 Hello 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 local maximum and the local minimum values, as the case may be. Sixth part is, function g is defined by gx is equal to x upon 2 plus 2 upon x, and x is greater than 0. First of all, let us understand that if we are given a function f defined on interval i, c belongs to interval i, such that f double dash c exists. Equal to c is a point of local maxima, f dash c is equal to 0, f double dash c is less than 0. This is equal to c is a point of local minima, if f dash c is equal to 0 and f double dash c is greater than 0. In this case, value of fc is the local maximum value of function f, fc is the local minimum value of function f. This is the key idea to solve the given question. Now, let us start with the solution. We are given gx is equal to x upon 2 plus 2 upon x, where x is greater than 0. Now, differentiating both the sides with respect to x, we get g dash x is equal to 1 upon 2 minus 2 upon x square. We can differentiate this term by applying quotient rule. And this term also by applying quotient rule, now we will find out all the values of x, where g dash x is equal to 0. So we can write, g dash x is equal to 0 implies 1 upon 2 minus 2 upon x square is equal to 0. Now, adding 2 upon x square on both the sides, we get 1 upon 2 is equal to 2 upon x square. Now, multiplying both the sides by 2x square, we get x square is equal to 4. Now, taking square root on both the sides, we get x is equal to plus minus 2. Now we know x is greater than 0. So we will make it the value of x is equal to minus 2. And we get x is equal to, now we know g dash x is equal to 1 upon 2 minus 2 upon x square. This we have already shown above. Now we will find out g double dash x by differentiating both the sides of this expression with respect to x. Differentiating both the sides of this expression with respect to x, we get g double dash x is equal to 0 minus 2 multiplied by minus 2 upon x cube. Simplifying, we get g double dash x is equal to 4 upon x cube. We know derivative of constant term is 0. And we can find the derivative of this term by quotient rule. On simplifying, we get g double dash x is equal to 4 upon x cube. Now let us find out value of g double dash x at x is equal to 2. So g double dash 2 is equal to 4 upon 2 cube, which is further equal to 1 upon 2. Clearly we can see this value is greater than 0. Now we get at x is equal to 2, g dash 2 is equal to 0. And g double dash 2 is greater than 0. So we get x is equal to 2 is the point of local minima. And we know local minimum value is given by g2. g2 is equal to 2 upon 2 plus 2 upon 2. Now simplifying further, we get g2 is equal to 1 plus 1, which is further equal to 2. So we get local minimum value is equal to, so our required answer is local minimum occurs at x is equal to 2 and local minimum value of the function is equal to 2. This completes the session. Hope you understood the session. Take care and have a nice day.