# Find the po

We have, g (x) = x3 – 3x

Differentiate w.r.t x then we get,

g’ (x) = 3x2 – 3

Now, g‘(x) =0

= 3x2 = 3 x = ±1

Again differentiate g’(x) = 3x2 – 3

g’’(x)= 6x

g’’(1)= 6 > 0

g’’( – 1)= – 6>0

By second derivative test, x=1 is a point of local minima and local minimum value of g at

x =1 is g(1) = 13 – 3 = 1 – 3 = – 2

However, x = – 1 is a point of local maxima and local maxima value of g at

x = – 1 is g( – 1) = ( – 1)3 – 3( – 1)

= – 1 + 3

= 2

Hence, The value of Minima is – 2 and Maxima is 2.

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