Need the extrema of x^3 - 4x^2 - 3x + 2 over the reals?

f(x) = x³ - 4x² - 3x + 2

f'(x) = 3x² - 8x - 3 = 0

3x² - 9x + x - 3 = 0

3x(x-3) + (x-3) = 0

(3x+1)(x-3) = 0

Extrema exist at

x = -1/3, 3

Those extrema are:

f(-1/3) = -1/27 - 4(1/9) - 3(-1/3) + 2 = (-1 - 12 + 27 + 54)/27 = 68/27

f(3) = 27 - 36 - 9 + 2 = -16

Maximum: (-1/3, 68/27)

Minimum: (3, -16)

The derivative is zero at x = 3 and -(1/3).

 d/dx of x^3 - 4x^2 - 3x + 2 = 3x^2 - 8x + 2 = 0

3x^2 - 8x - 3 = 0

You can use the quadratic formula, or if you're clever, you can multiply by 3.

9x^2 - 24x - 9 = 0

with a little imagination you can see a 27 - 3 combo to get 24

=> (9x + ?) (x - ?)

=> (9x^2 - 27x + 3x - 9)

=> (9x + 3) (x - 3)

One of those has to be zero, which only happens if x is 3 or -1/3

that is just an expression, it has no min or max.