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Stationary Points: Exercise 3

by Batool Akmal
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    00:01 So our examples are getting a little bit more interesting and moving up in difficulty. We can see that this function is longer. We now have an x³, x², x term and a number so this should be interesting to solve. Let’s go straight into it and remember what we’re finding is the stationary points and the nature of stationary points. Eventually, we’ll also just draw it just to see what it looks like.

    00:23 So our graph states y = x³ + 9x² + 27x + 31. Remember what happens at stationary points? So we can say that at stationary points, dy/dx = 0. So we differentiate this, still a straightforward equation to differentiate. So dy/dx, remember for each one, bring the power down, decrease the power by one. So that goes to 3x² + 18x + 27. At stationary points, this equals to 0.

    01:05 We're trying to solve this equation. Now, the easiest thing to do here before we start to factorize might be that we divide all by 3. Because the other side equals to 0, you can divide it by any number if it divides the whole equation. So we’re dividing the entire equation by 3. That then gives me x² + 6x + 9 = 0.

    01:28 We now factorize. The way to factorize something like this, because it’s a quadratic, is using two brackets.

    01:33 You want the first two numbers to multiply together to give you x², which is this and the last two numbers to multiply together to give you 9, but add together to give you 6x. So nice, easy factors here. Have a little think. Two numbers that multiply together to give you 9, add together to give you 6.

    01:52 They must be 3 and 3. If you look at the signs, they should both be +. So we have +3, +3.

    02:00 You can double check this equation. If you multiply those two numbers, you get x². Then you get 3x and another 3x, which adds together to give you 6x. Then these two numbers multiply together to give you 9. So, we can now say that either x + 3 equals to 0 from the first bracket or x + 3 equals to 0 from the second bracket. You’ll notice that both of them are the same. So we just need to solve one.

    02:26 So our answer or the stationary point occurs at x = -3. Let’s, from here, calculate our y values.

    02:36 To find y, we’re just going to substitute -3 into our equation, into our original y equation. So let’s put that all in here, + 31. Do each one of them one bit at a time. So remember any negative number cubed will still be negative. So that gives me -27. Here we have 9(9). So the inside value here 3² gives you 9.

    03:03 So 9(9) will give you 81. 27(-3) so if in doubt, do this on the side. You’re just doing some long multiplication. So 27(-3) should give you -81 as well. Then you have the + 31 at the end.

    03:22 The good thing is that + 81 – 81 cancels out. So you’re just dealing with -27 + 31 which gives you a positive answer of 4. Remember, there’s no shame in stopping and doing some long multiplication or long division on the side. We don’t have calculators here but it’s important that you get to the right answer no matter what method you use. We now say that our coordinates here are x = -3 and y = 4.

    03:53 So, just one stationary point, we have x and y. We’re saying that this stationary point occurs when x = -3 and y = 4, so here. From here, let’s see if we can figure out what kind of stationary point we get.

    04:15 We’re now going to look at the nature. We’ll have to do the second differential. But to do that, let me just get out my first differential first. So dy/dx is 3x² + 18x + 27. So I’m getting that from here before I equaled it to 0 or did anything else to it. We’ll now do our second differential so d²y/dx², which gives you 6x + 18. We now need to work this out at x = -3. So when we put x = -3 here, d²y/dx² gives you 6(-3) + 18 which gives you -18 + 18 to give you 0. So, you think what kind of point is this. If the second differential equals to 0, what kind of point are you expecting to get? It’s not greater than zero. It’s not less than zero. So it’s not a maximum and not a minimum as far as we know yet. But this is a point of inflection. Remember what I said before, a point of inflection could look like positive-flat-positive, negative-flat-negative. But it could also be a misleading kind which goes positive-flat-negative, we’re talking gradients now, or it could be negative-flat-positive.

    05:40 So this here is a maximum and a minimum point. So it could be, in some cases that it’s not actually a point of inflection. It’s a positive or negative, I mean it’s a maximum or minimum point but it looks like a point of inflection to start with. So these definitely need more investigation. Now, the method that I went over earlier is that you can go 1 point back and 1 point further to find out what the gradient is doing. So, you could go here to -2 and you could go here to -4. You can substitute them into your gradient to get an idea of what kind of gradient it is. So this gives you the gradient.

    06:20 You could have positive-positive. The numbers don’t matter. It’s just what kind of gradient, or you could have negative-negative. Like we discussed before, this is positive-negative, which wouldn't give you a point of inflection. It would give you maximum or we have negative-positive.

    06:37 I’ll also tell you another little way that you could do this. But if we stick to the method that we’ve just discussed where we use gradients, we’re just going to substitute the two values in. So we’re going to find dy/dx at x = -2 and at x = -4. Remember, if nothing happens at those points, go further. So rather than -2, go to -1. Rather than -4, go to -5. So let’s put this into our gradient. If we find our gradient here, that’s dy/dx. If we put -2 in first so that gives me 3(-2)² + 18(-2) + 27. That gives me -2² is 4, so I have 12. This will give me -36 + 27, which then gives me – 24 + 27 to give me an answer of positive 3.

    07:40 I have a positive gradient at -2. So around here at -2, I know that it has to go up because it's positive.

    07:49 That’s the only way it can join up. If I start to do 4 as well, so if I do dy/dx, I’ve got 3(-4)² + 18(-4) + 27.

    08:04 Do some long multiplication here if you need to. This gives me 3(16) on the inside.

    08:11 I’ve got 18(-4) which is -72 + 27. If we carry on here, that gives me 48 – 72 + 27 to give me a final answer of 3. So I’ve got a positive-positive gradient. So it’s looking like this. Before at -4, I’ve also got a positive gradient. So that’s what positive gradient looks like. If you join it up it would be like so. So I’ve got a curve that looks like this with a flat in the middle. Remember, if it was a negative gradient, negative gradients look like that. So it would go downwards like so.

    08:56 Now, another way that you could do this in case you didn’t want to use gradients is just take these points, -4 and -2, and substitute them into the original equation here. In that case, you’re finding coordinates.

    09:10 You’re not finding gradients. You’re actually looking for what the coordinate is when you put -2 in.

    09:15 So when you put -2 in, you should get some coordinate which is greater than 4.

    09:20 So you might get another coordinate there. You can actually just get an idea of where this graph is going after. Then you can put -4 in and then that should give you a y value of less than 4.

    09:32 So you’ll get some idea of what the graph looks like. You can just put numbers in to find coordinates or you could just calculate the gradient to find the kind of direction it’s going in. Both will give you an idea of what the graph is doing. That’s what points of inflections are really. They need more investigation.

    09:50 However you choose to investigate you graph is up to you.


    About the Lecture

    The lecture Stationary Points: Exercise 3 by Batool Akmal is from the course Stationary Points.


    Author of lecture Stationary Points: Exercise 3

     Batool Akmal

    Batool Akmal


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