### Video Transcript

Use the second derivative test to find, if possible, the local maximum and minimum values of 𝑓 𝑥 is equal to nine 𝑥 to the power of four minus two 𝑥 squared minus five.

Before we use the second derivative test to actually determine whether the points are maximum or minimum values, but before we actually do that, we have to find the values where actually we’re gonna have maximum and minimum points. And the way that we do that is to use a relationship that they share, which is that they both have a slope or 𝑚 that is equal to zero.

In order to determine those two points, we’re gonna first find the 𝑥-coordinates at the points where the slope is actually equal to zero. And to do that, we’re gonna differentiate our function to actually find the slope function. And if we differentiate 𝑓𝑥 is equal to nine 𝑥 to the power of four minus two 𝑥 squared minus five, we actually get 36𝑥 cubed minus four 𝑥.

And just to remind us gonna use the first term. The way we got that is we actually multiplied the exponent by the coefficient, so four by nine, which gives us 36. And then it’s 𝑥 to the power of four minus one cause we reduced the exponent by one. So we get 36𝑥 cubed.

Okay, great! So now we’ve got the slope function. We know that this is gonna be equal to zero at our maximum and minimum points because we’ve already said that they share the relationship that, at these points, the slope is equal to zero. So therefore, we get that zero is equal to 36𝑥 cubed minus four 𝑥. Then we actually divide each side by four just to make it easier to actually solve to find 𝑥.

So we’re left with zero is equal to nine 𝑥 cubed minus 𝑥. So then what we do is we actually take out 𝑥 as a factor because 𝑥 is a factor of both of our terms. So we get zero is equal to 𝑥 multiplied by, and then open parenthesis, nine 𝑥 squared minus one, close parenthesis.

And now what we can actually do is factor the expression in the parentheses. And we can do that because it’s a very special expression, because it’s actually the difference of two squares. And we know that because each of the terms in our expression are squares, because nine is a squared number, 𝑥 squared is squared, and one is a squared number. And we also have a negative. So therefore, we can see that it’s nine 𝑥 squared minus one is gonna be easily factored using the difference of two squares. So we can say fully factored.

We actually get zero is equal to 𝑥 multiplied by three 𝑥 minus one multiplied by three 𝑥 plus one. And we actually achieve that because we actually took the root of each of our terms. So root nine is three, root 𝑥 squared is 𝑥, and root one is one. And then we had one of our parentheses is positive. One of our parentheses is negative. And therefore it’s fully factored.

Okay, great! So now let’s solve for 𝑥. Well, first of all, we know that one of the solutions is 𝑥 is equal to zero. That’s because if we have zero multiplied by anything, then the result would be zero. And now we’re gonna find the other two solutions. And we do that by setting each of our parentheses equal to zero.

So we start with three 𝑥 minus one is equal to zero. We add one to each side. So we get three 𝑥 is equal to one. And then we divide by three to get our second solution, which is 𝑥 is equal to a third.

Now let’s move on to our third solution. So we have three 𝑥 plus one is equal to zero. So then we subtract one from each side. And we get three 𝑥 is equal to negative one. And then, finally, we divide by three to get our third solution, which is 𝑥 is equal to negative a third.

So now we actually have our three 𝑥 values that are turning points to possible maximum and minimum values. We now need to find out actually what would the value of our function be at each of these points. We’re gonna start with 𝑓 zero.

So when we substitute 𝑥 is equal to zero into our function, so we’re gonna get nine multiplied by zero to the power of four minus two multiplied by zero squared minus five, which gives us the result of negative five. So then we can move on to our next value of 𝑥. So 𝑥 is equal to a third.

So we’re gonna substitute that into our function. So we get nine multiplied by a third to the power of four minus two multiplied by a third squared minus five, which gonna be equal to nine over 81 minus two over nine minus five, which is equal to one over nine minus two over nine and then minus 45 over nine because we’ve actually converted our five into 45 over nine. So we get a value of negative 46 over nine. So that’s the value of our function when 𝑥 is equal to a third.

Then we move on to our final value of 𝑥. And that’s where 𝑥 is equal to negative a third. So that’s gonna be equal to nine multiplied by negative a third to the power of four minus two multiplied by negative a third squared minus five, which is gonna give us nine over 81 minus two over nine minus five, which is actually the same as the previous value of 𝑥. So therefore this is also gonna give a value of negative 46 over nine.

So now we’ve actually calculated our possible values. We now need to determine whether they’re local maximum or local minimums. And in order to do that, we’re gonna use our second derivative test. And the why we actually use that is if we take a look at this diagram here, well, we actually find the second derivative. It helps us find the concavity of a section of our function.

If our second derivative is actually positive at a point, then therefore we can say it’s concave up. And the way that helps us is actually with the second derivative test is that if it’s concave up, we know that it’s actually gonna be a local minimum point. If our second derivative at a point is negative, then therefore we know that this is actually gonna be a section that is concave down. And our second derivative test tells us, therefore, we know that this is gonna be a maximum point.

Okay, great! Now we know what we need to do. Let’s get on and actually find the second derivative of our function. In order to actually find the second derivative of our function, what we need to do is actually differentiate our slope function we found earlier, which was 36𝑥 cubed minus four 𝑥. And if we do that, we actually get 108𝑥 squared minus four. And we actually differentiate it a second time using the exact same rules as previously. So we actually multiplied the coefficient by the exponents give us 108. And then we reduced the exponent by one.

Okay, we’ve got the second derivative. Now what’s the next stage? Well, now what we’re gonna do is actually substitute our values of 𝑥 in to determine whether it’s concave up or concave down and thus a maximum or minimum point. So first of all, we start with the point where 𝑥 is equal to zero. So we can say that the second derivative is gonna be equal to 108 multiplied by zero squared minus four, which is just gonna give us an answer of negative four. And if we take a look, because it’s negative, then therefore it’s gonna be concave down. So we know that this is actually gonna be a maximum point.

So next, we’re gonna move on to when 𝑥 is equal to a third. So we get 108 multiplied by a third squared minus four, which is equal to 108 over nine minus four, which is 13 minus four, which is equal to nine. And we can see that actually this answer is positive. So therefore, at this point, we can see that actually it’s gonna be concave up. So it’s actually therefore going to be a minimum value.

Then finally, we move on to the value when 𝑥 is equal to negative a third. Well, actually if we look at this, it should be the same as the previous value. And that’s because when we found the values of the function when it was negative a third, they were the same. But let’s do the calculation anyway.

So we have 108 multiplied by negative a third all squared minus four, which will give us 108 over nine minus four. That’s because we had negative a third squared. So it makes it positive. So as expected, we got the same value as previously when 𝑥 was equal to a third, which was nine. So again, we’d say it’s concave up, which means it’s going to be a minimum point.

So now we’ve actually found all of our values. And we’ve determined whether they’re maximum or minimum. So we can bring it together and say the function has a local maximum value of negative five. And we actually got to this point because when 𝑥 was equal to zero, the value of our function was negative five. And then we used the second derivative test to actually show that it was a maximum because the value of the second derivative at this point was negative four.

And we have a local minimum value of negative 46 over nine. That’s because, at the 𝑥-coordinates of a third and negative a third, we found the value of our function to be both negative 46 over nine. And then again, we used the second derivative test to prove that it was a minimum, because at these points, the value of our second derivative was nine, which is positive and therefore a minimum value.