welcome to this lecture on solutions of exercise 6.4 question 14 to question 18 by dr rajesh singh welcome everyone today we are going to solve exercise for section 6.4 from the book barton and sherbert so in previous videos we have solved from question number one to question number 13. so in this video we will solve from question number 40 to question number 18. let us start with the first question which is question number 14. so in this question we are asked to determine whether the point x equal to 0 is a relative point of extrema of the following functions or not so for this we are going to use the relative extrema theorem which we did this is the theorem that we are going to use so according to this so basically what we are going to do is we will keep on finding the derivatives at x equal to 0 all the derivatives at x equal to 0 provided until we get the first derivative which is non-zero this is what we are going to do so we'll continue to find derivatives unless we get a non-zero derivative at zero so let us start so let us take the first part which is a let us start with part a a is f x equal to x cube plus 2 so basically our aim is to find the smallest n says that fn is not equal to zero so first we will find this and then we will apply this theorem to get our result so f dash x is 6 3 x square so f dash 0 is non-zero so sorry 0 f 0 is 0 so we will continue and we will find the second derivative f double dash x which is 6 x and f double dash 0 is also 0 so we'll find the third derivative f triple dash x is 6 so f triple dash 0 is non-zero so that smallest higher order derivative which is non-zero is f third derivative and because it is odd so f has neither a relative minimum nor negative maximum at x equal to zero so by higher order derivative test f does not have relative extrema at x equal to zero now let us take the part b so b is j x equal to sine x minus x so g x equal to x minus x so again we will find the first derivative which is cos x minus 1 so g dash 0 is so cos 0 is 1 so 1 minus 1 which is equal to 0 so first derivative is 0 now we will find the second derivative minus sine x and g double dash 0 is sine 0 which is 0 so now we will find the third derivative g triple dash x which is equal to minus cos x and so g triple dash 0 is equal to -1 which is non-zero so again in this case g does not have relative extrema at x equal to zero because this is odd so again by higher order negativity test higher order derivative test does not have relative extrema at x equal to zero now let us take the third third part the c part so we are given h x equal to sine x plus one by six x cube so hx equal to sine x plus 1 by 6 x cube so h dash x is cos x plus 3 x square by 6 so h 0 will be so h dash 0 is 1 which is non-zero so again it is non-zero and odd derivative so this by higher order to test h does not have relative extrema at x equal to zero now let us take the fourth part the d part so k x is equal to cos x minus one plus one by two x square kx equal to cos x minus 1 plus 1 by 2 now finally let us find the derivative so k dash x is minus sine x plus x so k dash 0 is minus sine minus 0 plus 0 which is equal to 0 so first derivative is 0 so we will find the second derivative so it is minus cos x plus 1 so k double dash 0 is again 0 so we will find the third derivative so it is third derivative is sine x so k triple dash zero is zero so we'll find the fourth derivative so fourth derivative is cos x so k k 4 0 is equal to 1 which is nonzero so by higher order test and it is positive it is positive which is strictly greater than 0 and this is even so by higher order derivative test k has relative minima at x equal to zero and if you see k 0 is cos 0 minus 1 plus 1 by 2 x square so this is cos 0 is 1 minus 1 so 1 by 2 it is 1 minus 1 which is equal to 0 so cos so 0 is the minimum value 0 is the minimum value of k okay now let us take so that we have completed question number 14 now let us take the question number 15. so in question number 15 we are given a continuous function which is continuous and close interval av and we are assuming that its second derivative exists on the open interval a b we also suppose so suppose that the graph of f and the line segment joining the points a f a and b f b intersects at a point x naught f x naught where a is less than x naught less than b so that there exists a point c belonging to a b says that the double derivative at that point is zero so we have to find a point at which the double derivative of the function is zero so basically if you look at this so it is so we are given some function so we are given some functions so this this is b this you can take it as a i'll make it a little so this you can take it as a this is b this is a f a this is b f b and so this is the line joining a f a and v f b and this is the point so it intersects the curve at the point x