# Question Video: Finding the Local Maximum and Minimum Values of a Polynomial Function Mathematics • Higher Education

Find, if any, the points (𝑥, 𝑦) where 𝑦 = 𝑥³ + 3𝑥³ − 16 has a local maximum or local minimum.

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### Video Transcript

Find, if any, the points 𝑥, 𝑦, where 𝑦 equals 𝑥 cubed plus three 𝑥 squared minus 16 has a local maximum or local minimum.

If we think about local maximum or local minimum points, we often think of a shape like this, which I’ve drawn in the sketch. And the key thing that they both have in common is that at these points both the maximum and the minimum will have a slope or 𝑚 which is equal to zero. And it’s this that’s gonna help us to solve the problem because it tells us what we need to do first. Because in order to actually deal with the slope, we want to first find the slope function.

And we find that slope function by differentiating the function that we have. So we’re gonna differentiate to find 𝑑𝑦 𝑑𝑥. So if we differentiate our function which is 𝑦 is equal to 𝑥 cubed plus three 𝑥 squared minus 16, we’re gonna get three 𝑥 squared plus six 𝑥. And to remind us how we did that, I’m just gonna take a look at the first term. Well, what we do is we actually multiply the exponent by the coefficient — so three multiplied by one — and then we reduce the exponent by one — so three subtract one. So we get three 𝑥 squared.

So that’s how you actually carry up differentiation. Okay, great, so now that we have the slope function. We can actually use a bit of information that we mentioned earlier that actually at the maximum and minimum points, this is gonna be equal to zero. And therefore, we can actually set it equal to zero to find the 𝑥-coordinate of these points. So we get zero is equal to three 𝑥 squared plus six 𝑥.

And now the first stage that we’re gonna do here is actually divide both sides by three. And we actually do that just to actually make it easier to solve. So we’ve done that and we’re left with zero is equal to 𝑥 squared plus two 𝑥. And now to actually solve and find 𝑥, what we actually do is we factor our expression. And as 𝑥 is a factor of both of our terms, we’re actually gonna take that outside the parentheses. So we get zero is equal to 𝑥 and then inside the parentheses we have 𝑥 plus two. And that’s because 𝑥 multiplied by 𝑥 gives us 𝑥 squared and 𝑥 multiplied by two gives us two 𝑥.

Okay, great, we factored it. Now, let’s find the values of 𝑥 which are actually the solutions to this equation. So we can therefore say that 𝑥 is equal to zero or negative two. So just to remind ourselves why, well, if we substitute 𝑥 is equal to zero into our equation, we’d have zero multiplied by zero plus two which is just equal to zero. So that will be one of our solutions. And if we had 𝑥 equal to negative two, then we’d have negative two multiplied by negative two plus two. Well, negative two plus two is just zero. So that will also be equal to zero.

So great, we’ve now found our solutions. Well, these are actually our 𝑥-coordinates. However, if we look back at the question, we want the points 𝑥, 𝑦. So therefore, we also want the 𝑦-coordinates. So we’re now going to find 𝑦-coordinates by substituting our values of 𝑥 back into our original function.

So first of all, if we substitute 𝑥 is equal to zero, so we get the function when 𝑥 is equal to zero. We’re gonna get zero cubed plus three multiplied by zero squared minus 16, which will therefore give us our 𝑦-coordinate of negative 16. So now, we’ll move on to when 𝑥 is equal to negative two. And if we do that, we actually get negative two cubed plus three multiplied by negative two squared minus 16, which is equal to negative eight because negative two cubed is negative eight plus 12 and then minus 16, which therefore gives us a 𝑦-coordinate of negative 12.

So now that we’ve actually found the 𝑥- and 𝑦-coordinates of our maximum and minimum points, we now need to determine which one is a maximum and which one is a minimum. And in order to do that, what we’re actually gonna look at is the second differential. And we do that because the second differential actually helps us to see the concavity of actual different parts of our function.

We can see actually from my sketch that when we have a concave up part of our function, then actually our second differential is going to be positive. And the reason why that’s useful in this question is because actually at that point we’re gonna find our local minimums. However, at the point when the function is actually concave down, we actually find that our second differential is going to be negative. And the reason this is useful is because this is actually going to be where our maximum point lies.

So therefore, we can see that we can actually use the second differential to determine whether the points that we found are gonna be maximum or minimum points. So in order to actually find our second differential, what we actually do is we differentiate our slope function — so our original 𝑑𝑦 𝑑𝑥. And if we do that, we actually get six 𝑥 plus six. And that’s because we actually differentiate in the same way we differentiate any expression. Because if we look at three 𝑥 squared, well, again, we multiply two — the exponent — by three, which is the coefficient. So we get six and then we reduce the exponent by one. So we’re left with 𝑥.

Okay, so great, we’ve got the second differential is equal to six 𝑥 plus six. Okay, so we’re now going to actually substitute our 𝑥-values into the second differential. And what this means is we’re going to be able to find out the concavity of that part of the function and therefore if it’s a local maximum or local minimum.

So if we start with 𝑥 equals zero, then the second differential is going to be equal to six multiplied by zero plus six, which gives us an answer of six. And this is actually positive. So therefore, we can say it’s going to be concave up at this part of the function. So therefore, we know that it’s going to be a minimum point.

So now, we can move on to 𝑥 is equal to negative two and we can actually substitute that into our second differential. So we’re gonna have six multiplied by negative two plus six which gives us an answer of negative six. So as it’s actually a negative, we therefore know that this is going to be concave down at this point. So it’s actually gonna be a maximum point.

So therefore, we can say that the function 𝑦 equals 𝑥 cubed plus three 𝑥 squared minus 16 has a local minimum point at zero, negative 16 and a local maximum point at negative two, negative 12.