 Hello and welcome to the session. In this session, we will discuss maxima and minima. Here we will see how we use the concept of derivatives to calculate the maximum and minimum values of various functions. Let f be a function defined on interval i. Then we have f is said to have a maximum value in i. If there exist a point c in i such that fc is greater than equal to fx for all x belongs to i. And this number fc is the maximum value of f in i and the point c is called the point of maximum value of f in i. Then f is said to have minimum value in i. If there exist a point c in i such that fc is less than equal to fx for all x belongs to i. And in this case the number fc is called the minimum value of f in i and the point c is the point of minimum value of f in i. Next is f is said to have an extreme value in i if there exist a point c in i such that fc is either a maximum value or minimum value of f in i. And in this case the number fc is called extreme value of f in i and the point c is the extreme point. Let's try and find the maximum and minimum value of the function given by fx equal to 2x minus 1 the whole square plus 3. Now we know that the perfect square cannot be negative so we have 2x minus 1 the whole square is greater than equal to 0. Now we add 3 on both the sides we get 2x minus 1 the whole square plus 3 is greater than equal to 0 plus 3. That is we get 2x minus 1 the whole square plus 3 is greater than equal to 3. Now this is fx so we get fx is greater than equal to 3 or we can say that 3 is less than equal to fx and we know when fc is less than equal to fx then fc is the minimum value of f in i. So in this case fc is the number 3 so we have 3 is the minimum value of the function fx. Next we have let f be a real valued function let c be an interior point in domain of f. Then we have c is called the point of local maxima if there is an edge greater than 0 such that fc is greater than equal to fx for all x belongs to open interval c minus h comma c plus h. And the value fc is called the local maximum value of f then point c is called the point of local minima if there is an edge greater than 0 such that we have fc is less than equal to fx for all x belongs to open interval c minus h c plus h then the value fc is called the local minimum value of f. The above discussion leads us to a very important result it says let f be a function defined on an open interval i suppose c belongs to i be any point if f has a local maxima a local minima at x equal to c then either f dash c is equal to 0 or f is not differentiable at c and the converse of this result need not be true that is a point at which the derivative vanishes need not be a point of local maxima or local minima. So we have a point c in the domain of a function f at which either f dash c is equal to 0 or f is not differentiable is called a critical point of f. Now we shall give a working rule for finding points of local maxima or points of local minima using only the first order derivatives. So let's discuss the first derivative test. Let f be a function defined on an open interval i let f be continuous at a critical point c in i then we have if f dash x changes sign from positive to positive. To negative as x increases through c then the point c is called the point of local maxima. Next we have if f dash x changes sign from negative to positive x increases through c then we have the point c is called the point of local minima. Next is if f dash x does not change sign x increases through c then neither c is a point of local maxima nor a point of local minima such point c would be the point of inflection. One more important thing is that if c is a point of local maxima of f then fc is the local maxima value of f. And if c is the point of local minima of f then fc is the local minima value of f. Let's try and find the local maxima or local minima of the function given by fx equal to 1 upon x square plus 2. First let's see what is f dash x this is equal to minus 2x upon x square plus 2 the whole square. And for local maxima or minima we have f dash x equal to 0 that is minus 2x upon x square plus 2 the whole square is equal to 0 which implies that x is equal to 0. So x equal to 0 is the only critical point which could possibly be the point of local maxima or local minima of f. Now when we take x negative that is x less than 0 then we have f dash x to be positive. And when we take x to be positive that is greater than 0 then f dash x comes out to be negative that is less than 0. As you can see that f dash x changes sign from positive negative. So we get x equal to 0 is the point of local maxima and local maximum value is given by f of 0 which comes out to be equal to 1 upon 0 square plus 2 and that is equal to 1 upon 2. So this is the local maximum value of the function f. So this is how we use the first derivative test to examine the points of local maxima and local minima of a given function. This completes the session hope you have understood maxima and minima, local maxima and local minima and the first derivative test.