*Question: find and classify all the critical points of the function f(x)=2x^3+3x^2-12x+2.*

f = 2x^3 + 3x^2 - 12x + 2

differentiating,

f' = 6x^2 + 6x - 12

Critical points, or stationary points, are those points on the curve of f(x) where the slope is zero, i.e. the derivative of f(x) is zero.

Setting f'(x) = 0,

6x^2 + 6x - 12 = 0

x^2 + x - 2 = 0

(x - 1)(x + 2) = 0

x = 1, x = -2

2nd derivative, f''(x)

f''(x) = 12x + 6

at x = 1

f''(1) = 12 + 6 = 18 > 0

Since f'' > 0,then the turning point is a minimum.

at x = -2

f''(-2) = -24 + 6 = -18 < 0

Since f'' < 0,then the turning point is a maximum.

f(x) = 2x^3 + 3x^2 - 12x + 2

at x = 1, f(1) = 2(1) + 3(1) - 12(1) + 2 = 2 + 3 - 12 + 2 = -5

at x = -2, f(-2) = 2(-8) + 3(4) - 12(-2) + 2 = -16 + 12 + 24 + 2 = 22

**The critical points are: (1, -5) min, (-2, 22) max**