# Stationary Points: Exercise 2

by Batool Akmal

Questions about the lecture
My Notes
• Required.
Learning Material 2
• PDF
DLM Stationary Points Exercise Calculus Akmal.pdf
• PDF
Report mistake
Transcript

00:01 So let’s look at our second example. We are now given a function which says y = x³ + 3x² – 1.

00:12 Remember, it’s part of the same question that is asking us to find stationary points and the nature of the stationary points. So let’s begin. We’ll start with saying that at stationary points, dy/dx = 0.

00:25 At the start, I said that you will do this some many times that you will actually remember that the gradient is zero at stationary points. I hope you’re at the point now where you just know this.

00:35 Let’s differentiate our graphs. So we have dy/dx = 3x² + 6x. Remember that this is the general gradient of this curve. If you are looking for equations of tangents and normals next to the curve at this point, you could just substitute your x values here and work from there. But because we’re looking at stationary points, we have to equal it to zero in order to find out what those points actually are.

01:03 We now solve this equation. So there’s different ways of solving this equation. The most obvious one I can see is to factorize. So we can factorize this equation by taking a common factor.

01:14 They both have a common factor of 3x. So you can take a 3x out, leaving you with x + 2 on the inside.

01:23 Another thing you could have done perhaps is at this point, you could have divided it by 3 to make lives easier for yourselves before you started it but either way, we’ll get the same answer.

01:33 We didn’t divide by 3. We’re just factorizing it by 3x. Now that it’s factorized, you can split it into two options.

01:42 You can say that either 3x = 0 or x + 2 = 0. This will give you an answer of x = 0.

01:51 This gives you an answer of x = -2. So in our solution so far, we know that we have two stationary points.

02:00 We’ll have to find exactly where they are and in order to do that, we’ll have to find the y coordinates.

02:06 So, to find your y, we’re going to substitute x = 0 and x = -2 into the original equation, into y which is here. So if I start doing that, if I put in x = 0 first, I’ll get y = 0³ + 3(0)² – 1.

02:30 Just gives me -1. So my first coordinate is (0, -1). For my second coordinate, I’m going to put x = -2 into my equation. So I have y = (-2)³ + 3(-2)² – 1. Remember any negative number cubed will still be negative.

02:55 So 2³ is 2 * 2 * 2 which gives you -8. We then have + 3(2)², which is 4, – 1. That gives me -8 + 12 – 1.

03:09 So you could do -8 + 12 first. You can do this in any order that you’d like. So that gives you 4 – 1 to give you a point 3. So if we now decide or evaluate what our second coordinate is, that will be (-2, 3). Moving forward, we now know that we have two stationary points.

03:31 So XY, my first stationary point is at x = 0, y = -1. This is -1 here. That’s my first stationary point.

03:42 My second stationary point is when x = -2 and y = 3. So say that’s here. So that point here.

03:52 We have our two stationary points. Not quite sure what this graph looks like but we can find out by calculating the nature. We’ll now do the nature of our stationary points. Remember, to do the nature you need to do the second differential. I’ll start by just getting my dy/dx which is 3x² + 6x, just as it is here. We now do the second differential on this. So d²y/dx², so second differential means just differentiate this again. Bring the power down, 6x + 6. Now we find out what the second differential is doing at the two x values here, so this x value and this x value.

04:37 We can say that at x = 0, we have d²y/dx² = 6(0) + 6 which gives you an answer of 6.

04:48 This gives you an answer greater than zero. Remember, think back at what stationary points are doing.

04:55 What happens when a stationary point is greater than zero and it’s positive? It will look like this which gives you a minimum value. Second point, when d²y/dx² = -2, we can do 6(-2) + 6 which gives you -12 + 6 to give you an answer of -6. This is less than zero. Think through your stationary points. What are all three stationary points doing? Hopefully you remember, when you get a negative answer, you get a point that looks like this which has to be a maximum point.

05:31 Let's put it all together in our graph now. At x = 0 which is here, you had a minimum point. At x = -2, which is here, you had a maximum point. So you can just join this graph together and you’ve got your curve drawn out. So x³ + 3x² – 1 looks like this graph here. We know the gradients.

05:56 We know the points where it curved, so where the gradient equals to zero. So we very accurately represented this graph that was given to us algebraically on to its geometric interpretation.

### About the Lecture

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

### Included Quiz Questions

1. (1, 5) and (3, 1)
2. (5, 5) and (3, 1)
3. (5, 5) and (1, 1)
4. (2, 5) and (3, 0)
5. (0, 5) and (-3, -1)
1. One maximum point and one minimum point
2. Two maximum points
3. Two minimum points
4. Two points of inflections
5. One maximum point and one point of inflection
1. (1, 5)
2. (3, 1)
3. (-3, -1)
4. (-1, -5)
5. (0, 0)
1. (3, 1)
2. (1, 5)
3. (-3, -1)
4. (-1, -5)
5. (0, 0)

### Author of lecture Stationary Points: Exercise 2 ### Customer reviews

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