 Hi and welcome to the session. Let us discuss the following question. Question says, find the local maxima and local minima if any of the following functions. Find also the local maximum and local minimum values as the case may be. Second part is, function g is given by gx equal to x cube minus 3x. First of all let us understand that if we are given a function f defined on local interval i then let function f be continuous at critical point c in interval i. Then if f dash x is greater than 0 for x less than c and f dash x is less than 0 for x greater than c then c is the point of local maxima and if f dash x is less than 0 for x less than c and f dash x is greater than 0 for x greater than c then c is the point of local minima. This is the key idea to solve the given question. Now let us start with the solution. We are given gx equal to x cube minus 3x. Now differentiating both sides with respect to x again g dash x equal to 3x square minus 3. Now to find maxima and minima points we put g dash x equal to 0. Now this implies 3x square minus 3 is equal to 0. We can say 3 multiplied by x square minus 1 is equal to 0. This implies 3 multiplied by x minus 1 x plus 1 is equal to 0. So we get x is equal to 1 of minus 1. First of all let us discuss for x equal to minus 1. So we can write for x less than minus 1 g dash x is greater than 0 and for x greater than minus 1 g dash x is less than 0. Now g dash x changes its sign from positive to negative as x increases through minus 1. So we get x equal to minus 1 is the point of local maxima. Now we know local maxima value is given by g minus 1 which is equal to minus 1 cube minus 3 multiplied by minus 1 which is further equal to minus 1 plus 3. Now we can say g minus 1 is equal to 2. Now let us discuss for x equal to 1 for x less than 1 g dash x is less than 0 for x greater than 1 g dash x is greater than 0. So g dash x changes its sign from negative to positive as x increases through 1. So therefore we get x is equal to 1 is the point of local minima. Now let us find out local minima value. Local minima value is given by g1. g1 is equal to 1 cube minus 3 multiplied by 1. This is equal to minus 2. We get g1 equal to minus 2. So we can write local minima value is equal to minus 2. Now this is our required answer. This completes the session. Hope you enjoyed the session. Have a nice day.