Find the coordinates of 2 stationary points for each function, determine the nature of the stationary points

1. y= 2 + 4x - 2x^2

2. y= x^3 - 3x

3. y= 2x^3 - 15x^2 + 36x - 4

1. y= 2 + 4x - 2x^2

y' =4-4x =0
4x=4
x=1
y = f(1) = 2+4(1)-2(1)^2 = 4
(1,4) is the stationary point

y'' = -4 < 0 , so y=f(x) has a local maximum at x=1

2. y=x^3 -3x
y' = 3x^2-3 = 0
3(x^2-1)=0
x^2=1
x= 1 and -1
f(1)= 1^3 -3(1) = -2
(1,-2) is a stationary point
f(-1) = (-1)^3 -3(-1) = -1+3 =2
(-1,2) is another stationary point

y'' = 9x
At x=1, y'' = -9 <0, so y=f(x) has a maximum at x=1
At x=-1, y''= -9(-1) = 9 >0, so y=f(x) has a minimum at x=-1


3.
y=2x^3-15x^2+36x-4
y' = 6x^2-30x+36 = 0
x^2-5x+6=0
(x-3)(x-2)=0
x=2, 3
f(2) = 24
(2,24) is a stationary point
f(3) = 23
(3,23) is a stationary point

y'' = 12x-30
At x=2, y'' = 12(2)-30 = -6 < 0, so f(x) has a relative maximum
At x=3, y'' = 12(3)-30 = 6 > 0, so f(x) has a relative minimum

Can someone help me out with the other question i posted

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First, find the derivatives:
1. dy/dx=4-4x
2. dy/dx=3x^2-3
3. dy/dx=6x^2-30x+36
Now, set these derivatives equal to 0 and solve for x (these are your stationary points):
1. 0=4-4x, x=1
2. 0=3x^2-3, x=+1, x=-1
3. 0=6x^2-30x+36, x=2, x=3
Then look to the left and right of these points to determine their nature.
1. x=1 is the maximum point
2. x=1 is the minimum point, x=-1 is the maximum point
3. x=2 is a relative maximum, x=3 is a relative minimum