, not 2.
5. (Original post by RDKGames)
In that case check the factorisation. There should not be a 7.

Also, where did you get (0,72) as a stationary point when x=0 isn't even a solution to the differential??? Plus there are 3 roots so there should be 3 stationary points, not 2.
so you get x=2
x=-3
x=7/5
6. (Original post by Custardcream000)
so you get x=2
x=-3
x=7/5
Nope. First two are correct but not he last one. Check your factorisation.
7. (Original post by RDKGames)
Nope. First two are correct but not he last one. Check your factorisation.
x= -1
8. (Original post by Custardcream000)
x= -1
thanks for your help

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Updated: September 18, 2016
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