naught f x naught so this is the point x naught f x naught so f x naught is here so we have to show that there exists some point in the open interval a b says that the double derivative of the point is equal to zero so what we will do is we will apply because the functions that is we will apply mean variable theorem to now so we will apply mean value theorem two functions on this interval uh a to x naught and x naught to b and then we will use uh the property one of the properties let us see let us start with this proof so because f is given to be continuous on close interval ap and whatever double derivative exists so that means f dash is not also existing so since f is continuous on close interval a b and f dash f is differentiable because it is given the double derivative exists so that means f is also differentiable on open interval a b so by mean value theorem there exists c 1 and c 2 c 1 lying between a a a signal belonging to a to x naught and c 2 belonging to x naught to b such that f dash c 1 is equal to f x naught minus f a upon x naught minus a and f dash c 2 is equal to f b minus f x naught upon x naught minus b and since f x naught minus f a upon x naught minus a and f b minus f x naught upon x naught minus b is or we can say since because if the slope of this line the slope of then slope of line segment joining afa and bfb is given by is equal to so because these are points on the line so we can have this this is also slope of the line and this is also the slope of the line segment joining the points afa b so these two are equal so that implies f dash c 1 is equal to f 2 so this implies f c 1 is equal to f c 2 so applying then f dash is continuous on close interval c 1 to c 2 and f dash is differentiable because double derivative exists c 1 2 c 2 so by also f c 1 is equal to f c 2 so by rows theorem by rows theorem there exists c line between c1 c2 says that f double dash c is equal to zero the derivative of this so that implies since c one c two is contained in a b so this implies there exists c belonging to a b such that f double dash c is equal to zero hence proof so basically what we have done in this question is first we have applied mean value theorem uh to the interval a to x naught and to the interval x naught to a b or to f on the interval to x naught and x naught to be so we got these two derivatives and then we observed that this and this both are slope of this line and because this line is same so both the slopes are equal so the slopes are equal so that means this derivative and this derivative is equal so now we apply row 0 to the function f dash on the interval c 1 2 c 2 to get our desired result so this computes question number 15 now let us move to question number 16. so question number 16 is again a simple question you just start with double derivative applied the limit definition of derivative to get this formula so let us start the solution so we will start with f double dash is simple so we will apply the limit derivative formula so this is f dash a minus f dash a minus h basically change in a upon it so this is the formula for derivative then again apply this formula so now we want so now apply to this so this is limit h tends to zero f of a plus h minus f of a upon h again apply to this so there will be change so we will we can have like this f of a minus f of a minus h so in this case we are adding or let us just move this slightly on this side so that we can write it properly okay so this is f of a minus h upon h now you can take the limit outside and just organize this so you will get c so the f a f a becomes 2 f a so you you just get the desired formula so f of a plus h minus 2 f a plus f of a minus h upon x square and you can take this limit x tends to zero outside and you can club it with this because h is tending to zeros anyway so we get the desired formula okay this completes question number 16. now let us move to question number okay so there is one more thing in question number 16 so we have to give an example where this limit exists but the function does not have a second derivative at a so this is also there in this question number 16 this thing paste it so we have we also have to give an example where this limit exists but the function does not have a derivative second derivative at a so this side this limit exists but we will show that it is not the second derivative does not exist at that point so let us give the example example is the function that we have done in one of our classes so it is the function f x equal to all i have seen this function x square sine 1 by x if x is not equal to 0 and 0 if x is equal to 0 so we know about this function that its derivative exists at every point but the derivative is not continuous and hence cannot be differentiable so we are going to use this that fact so first we observe that this limit exists so we will work