 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. Seventh part is gx is equal to 1 upon x square plus 2. First of all let us understand that if we are given a function f defined on interval i, it belongs to interval i such that f double dash c exists. f dash c is equal to 0 and f double dash c is less than 0. Then x is equal to c is a point of local maxima. f dash c is equal to 0 and f double dash c is greater than 0. Then x is equal to c is a point of local minima. fc is the local maximum value of function f and here fc is the local minimum value of function f. This is the key idea to solve the given question. Let us now start with the solution. We are given gx is equal to 1 upon x square plus 2. Now differentiating both the sides with respect to x we get g dash x is equal to minus 2x upon x square plus 2 whole square. We have applied quotient rule to find the derivative of this term. Now we will find out all the critical values of x or we can say we will find all the points at which g dash x is equal to 0. So we will put g dash x is equal to 0. Now this implies minus 2x upon x square plus 2 whole square is equal to 0. Now multiplying both the sides by x square plus 2 whole square we get minus 2x is equal to 0. Biding both the sides by minus 2 we get x is equal to 0. Now to find the value of g double dash x at x is equal to 0. First of all we will find out g double dash x we know g dash x is equal to minus 2x upon x square plus 2 whole square. This we have already shown above. Differentiating both the sides with respect to x again we get g double dash x is equal to x square plus 2 whole square multiplied by minus 2 minus minus 2x multiplied by 2 multiplied by x square plus 2 multiplied by 2x x square plus 2 raised to the power 4. We can find derivative of this term by applying quotient rule. Here we have applied quotient rule. Now simplifying we get g double dash x is equal to minus 2 multiplied by x square plus 2 whole square plus 8x square multiplied by x square plus 2 x square plus 2 whole raised to the power 4. Now this implies g double dash x is equal to clearly we can see in these two terms x square plus 2 is common. So taking x square plus 2 common we get x square plus 2 multiplied by minus 2x square minus 4 plus 8x square upon x square plus 2 whole raised to the power 4. Now simplifying further we get g double dash x is equal to we know 8x square minus 2x square is equal to 6x square 6x square minus 4 upon x square plus 2 whole cube. We know this x square plus 2 will get cancelled by x square plus 2 and we will get x square plus 2 whole cube in that denominator. Now we will find value of g double dash x at x is equal to 0. So we can write g double dash 0 is equal to 6 multiplied by 0 square minus 4 upon 0 square plus 2 whole cube. Now this is equal to minus 4 upon 8 simplifying we get minus 1 upon 2 we know 4 one is 4 and 4 two is 8. So we can cancel common factor 4 from both numerator and denominator and we get g double dash 0 is equal to minus 1 upon 2. Now from above discussion we get at x is equal to 0 g dash 0 is equal to 0 and g double dash 0 is equal to minus 1 upon 2 which is less than 0. So this implies equal to 0 is a point of local maxima and we know local maximum value is given by g 0, g 0 is equal to 1 upon 0 square plus 2 which is further equal to 1 upon 2. So clearly we can see x is equal to 0 is a point of local maxima and local maximum value is equal to 1 upon 2. So our required answer is local maximum occurs at x is equal to 0 and local maximum value is equal to 1 upon 2. This completes the session. Hope you understood the session. Take care and have a nice day.