hi welcome to this chord mask video on stage 3 points in this video we're going to start off by looking at how to use differentiation to find the coordinates of stationary points and then in the second part of the video we're going to look at how to use the second derivative to show whether a turning point or stationary point is a maximum or a minimum so first of all say string points are sometimes called turning points and the points on the graph where the graph turned so to speak so as you can see this parabola has got a turning point or a stationary point at the top here and this would be a maximum and the reason it's a maximum is because the gradient is positive to begin with at this particular point the gradient is zero so dy by dx is equal to zero at this point and then the gradient is n negative so the graph turns at this point or it's constitutionally because the gradient is equal to zero there are maximum turning points called maxima if there's more than one there's minimum turning points called a minima if there's more than one and there's also points of inflection now in this video we're going to focus on maxima and minima i've watched the video on club mavs if you want to learn more about points of inflection so first of all let's have a look at how to find those coordinates of the stationary points and just to remember dy by dx is equal to zero so in other words when you differentiate you know that the gradient function will be equal to zero at the particular turning points of those stationary points so let's have a look at our first example now so first example says find the coordinates of the minimum point off the graph y equals x squared minus six x plus four now i've started off with this one this is quite a nice one we could have done this question by using completing the square but we're going to use differentiation because it's a fantastic approach to finding the coordinates of these points really easily so we've got our y is equal to x squared minus 6x plus 4. so we're going to differentiate to get d y by dx so this gradient function so let's differentiate so bring the power down to x and reduce the power by one so 2x to the power of 1 or just 2x and then differentiate minus 6x so i'll be minus 6 and the differentiating 4 would be 0. so the gradient function is equal to 2x minus 6. and at this particular minimum point here the gradient is equal to 0 so we know that 2x minus 6 is equal to 0 at this point at the minimum point so now let's solve this equation add 6 to both sides 2x equals 6 and dividing by two we get x equals three so the x coordinate of this point is equal to three so we've got three and then something so we just need to find what something will be now obviously we know that x is equal to three so if we substitute that back into our original y equals x squared minus x x plus four we'll find the y coordinate so y equals three squared subtract six times three plus four so whenever we work that out we get three squared is nine six times three is eighteen so minus eighteen plus four so nine take away eighteen would be negative nine or minus nine plus four would be equal to minus five so the y coordinate of this point is equal to minus five so the coordinates of this minimum point are equal to three minus five that's it so to find the coordinates of a minimum point of a graph we can just differentiate it put it equal to zero and then find our x value and then find our y value so let's have a look at our next question now so this time we've got a cubic and we've got two turning points we've got this turning point here which is a maxima or maximum and then we've got this minimum turning point here and we've been asked to find the coordinates of both the stationary points of this graph y equals x cubed minus 3x squared minus 24x so again we're going to differentiate so we're going to get dy by the x and when we differentiate we get bring the 3 down to 3 x squared reducing the pi by 1. subtract bring the two diodes that's going to be 6 x to the power of 1 and then differentiate minus 24 x would be just b minus 24. now we know these two particular points we know the gradient is equal to zero so what we know is that d y by dx is equal to zero so we're going to write zero equals three x squared minus six x minus twenty four so now we've got zero equals three x squared minus six x minus twenty four now what's fantastic is this right hand side can all be divided by three actually both sides of the equation could be divided by three so we would get 0 equals x squared subtract 2x subtract 8. now we can factorize this so 0 equals brackets brackets brackets brackets so x and x and minus 4 and plus two and just checking minus four times two is minus eight and minus two plus fantastic that means that x equals four or x equals negative two so if we look back at the graph here we know that this negative two will be for this first turning point here this maximum so minus two and then we'd have to find the y coordinate and this four would be for the second coordinate this minimum point here so this is b4 and something else okay so let's find the y coordinates of these turning points are these stationary points so let's start off with x equals four so whenever x equals four y will equal four cubed minus three times four squared minus 24 times four and when we do that we get the answer would be equal to negative 80. i might do it on the calculator and the next one x is equal to negative 2 so again we're going to substitute that into our y equals our equation of the curve so we get y equals make sure you put your negative 2 in brackets a negative 2 cubed minus 3 times negative 2 squared minus 24 times negative 2. and when we do that we get the answer of 28. so that means that this maximum point will have the coordinates a negative 228 and the coordinates of this minimum point are 4 negative 80. and that's it so to find the coordinates of these stationary points all we've done is differentiated the equation of the curve got d y by dx put that equal to zero solve that to get the x coordinates and put those back in to get the y coordinates okay now what