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Stationary points

Watch this video for an introduction to stationary points, including what they are and how they link to the topic of differentiation.
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In this video, we are going to look at stationary points.
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We’re initially going to look at finding the second derivative. The second derivative is defined as d squared y over dx squared, and it is just finding the derivative with respect to x of the first derivative. We’re going to go for an example of this now. We want to find the second derivative of y is e to the power of negative 2x. So if y is e to the kx, just as a reminder we know that the first derivative is k multiplied by e to the kx. So in our case, k would be negative 2. So the first derivative would be negative 2 lots of e to the power of negative 2x.
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Next, to find the second derivative, we just take the derivative of this first derivative. Using this rule in the purple box again, we have k is negative 2, so we will have negative 2 lots of negative 2 e to the power of negative 2x. And negative 2 multiplied by negative 2 is positive 4, so our second derivative is 4 e to the power of negative 2x. And this is our first example.
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We’re now going to look at finding stationary points. So we have this graph of y equals x squared. The stationary point is found at 0, 0, it looks like in this case, where we can see that the tangent is there and the gradient of the tangent appears to be 0. At the point where the tangent’s gradient is equal to 0, this is a stationary point. And as you can see here, it looks like that stationary point is a minimum of the function. We can find a stationary point by solving when is the first derivative equal to 0. We’re going to look at the graph that we had, and we’re going to solve this using the equation.
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So we want to find the stationary point of y equals x squared. From our definition again, we can find the stationary point by solving when is the first derivative equal to 0. So we can find the first derivative. If y is x squared, then the derivative of y with respect to x is 2x. We want to solve when is that equal to 0. That is equivalent to solving when is 2x equal to 0, which has one solution, which is when x is equal to 0. And, of course, when x is equal to 0, y is x squared, so y is 0 squared, which is 0. So our stationary point in this case is 0, 0.
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And as we can see from the graph, that 0, 0 is where we said the stationary point was.
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We’re now going to look at finding maxima and minima. We can determine if a stationary point is a maximum or a minimum by using the second derivative. If the second derivative is greater than 0, then the point x, y is a local minimum. And instead if the second derivative is less than 0, then x, y is a local maximum. We’re going to look at an example of this now. We’ve got a curve y equals 4x cubed add 2x squared. And we want to first identify any stationary points and then determine whether the stationary points are a minimum or a maximum.
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We’re going to use this rule, which is if y is x to the power of n, then the first derivative of y with respect to x is n multiplied by x to the power of n subtract 1. When we find the first derivative of 4x cubed, n would be 3 in this case, so it would have 3 lots of 4x squared. And when we find the derivative of 2x squared, n is 2, so when we find the first derivative, it’s 2 lots of 2x to the power of 1. That can be simplified to 12x squared add 4x. So we’ve found the first derivative. Now to find any stationary points, we find when does the first derivative equal to 0.
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So we’re solving when is 12x squared plus 4x equal to 0. We can take a factor of 4x out of both of the terms so that we get 4x multiplied by 3x plus 1 is equal to 0. For the first one, when 4x is equal to 0, x is equal to 0 and y is 4x cubed add 2x squared. So if we replace x with 0, then we get that the y-coordinate is 0 when the x-coordinate is 0. And when we solve when 3x plus 1 is equal to 0 instead, we would get x equals negative 1/3, and the y-coordinate– by substituting it back into the formula for y, we get that the y-coordinate is 2/27.
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So we have two stationary points: 0, 0 and negative 1/3, 2/27. The next step is to determine whether the stationary points are a maximum or a minimum. We now have got to find the second derivative. We know the first derivative. We found that in the previous question. So the second derivative will be 2 lots of 12x to the power of 1 plus– and 4x, find the derivative of that, will just get us 4. So for our first coordinate, when x, y is 0, 0, the second derivative when we substitute x equals 0 in will be equal to 4. Using our rules, we see that if the second derivative is greater than 0, then x, y is a local minimum.
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So 0, 0 is a minimum. And for the second stationary point, when x, y is negative 1/3, 2/27, we find the second derivative and substitute the value of x in. So it will be 24 multiplied by negative 1/3 plus 4, which gets us negative 4. And then using our rules, the second derivative is less than 0, so x, y is a local maximum in this case.
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So in the comments below this video, if you would like to discuss, can you think of an example of a function y where there is a point such that the first derivative is equal to 0 and at that point the second derivative is also equal to 0? And in the comments below, from that function can you also say, what does it mean? What is that stationary point? Is it a maximum? Is it a minimum? Or is it something completely different? Thank you for listening.

In this video, Dr Lisa Mott gives an introduction to stationary points, including what they are and how they link to the topic of differentiation.

Discuss in the comments below

  • Can you think of an example of a function y where its first and second derivatives are both zero?
  • What kind of stationary point is this?
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