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

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.