 Hello and welcome to the session. Now we shall discuss second derivative test in this session. This is another test to examine local maxima and local minima of a given function. This test is easier to apply than the first derivative test. Let f be a function defined on an interval i and c belongs to i. Let f be twice differentiable at c. Then we have x 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. Then sc is the local maximum value of f. Next is x 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 fc would be the local minimum value of f and in case if we have f dash c is equal to 0 and f double dash c is equal to 0 then second derivative test fails and in this case we go back to the first derivative test and find whether c is a point of local maxima local minima or a point of inflection. As f is given to be twice differentiable at c we mean that the second order derivative of f exists at c. Let's try and find out the local maxima and local minima of the function given by fx equal to xq minus 6x square plus 9x plus 15. From here we have f dash x equal to 3x square minus 12x plus 9 for critical points we have f dash x equal to 0 that is 3x square minus 12x plus 9 equal to 0 or we can say x minus 3 into x minus 1 is equal to 0 which gives x equal to 1 or 3. Now to check if x equal to 1 or x equal to 3 are points of local maxima or local minima we need to find f double dash x which is equal to 6x minus 12. Now for x equal to 1 we have f dash 1 is equal to 0 then f double dash 1 would be equal to minus 6 which is less than 0. So we have x equal to 1 is the point of local maxima and f of 1 which is equal to 1q minus 6 into 1 square plus 9 into 1 plus 15 which comes out to be equal to 19 is the local maximum value of f. Next let's consider for x equal to 3 we have f dash 3 is equal to 0 then f double dash 3 would be equal to 6 into 3 minus 12 and that is equal to 6 which is greater than 0. So we say that x equal to 3 is point of local minima and now f of 3 equal to 3q minus 6 into 3 square plus 9 into 3 plus 15 which comes out to be equal to 15 is the local minimum value of f. So this is how we use the second derivative test to examine local maxima and local minima of a given function. Now we have very important results regarding absolute maximum and absolute minimum values of a function on a closed interval i. The first result says let f be a continuous function on an interval i equal to closed interval a b then f has the absolute maximum value f attains it at least once in i and so f has the absolute minimum value and attains it at least once in i. Next result says let f be a differentiable function on a closed interval i let c be any interior point of i then we have f dash c equal to 0 if f attains its absolute maximum value at c then f dash c equal to 0 if f attains its absolute minimum value at c. In view of the above results that we have just discussed we have working rule for finding absolute maximum or absolute minimum values of a function in a closed interval a b. Consider the function fx equal to x cube where x belongs to closed interval minus 2 2 we will find the absolute maximum value and the absolute minimum value of this function by using the working rule that we will discuss now. In the first step we find the critical points of the function f in the given interval first let's see what is f dash x this is equal to 3x square now for the critical points we have f dash x equal to 0 that is 3x square equal to 0 which implies that x equal to 0 so x equal to 0 is the only critical point for the given function. Now we shall find the values of f at the end points of the interval that is minus 2 2 and the critical point that is 0 so we have f of 0 would be equal to 0 then f of minus 2 would be equal to minus 2 the whole cube that is equal to minus 8 then f of 2 equal to 2 cube and that is equal to 8. Then we identify the maximum and minimum values of f out of the values calculated in this step that is from these three values we see which is the maximum value so we get that 8 is the maximum value and minus 8 is the minimum value thus we get absolute maximum value of f on the closed interval minus 2 2 is 8 and absolute minimum value of f on the closed interval minus 2 2 is minus 8 so this is how we find the absolute maximum and absolute minimum value of a given function in a closed interval a b. This completes the session hope you have understood the second derivative test and how we find the absolute maximum and absolute minimum values.