at this so a equal to zero so f of a plus h minus 2 f a plus f of a minus h by h is equal to h square so this is equal to f of h minus 2 f 0 plus f of minus h upon h square this is x square sine 1 by h minus 0 plus h square sine minus 1 by h by h square now sine 1 sine minus 1 by h is minus sine 1 by n so these two terms will cancel out each other and you will have 0 by s square which is equal to 0 so when you will take the limit x tends to 0 f of a plus h minus twice of f a plus f of a minus h by h square so you will get limit h tends to zero zero so this is equal to zero hence limit exists now let us show that a double derivative of f at zero does not exist so for that observe that f dash x f 0 when we know that f dash x is so you can easily check it out f dash axis so just differentiate this you will get for x not equal to 0 it will be 2x 2x sine 1 by x and then minus plus x square cos 1 by x into minus x square so it is you will get minus cos 1 by x when x is not equal to 0 and for x equal to 0 you can easily calculate using the limit formula so limit and x tends to 0 f x minus f 0 by x minus 0 this is limit x tends to 0 x square sine 1 by x by x so this is limit x tends to 0 x sine 1 by x which is equal to 0 so f 0 is 0 when x is equal to 0 now this function is not continuous at x equal to 0 it is easy to see so the function f dash is not continuous at x equal to 0 and hence f double dash 0 does not exist so we are able to show that all the limit exists this limit exists but the function is double derivative does not exist so if the double derivative exists then it has to be equal to this limit but this limit can exist without double derivative being into existence so this completes question number 16. now let us move to question number 17 so in question number 17 we are given an open interval i and we are also given that the double derivative is positive on the open interval i and for any point c belonging to i we have to show that the part of the graph of f on i is never below the tangent line to the air graph at c fc so basically what um it means is that our graph is of this form so graph is always of this form so this is the tangent so this is the point cfc so tangent is always below the a graph so what or let me just show you so we this situation is not there basically this situation is not there so we have to show that this situation is there that is what we have to show basically is if you look at this so we have to show that our at point c belonging to i so this is some c this is tangent at c f c so we have to show that this so for any point in this neighborhood in around c so if you pick any point around x x any point x so that means our f x so this is f x and this is the image of x on the line so this this effect should be greater than this point so that is what we have to show so so if this if the equation of this line is gx and if the equation of this line is f x so basically we have to show that f x is greater than equal to g x j x is the equation of the line okay so this is what we have to show so equation of the tangent is equation of tangent to f at c fc is given by so slope point form we will use so cfc is the point and the slope is f dash c so it will be so slope point form so we have y minus fc upon x minus c is equal to f dash c so y is equal to fc plus y is equal to fc plus f dash c x minus c so now to prove this is the equation of the tangent to the curve f at c so we have to basically show that that our f x is greater than equal to y so that means we have to show that f x is greater than equal to this because what we are asked is whereas the part of the graph of f on i is never below the tangent line to the curve at c f c so basically we have to show that our f x is greater than equal to y so now claim f x is greater than equal to f c plus f dash c x minus c for every x in i so this is what we need to show uh to prove this theorem to prove this in this example part of the graph of f is never below the tangent line so this is what we need to show so what we will do is we will apply taylor's theorem for this at c so because it is x minus c and since it is the first derivative so we will take n equal to 1 so that x naught equal to c using x naught equal to c and for any x belonging to i by taylor's theorem for n equal to 1 so we are using taylor's theorem for n equal to 1 so what we will have is f x is equal to fc plus f dash c x minus c plus f double dash c x minus c square by 2 factorial now if you see this that now this double derivative is given to be positive and this is square so this is positive and so this term is positive so which is greater than equal to fc plus f dash c x minus c since f double dash x is c is positive because we are given that derivative is positive okay so this c is different this c is different ac lies between c and x so we can write it as d okay there