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

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We see that the domain has no endpoints and can be differentiated at all values of x. Therefore to find the extremes we have to differentiate the function and equate the derivative to 0.

y = x^3 – 3x +2

We use the power theorem to get

y’ = 3x^2 – 3

Equating y’=0

=> 3x^2 – 3 =0

=> 3x^2 = 3

=> x^2 = 1

=> x = 1 or x =-1

Now y’’ = 6x. This is negative at x= -1 and positive at x=1.

**Therefore the local minimum is y (1) = 0 and the local maximum value is at y (-1) = 4.**

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.