Stationary Points: Example 2

by Batool Akmal

My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Stationary Points Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake

    00:01 So we did a fairly straightforward example, a curve x² at the origin, (0, 0). Let’s hope that this is a little bit more interesting. So now, let’s look at our second example. We have y = x³ – 3x + 2.

    00:18 Imagine you know nothing about this curve and we want to sketch it. So let’s start with the basics.

    00:24 We’ll talk about the gradients of this curve, where its stationary points are, what kind of stationary points these are, and actually sketch this entire curve. So the first thing you need to remember that at stationary points, dy/dx = 0. Depending what kind of function this is, you use different methods of differentiation. You’ve seen so far that the graphs that we’ve had are fairly straightforward.

    00:52 So it’s not a function of a function, it’s not two functions multiplying with each other or a quotient rule.

    00:58 They’re fairly straightforward questions that we can just differentiate. But remember, keep your minds open to all different types of methods of differentiation. Let’s differentiate this. So we can now say that dy/dx = 3x² – 3. Remember that 2 is a constant so that just goes away. Each time when I’m differentiating, my head is repeating, "Bring the power down. Decrease the power by one." So that’s just what we’re doing every time we differentiate. Also remember that this now needs to equal to zero because if we want to find stationary points, we’re telling our equations. We’re saying, "Calculate what’s happening when the gradient equals to zero." So if we just separate this out, I’ve got 3x² – 3 = 0. Any way of solving this really, so you can leave the 3x² here equals to 3, x² = 3/3 which is 1. So x = √1 and remember that that gives you ±1. So that’s exciting. We’ve got two stationary points. So either x = 1 or x = -1. We now need to know our y coordinates if we really are to sketch this on a graph. So let’s just do that first. Let’s find our y values. We’re going find y when x = 1 and also when x = -1. So we can get the coordinates of the stationary points.

    02:24 Remember, to find y, we just substitute it into the original equation. So for 1, I have y = 1³ – 3(1) + 2.

    02:35 That gives me 1 – 3 + 2. So that’s -2 + 2 which equals to zero. So our first coordinate will be (1, 0).

    02:47 X is 1 and y is 0. The next one, we now substitute -1 in. So I have -1³ - 3(-1) + 2 if I do this calculation here.

    03:00 Any negative number cubed will still be negative. So that gives me -1. That becomes a +3.

    03:07 Then we have +2. That gives me 2 + 2, so my answer is 4. So my second coordinate is (-1, 4).

    03:17 So those are the coordinates of my two stationary points. Let’s just see if we can start to put it together on a graph. So my first stationary point was when x is 1 and y is 0. So we’ve got a stationary point here.

    03:34 Let’s call that (1, 0). So x is 1, y is 0. Our second stationary point is when x is -1. Let’s call that -1 here and y is 4. So if we put 4 there, we’ll say we have another stationary point here. We don’t know what kind of stationary points these are yet but we can find out. So we can find the nature of these stationary points by doing the second differential. Then we’re going to find out what the second differential is doing at x = 1 and x = -1. So to find the nature, we will now look at the second differential. So let’s get our first differential first. Dy/dx, we worked out earlier from here.

    04:16 You’re not equaling it to zero anymore. It’s 3x² – 3. D²y/dx² is just going to be bring the power down, 6x.

    04:29 We now find the second differential at 2 different values. So we’re going to say at x = 1, d²y/dx² is going to be 6 and replace this x with 1, so we get 6. Then when x = -1, d²y/dx² is going to be 6(-1) which gives me -6. So this is a little bit more interesting now. This is greater than zero. So at x = 1, we get a second differential which is positive. It doesn't really matter what number or what value it is, it’s more about whether it’s positive or negative or zero. So because it’s greater than zero, we can say this gives you a minimum point, just write that here. In this case, because we get a negative gradient, so that’s less than zero, it’s going to be a maximum point. So I always tend to draw or sketch my little point first in order to evaluate whether it’s a maximum or minimum. So let’s go to x = 1 here on my graph.

    05:33 I’m saying that that here is a minimum point. I’ve said that this point here is a maximum point.

    05:40 Then you can just join them up together and you’ve got a cubic curve. So what you’ve done here is technically what your calculators and computers are doing but you are just going through the whole process and doing this from scratch. So the beauty about this all is that you can now look at a graph, fairly complicated graph, and you will know exactly what it looks like graphically or you can look at the graphic interpretation of a graph. You can actually find out what the equation of this graph would be, which again is completely useful in any field which has a little bit of mathematics in it.

    About the Lecture

    The lecture Stationary Points: Example 2 by Batool Akmal is from the course Stationary Points.

    Included Quiz Questions

    1. Maximum
    2. Minimum
    3. Point of inflection
    4. Critical point
    5. Saddle point
    1. (-2, -10)
    2. (-10, -2)
    3. (2, 10)
    4. (0, 0)
    5. (-2, 2)
    1. For x = -3/5 maximum and for x = 1/3 minimum
    2. For x = -3/5 minimum and for x = 1/3 maximum
    3. For x = 3/5 minimum and for x = -1/3 maximum
    4. For x = 3/5 maximum and for x = -1/3 minimum
    5. For x = -1/5 maximum and for x = 3 minimum

    Author of lecture Stationary Points: Example 2

     Batool Akmal

    Batool Akmal

    Customer reviews

    5,0 of 5 stars
    5 Stars
    4 Stars
    3 Stars
    2 Stars
    1  Star