# Question Video: Finding Where the Local Maximum and Minimum Values for a Polynomial Function Occur

Determine where the local maxima and minima are for 𝑓(𝑥) = 𝑥⁴/4 − 2𝑥² + 5.

06:23

### Video Transcript

Determine where the local maxima and minima are for 𝑓 of 𝑥 is equal to 𝑥 to the power of four over four minus two 𝑥 squared plus five.

So the first thing we need to do is understand what local maxima and local minima are. So I’ve drawn here a sketch. And on this sketch, we’ve got three points that I’ve noticed. So we’ve got the local maxima and the local minima, so two local minima and a local maxima. And as you can see here, these are the points on our function where the slope is equal to zero. So, therefore, if we know that at these points the slope is gonna be equal to zero, the first thing we need to do is find the slope function of our function. And we find that slope function by differentiating our function. So we’re gonna differentiate the function 𝑓 of 𝑥 is equal to 𝑥 power of four over four minus two 𝑥 squared plus five.

Well, when we differentiate, the first term is going to be 𝑥 cubed. We get that just to remind ourselves of how we differentiate by multiplying the exponent by the coefficient. So in this case, it’d be four by a quarter cause it’s 𝑥 to the power of four on four. That gives us one. So we’ve got single 𝑥. And then we reduce the exponent by one. So four minus one gives us three. So we get 𝑥 cubed. And then when you get minus four 𝑥 and that’s because, again, we had two multiplied by two which gives us four. And then we’ve reduced two by one, so we get one. So we just get 𝑥 on its own or 𝑥 to the power of one. And then if we differentiate positive five, we just get zero because if we differentiate a constant, we get zero.

So now, what we’re gonna do is use the information that we had from earlier. That was that the slope is gonna be equal to zero at our local maxima and local minima points. So, therefore, we’re gonna set our slope function equal to zero. And the reason we’ve done that is that we can find the 𝑥-values where our local maxima and minima points are. So we’ve got 𝑥 cubed minus four 𝑥 equals zero. So what we’re gonna do now is solve to find our values of 𝑥.

So the first thing we’re gonna do is factor. We can factor because there’s an 𝑥 term in both of our terms. So we can take 𝑥 outside of the parentheses. So then we have 𝑥 multiplied by 𝑥 squared minus four is equal to zero. So if we take a look at this, we can see that 𝑥 squared minus four is, in fact, the difference of two squares. So, therefore, we know how to factor this. So when we do that, we get 𝑥 multiplied by 𝑥 plus two multiplied by 𝑥 minus two is equal to zero.

So what we’ve got here are three values that are all multiplied together. So 𝑥, 𝑥 plus two, and 𝑥 minus two equal zero. And when we multiply these together, like we said, it equals zero. So, therefore, at least one of these values must be equal to zero. So either 𝑥, 𝑥 plus two, or 𝑥 minus two must be equal to zero. So, therefore, our first possible value of 𝑥 is 𝑥 is equal to zero because that’ll be zero multiplied by the rest. So it’ll give us a result of zero. Then next, we’ve got 𝑥 plus two is equal to zero. So we know that this value must be equal to zero. So in order to make that happen, what’s 𝑥 gonna be?

Well, 𝑥 must be negative two. Then finally, we could have 𝑥 minus two is equal to zero. And what would 𝑥 be to make this true? So, therefore, the other possible value of 𝑥 is positive two. So now, what we have are three possible values of 𝑥 at 𝑥 equal zero, 𝑥 equals negative two, and 𝑥 equals two. So these are the three values of 𝑥 where we have our local maxima and minima points. So now, what we want to do is we want to work out whether they are, in fact, maxima or minima points.

In order to do that, what we’re gonna do is use the second derivative test. So first of all, we need to differentiate the first derivative, so differentiate 𝑥 cubed minus four 𝑥. So when we do that, we get three 𝑥 squared minus four, just using the same method as before. So for the first term, it would be three multiplied by one because it’s the exponent multiplied by the coefficient. And then we will reduce the exponent by one. So great, we know what the second derivative is.

So now, what do we do? Well, we now substitute in our 𝑥-values. So 𝑥 is equal to zero. 𝑥 is equal to negative two. And 𝑥 is equal to two. And that’s because our second derivative test tells us that when we substitute in our 𝑥-values, if the second derivative is greater than zero, then it’s going to be a local minima. However, if the second derivative is less than zero, it’s going to be a local maxima. So we are gonna start off with 𝑥 equals zero.

So when we do that, we get three multiplied by zero squared minus four. So, therefore, we get a value of the second derivative of negative four because this is less than zero. Therefore, it’s going to be a local maxima. And then if we substitute in 𝑥 equals negative two, we’re gonna find the second derivative is equal to three multiplied by negative two all squared minus four. This will give us a result of eight. And, therefore, as this is greater than zero, this is gonna be a local minima.

So now, when we substitute in 𝑥 equals two, we’re gonna expect the same result as when we substituted in 𝑥 equals negative two. And that’s because we squared the negative two. And we’re gonna square the two. And if you square each of these, you get positive four. So let’s have a look. The second derivative is gonna be equal to three multiplied by two squared minus four which gives us eight. So, therefore, this is gonna be a local minima as well. So, therefore, we can say that the function 𝑓 of 𝑥 is equal to 𝑥 to the power of four over four minus two 𝑥 squared plus five has local minima points at 𝑥 equals negative two and two and a local maxima at 𝑥 equals zero.

And what we can do to double-check this is take a little look at our sketch because our sketch was a sketch of a graph with a coefficient of 𝑥 of four. And it’s a positive 𝑥 to the power of four graph. And we can see that there we’ve got two local minima and one local maxima. And that’s what we found when we found our local minima and local maxima for our function.