Q. 3 B

# Find the local ma

Answer :

g(x) = x3 – 3x

g’(x) = 3x2 - 3

Now, g’(x) = 0

3x2 - 3 = 0

3x2 = 3

x = � 1

g’’(x) = 6x

Now, g’(1) = 6>0

and g’(-1) = -6 < 0

Then, by second derivative test,

x = 1 is point of local maxima and local minima of g at x = 1 is

g(1) = 13 – 3 = 1-3 =-2

And,

x = -1 is point of local maxima and local maximum value of g at x = -1 is

g(-1) = (-1)3 – 3(-1) = -1+3 = 2

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