Stationary Points: Example 3

by Batool Akmal

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    00:01 Let’s look at example 3 now. So we’re being asked to find the coordinates of the stationary point and their nature. We’ll do the whole thing, so we’ll sketch it just because it’s fun and of this graph y = 3x³ - 5. So again, it’s the same method for calculating the stationary points. We can say that at stationary points dy/dx = 0. So dy/dx = 0. This is just our theory. It’s important that you know this.

    00:33 You’re understanding that the derivative at stationary points equal to zero. If we now differentiate this function, dy/dx, bring the power down, decrease the power by one. So that gives me 9x².

    00:47 Now, remember that at stationary points, this equals to zero. So you can now say that 9x² = 0.

    00:54 You take the 9 over. That also gives you zero. So basically, x in our case is zero. The √0 is also zero.

    01:01 So there’s no plus or minuses involved. We’ve got an x coordinate of zero. We can find our y coordinate now.

    01:08 We can say y is going to be 3(0)³ – 5. Remember how to do that. We’re just putting it in the original function.

    01:16 So that gives me a coordinate of -5. Therefore your stationary points, or one stationary point occurs at x = 0 and y = -5. So, if we just do a little graphical interpretation of where we’re going with this, we have x = 0 and y = -5. So let’s just call this point -5 here. I have a stationary point here.

    01:43 We now need to decide what kind of stationary point we are getting. So let’s do the nature of this stationary point. So we’re going to do the second differential for which I need the first differential. So dy/dx = 9x². When we differentiate this the second time, d²y/dx², that gives me 18x. Now, we need to find out what the second differential is at x = 0.

    02:17 D²y/dx² is going to be 18(0) which is 0. So this is a breakthrough. We’ve just discovered that this gives us a second differential of zero. Now think hard. Try and remember what kind of point you get when you equal the second differential to zero. I hope you've remembered that we have a point of inflection here. So we can say here that we have a point of inflection. Now, in order to sketch this, because we don’t really know what kind of point of inflection this is. Remember it could be this or it could be this. In some cases, it could even be something misleading. So it might not be a point of inflection because it could be flat and then have a positive gradient and a negative gradient after, which would make this a maximum point or it could do something like this, which gives us a minimum point. So it could just be that it’s flat for a longer time. So it’s just misleading us for a little while. Now, in order to check which one it is, a point of inflection would look like that.

    03:30 But in order to check what kind of point this is from here on, we now need to use some gradients.

    03:36 I need to find out what this graph is doing. So if this point here is 0, and this is x is 1, and x is -1, I can find out what the graph is doing, what the gradient is doing at these points here. So if the point of inflection is at 0, I can go 1 point further and 1 point backward and find out what the gradient is doing. That will give me some idea of what this graph is doing. If it’s still zero at those points, you can go 2 points further. So you could go to 2 and -2. What you’re trying to do is really investigate this curve. You’re trying to find out what kind of gradients you’re getting a little bit after the stationary point and a little bit before the stationary point to get an idea. So, if we pick in our case, because our stationary point, so remember that the stationary point was at x = 0, let’s go and investigate dy/dx at x = 1 and at x = -1. So I’m just going 1 point forward and 1 point back.

    04:39 If that didn’t work, if you still got 0, you can go 2 points further or 3 points further or 5 points further, whatever’s easier. But 1 and -1 are fairly easy numbers so I’m going to start with those.

    04:51 Remember that I’m putting it into the gradient, not into the original function and not into the second differential either. So if I take my gradient, my gradient dy/dx was 9x². At x = 1, that becomes 9(1)², which is just 9. So at x = 1, I have a positive gradient. So this is a positive value. Let’s just call this greater than zero. This is the gradient again. It’s not the second differential, so keep that clear.

    05:27 At x = -1, you can have 9(-1)² which also gives you 9. So that’s also positive. So that’s nice because we’ve got a nice, easy point of inflection. What we’re saying is that the gradient is positive before the stationary point and the gradient is positive after the stationary point. That looks like this.

    05:48 So remember, positive gradient looks like that. So that’s positive. Negative gradient would look like this.

    05:54 Done two lines just to make sure you understand. But if we have a look at our gradient here, at 1, we have positive gradient. So we want the line to go up. You can extend it. At -1, we have a positive gradient as well. So our graph would look like that. So, at 3x³ – 5 would look like this.

    06:18 For anyone that knows what an x³ graph looks like, just the basic x³ graph, if I just do that here, not here. If I do that further down here, so an x³ graph usually looks like that. That’s sort of what we’ve done. But the -5 at the end of our function has moved it down to -5. So that graph has moved down by -5 units to look like this. I should put my x and y there. So our graph 3x³ – 5 crosses at -5 on the y-axis. It follows the same cubic shape as they generally do.

    06:53 So, a lot of stationary points that we’ve just looked at now. Now, it’s your turn to practice some of these questions. Remember the theory behind stationary points and when they occur.

    07:03 Also remember how you can find out the nature of those. It’s just two sections for each question that you need to be aware of. Then you can obviously draw it to show yourself what the graph looks like.

    07:15 The last question is an application of everything that we’ve looked at. So it’s still a stationary points question but we’re looking at a real life scenario. We usually call this optimization.

    07:26 But I’ll let you try it first and then we’ll get together and go through it together.

    About the Lecture

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

    Included Quiz Questions

    1. 0
    2. 1
    3. 2
    4. 3
    5. 4
    1. (1, 13)
    2. (-1, 13)
    3. (-2 , -14)
    4. (2 , -14)
    5. (-2 , 14)
    1. (0, 0)
    2. (1, 1)
    3. (-1, 1)
    4. (-3, 3)
    5. (3, 3)
    1. If f'(x) = 0, f''(x) = 0 and (f'(x) > 0 on both sides of the point or f'(x) < 0 on both sides of the point) then it is a point of inflection
    2. If f'(x) = 0 and f''(x) > 0 then it is a point of inflection
    3. If f'(x) = 0 and f''(x) < 0 then it is a point of inflection
    4. If f'(x) < 0 and f''(x) < 0 then it is a point of inflection
    5. If f'(x) = 0, f''(x) = 0 and (f'(x) > 0 on the left side of the point and f'(x) < 0 on the right side of the point) then it is a point of inflection

    Author of lecture Stationary Points: Example 3

     Batool Akmal

    Batool Akmal

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