Fnd the values of x where the function y=x^3-3x^2+2 reaches a maximum and minimum, values of the function. Sketch the function.

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You should know that the function rea = 0ches its maximum or minimum at values of x that represent the zeroes of f'(x).

You need to solve for x the equation f'(x) = 0 such that:

(x^3-3x^2+2)' = 0 => 3x^2 - 6x = 0 => x^2 - 2x =...

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You should know that the function rea = 0ches its maximum or minimum at values of x that represent the zeroes of f'(x).

You need to solve for x the equation f'(x) = 0 such that:

(x^3-3x^2+2)' = 0 => 3x^2 - 6x = 0 => x^2 - 2x = 0

Factoring out x yields:

x(x-2) = 0 => x = 0 and x-2 = 0 => x = 2

You should know that the expression x^2 - 2x is negative if x in (0,2) and it is positive if x in (-oo,0) or (2,oo).

Hence, the function reaches its maximum at x = 0 and it reaches its minimum at x = 2 as the graph below shows:

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