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