 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 the local minimum values as the case may be. 8th part is fx is equal to x multiplied by under root 1 minus x where x is greater than 0. First of all let us understand that if we are given a function f defined on interval i and c belongs to interval i such that f double dash c exists then x is equal to c is the point of local maxima if f dash c is equal to 0 and f double dash c is less than 0. This is equal to c is the point of local minima if f dash c is equal to 0 and f double dash c is greater than 0. In this case maximum value is given by fc and in this case minimum value is equal to fc. This is the key idea to solve the given question. Let us start the solution. We are given fx is equal to x multiplied by square root of 1 minus x where x is greater than 0. Now differentiating both the sides with respect to x we get f dash x is equal to x multiplied by 1 upon 2 multiplied by under root 1 minus x multiplied by minus 1 plus square root of 1 minus x multiplied by 1. To find the derivative of this term we have applied product rule. Simplifying we get f dash x is equal to minus 3x plus 2 upon 2 multiplied by square root of 1 minus x. We will add these two terms by taking their LCM and simplifying we will get this term. Now we will find all the points at which f dash x is equal to 0. So we will put f dash x is equal to 0. Now this implies minus 3x plus 2 upon 2 multiplied by square root of 1 minus x is equal to 0. Now multiplying both the sides by this term we get minus 3x plus 2 is equal to 0. Now this implies minus 3x is equal to minus 2. Subtracting 2 from both the sides we get minus 3x is equal to minus 2. Now dividing both the sides by minus 3 we get x is equal to 2 upon 3. Now we get at x is equal to 2 upon 3 f dash x is equal to 0. Now to find the value of f double dash x at x is equal to 2 upon 3. First of all we will find out f double dash x we know f dash x is equal to minus 3x plus 2 upon 2 multiplied by square root of 1 minus x. This we have already shown above. Now differentiating both the sides with respect to x again we get f double dash x is equal to 2 multiplied by square root of 1 minus x multiplied by minus 3 minus 3x plus 2 multiplied by 2 multiplied by 1 upon 2 2 multiplied by square root of 1 minus x multiplied by minus 1 square of 2 multiplied by square root of 1 minus x and we have applied quotient rule to find the derivative of this term. Now this implies f double dash x is equal to minus 6 multiplied by square root of 1 minus x plus minus 3x plus 2 upon square root of 1 minus x upon 4 multiplied by 1 minus x. Now this implies f double dash x is equal to minus 6 multiplied by 1 minus x plus minus 3x plus 2 upon 4 multiplied by 1 minus x raised to the power 3 upon 2. Now simplifying further we get f double dash x is equal to 3x minus 4 upon 4 multiplied by 1 minus x raised to the power 3 upon 2. Now let us find out value of f double dash x at x is equal to 2 upon 3. So f double dash 2 upon 3 is equal to 3 multiplied by 2 upon 3 minus 4 upon 4 multiplied by 1 minus 2 upon 3 raised to the power 3 upon 2. 3 and 3 will cancel each other and we know 2 minus 4 is equal to minus 2 minus 2 upon 4 multiplied by 3 minus 2 upon 3 raised to the power 3 upon 2. We can subtract these two terms by taking their LCM. Now we will cancel common factor 2 from numerator and denominator both and we get f double dash 2 upon 3 is equal to minus 1 upon 2 multiplied by 1 upon 3 raised to the power 3 upon 2. Now further we can see this value is less than 0. Now we get f dash 2 upon 3 is equal to 0 and f double dash 2 upon 3 is less than 0. Now this implies x is equal to 2 upon 3 is a point of local maximum and local maximum value is equal to f 2 upon 3. f 2 upon 3 is equal to 2 upon 3 multiplied by square root of 1 minus 2 upon 3. This is further equal to 2 upon 3 multiplied by square root of 3 minus 2 upon 3 we get. f 2 upon 3 is equal to 2 upon 3 multiplied by square root of 1 upon 3. Now this can be further written as 2 upon 3 multiplied by square root of 1 upon square root of 3. Now we know square root of 1 is equal to 1 so we can write it as 2 upon 3 root 3 rationalizing we get f 2 upon 3 is equal to 2 root 3 upon 9 so we get local maximum value is equal to 2 root 3 upon 9. So our required answer is local maximum occurs at x is equal to 2 upon 3 and local maximum value is equal to 2 root 3 upon 9. This is our required answer. Please the session hope you understood the solution take care and have a nice day.