Answer :

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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