# Find the stationary points of cubic curve y=(x-2)(x-1(x+4)?

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### 1 Answer

We'll recall what stationary points are. They are the zeroes of the first derivative of the function and they can be: maximum, minimum or inflection points.

We'll calculate the 1st derivative of the given function using the product rule:

y' = (x-2)'*(x-1)*(x+4) + (x-2)*(x-1)'*(x+4) + (x-2)*(x-1)*(x+4)'

y' = (x-1)*(x+4) + (x-2)*(x+4) + (x-2)*(x-1)

We'll remove the brackets:

y' = x^2 + 3x - 4 + x^2 + 2x - 8 + x^2 - 3x + 2

y' = 3x^2 + 2x - 10

Now, we'll determine the zeroes of the 1st derivative:

3x^2 + 2x - 10 = 0

We'll apply quadratic formula:

x1 = [-2+sqrt(4 + 120)]/6

x1 = (-2+2sqrt31)/6

x1= (-1+sqrt31)/3

x2=(-1-sqrt31)/3

**The function will have a maximum point at x = (-1-sqrt31)/3 and a minimum point at x = (-1+sqrt31)/3.**