exists d lying between c and x such that now f dash because we are given the derivative is positive everywhere on i so f double dash t is positive okay so this is given to us given fact so hence our f x is greater than equal to fc hence food so this is question so this completes question number 17 now let us move to question number 18 which is the last question that we are going to do today again this question is a simple application of taylor's theorem so in this question we are given an interval i and c is any point in i we are given two functions f and g defined on i says that fn and gn exist and are continuous on i we are also given that fkc is equal to zero and gkc is equal to zero for k equal to zero to n minus one so that means f f c zero f dash is 0 f double dash is 0 and f n minus 1 c is also 0 and we are also given that g 0 is 0 g dash a 0 and up to g n minus 1 is 0 but we are also given that g and c is not equal to 0 so what we will do is we will apply for stella's theorem to f and taylor's theorem to g and then we will take the limit and use the continuity of fn and gn to get this desired result so let us start so so because we want to use at c so let x not equal to c and for any x belonging to i by taylor's theorem there exists d1 and d2 line between c and x such that f x is equal to f 0 plus f dash 0 sorry fc because x naught is equal to c so fc plus f dash c x minus c plus f double dash c x minus c square by 2 factorial plus so we have we will apply it in n minus 1 so and we are taking up to n minus 1 basically so fn minus 1 c x minus c to the power n minus 1 by n minus 1 factorial and then remainder after n term so r n minus 1 so this is fn c so this will be d so fn d1 x minus c to the power n by n factorial and similarly g x will be g of c plus g dash c x minus c plus so on g n minus 1 c x minus e to the power n minus 1 by n minus 1 factorial and the remainder after n term so this is g n d2 so these two are different a1 d2 are different and they lie between x and c they lie between c and x x minus c to the power n by n factorial so since f0 sorry fc equal to f dash c equal to fn minus 1 c is equal to 0 and g c equal to j dash c equal to this it is given to us g n minus 1 c is equal to 0 so we get f x so all the terms with all these terms are 0 all these terms this is 0 this is 0 all these terms are 0 so you'll be only left with remainder and again here all these terms are zero so we'll be left with remainder so f x is equal to f x is equal to f n d 1 x minus c to the power n by n factorial and g x is equal to fn d2 x minus c to the power n by n factorial now because d1 and d2 lie between x and c and we need to find limit extends to c so that means our d1 and e2 are very close to c so since gnc is not equal to zero so we can assume that our g and a2 is also not equal to zero or what we can do is we can find the limit so limit x tends to c f x is equal to we have to find we cannot find the limit first otherwise it will become zero so divide f x so sorry this is g and d two now gnc is not equal to 0 and gn is continuous on i now glc is not equal to 0 and g n is continuous and i so there exists delta neighborhood v delta of c such that g n x is not equal to zero for all x belonging to e delta neighborhood of c so let so for any now so what we will do is so for x belonging to v delta of c we take x to be there f x by t x f x by j s because if x belongs to the delta of c then this t 2 will also belong to x direct of c because we delt of c because t 2 lies between c and x so then we can divide because now we are ensured that this is not equal to 0 so we can divide so we can divide these two terms so this will be fn a1 upon x minus c by n by n factorial upon g n t to x minus c n by n factorial x belonging to v delta of c implies d 2 belongs to v delta of c and this implies g and d2 is not equal to zero so then we can divide these two so after dividing these two gets cancelled out so we are left with f x by g x equal to fn d1 by gn d2 now we can we will take the limits limit x tends to c f x by g x is equal to limit x tends to c f and d one by g n two now because these two are continuous functions so we can take the limit inside so so limit x tends to c f x by g x is equal to limit or we can write like this fn limit x tends to c d one and g n limit x tends to c d2 now since d1 d2 lies between x and c so limit so x tends to c implies so x tends to c implies d 1 tends to c and d 2 also tends to see so using this they get limit x tends to c f x by g x is equal to so this will be fnc and this will be g and c so hence proved so this much is there in this video thank you everyone have a good day thank you for watching this video to stay updated please subscribe to our youtube channel cosmos learning happy learning through cosmos learning to watch more click on any of these cards thank you once again