Find all the maxima and minima of the function y = x^3 – 3x +2 for x lying between –inf. and + inf.

y = x^3-3x+2.

The maximum and
minimu of  y is determined by the solution  x =c of  y'(x) = 0, for whicg y"(c) 
< 0 or y"(c) >0.

y'(x) = (x^3-3x+2)' =
0

3x^2-3x = 0

3x(x-1).
Therefore x= 0 or x = 1.

y"(x) = (3x^2-3x)' = 6x
-3.

y"(0) = 9*0-3 = -3 which is negative. Therefore at x=
0. y = (0^3)-(3*0)+2 = 2 is maximum.

y"(1) = 6*1-3 = 3
>0. Therefore y = 1*3 - 3*1+2 = -1 is the minimum.