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.
