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


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

If the sum of theMathematics - Board Papers

A metal box with Mathematics - Board Papers

Show that aRD Sharma - Volume 1

Find the local maMathematics - Board Papers

Prove that the seMathematics - Board Papers

Prove that the raMathematics - Board Papers

Prove that the leMathematics - Board Papers