we're going to do is we're now going to look at how to use the second derivative to show whether a stationary point is a maximum or minimum so for maximum turning points the second derivative will be less than zero and for minimum turning points the second derivative is greater than zero so if we have a look at our last example again so our last example we had finding the coordinates of the stationary points of this this cubic graph so we have the cubic graph and we had our maximum turning point at minus 228 and our minimum turning point at 4 negative 80. so if we differentiate again i get d2y over dx squared we get bring in the two down six x and reduce the power by one one and then differentiate minus six x would be minus six and differentiating minus 24 will be zero so our second derivative our d2y over dx squared is equal to six x subtract six now if we substitute in this value of minus two for this maximum point so for this maximum point here x is equal to minus two and the second derivative if we substitute r minus two into that well six times minus two is minus 12 and take away 6 will be minus 18. so we've got for this maximum point the second derivative is negative and if we have a look at this for maximum points the second derivative should be less than zero so that's correct and for our minimum turning point we have x is equal to four so if we substitute four into this well whenever x is equal to four we get the second derivative is equal to well six times four is twenty-four t equals six is eighteen and that's positive and just checking for minimum turning points the second derivative should be positive and that's correct so let's have a look at a typical question now where we're asked to find the coordinates of the stationary points but also the nature of the stationary points as well so the question says find the coordinates in nature of the stationary points off the graph y equals x subtract x squared subtract x cubed so to find the coordinate association points we're going to differentiate so we're going to find d y by dx so d y by dx and we know that's obviously equal to zero so let's differentiate it and get what that is and put it equal to zero so differentiating one x well i'll just be one differentiate minus x squared we'll bring the two down so it'll be minus two x reduce the pi by one so that'll be minus two x and then differentiate minus uh x cubed so bring the three down so minus three x squared now we know that d y by d x is equal to zero at stationary point so we're gonna write zero equals one minus two x minus three x squared now i'm gonna bring everything over to the left hand side because i wanna make this x squared positive so it's going to be 3x squared plus 2x minus 1 equals zero and i'm not going to factorize it so factorizing would give me putting in my brackets where i've got minus i've got 3x squared plus 2x minus 1 so i'm going to put a 3x and an x in the brackets we know the two numbers are going to times together to be minus 1 but we want a 2x in the middle so i'm going to put the plus 1 there so that would give me 3x and then putting a minus 1 there will be then bring me down to the 2x that we want so then just solving this will give us then that x equals a third or x equals minus 1. so we're going to have a turning point at the point on the graph whenever the x coordinate is equal to a third we're going to have another turning point on our stationary point whenever x is equal to minus 1. so what we're now going to do is we're going to substitute these x values into the equation of the curve to get the y values so whenever x is equal to minus 1 we're going to substitute minus 1 in here and work that out i'm going gonna sort of cheat now i've already worked it out so y is equal to minus one so the coordinate b minus one minus one there's a stationary point there i would also substitute in x is equal to a third into this equation so we'd do a third minus a third squared minus a third cubed and that would give us the other stationary point and that is equal to y will be equal to five twenty-sevenths and that will be that the coordinate will be one third and five twenty-seven so be a stationary point there as well so now we've been asking the question to find the nature of the stationary points so we're going to find the second derivative we're going to differentiate again to get d2y over dx squared so let's do that so d2y over dx squared so differentiating one will be zero differentiating minus two x so to be minus two and differentiating minus three x squared will bring the two downs it'll be minus six x and what we're now gonna do is we're gonna substitute in these x values here and here this x value of minus 1 an x value of a third into our d2y over dx squared defined if they're positive or negative remember if it's positive it's a minimum if it's a negative it's a maximum so let's substitute those in so whenever x is equal to minus one d two y over d x squared would be equal to minus two minus six times minus one will be minus six so we've got minus two minus minus six that's equal to four so we have got at x equals minus one we've got d2y the second derivative is bigger than zero so that means that because it's bigger than zero that's a minimum turning point so we know that minus one minus one is a minimum so i'm just going to write min the same a bit of time there and then for our next coordinate whenever x is equal to one third we're gonna have d2y over dx squared is equal to minus two minus six times a third six times a third will be two because the third of six is two so we've got minus two minus two so it's gonna be minus four so that's negative so we know it's a maximum d2y over dx squared is less than zero so therefore the point one third five twenty sevenths is a maximum is a max that's it so if you wanna find the coordinates of stationary points use the fact that d y by dx is equal to zero to help you and if you want to find if a stationary point is a maximum or a minimum use this point that the second derivative d2y over dx squared is negative for maximum and d2y by dx squared is positive for a minimum that's it