Video Transcript
Find, if any, the points 𝑥, 𝑦
where 𝑦 equals negative 𝑥 minus four over 𝑥 add two has a local maximum or local
minimum.
Let’s say we have a function 𝑦
equals 𝑓 of 𝑥 that looks like this. Then, we call these high points
local maximums and these low points local minimums. Notice how these maximum and
minimum points occur where the slope is flat. But how can we identify these
points?
Well, at the points where the slope
is flat, the slope is equal to zero. And the points where the slope is
equal to zero, the derivative is equal to zero. That is, d𝑦 by d𝑥 is equal to
zero. So we’re going to look for the
points where the derivative of 𝑦 equals negative 𝑥 minus four over 𝑥 add two is
equal to zero.
Firstly, we can rewrite 𝑦 in its
negative exponent form as 𝑦 equals negative 𝑥 minus four multiplied by 𝑥 add two
to the power of negative one. Writing it this way is going to
help us to differentiate it. We’re going to differentiate this
with respect to 𝑥. So let’s do this term by term,
starting with the term negative 𝑥. We know that this differentiates to
negative one.
Now, because negative four
multiplied by 𝑥 add two to the negative one power is a function of a function,
we’re going to differentiate this using the chain rule. The chain rule says that if 𝑦
equals 𝑓 of 𝑢 and 𝑢 equals 𝑔 of 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢
multiplied by d𝑢 by d𝑥. But so we don’t confuse the 𝑦 in
the formula with the 𝑦 in the question, I’m going to rename the 𝑦 in the formula
with 𝑧.
So, if we start by letting 𝑢 be
our inner function, which in this case is going to be 𝑥 add two, then we can call
𝑧 four 𝑢 to the negative one power. We then need to find d𝑧 by d𝑢 and
d𝑢 by d𝑥. Well, d𝑢 by d𝑥 is the derivative
of 𝑢 with respect to 𝑥. So it’s the derivative of 𝑥 add
two. We know that 𝑥 differentiates to
one and two is a constant, so that differentiates to zero. So d𝑢 by d𝑥 is equal to one. d𝑧
by d𝑢 is a derivative of four 𝑢 to the negative one power with respect to 𝑢.
Using the power rule of
differentiation, which tells us to multiply by the power and then subtract one from
the power, gives us that d𝑧 by d𝑢 is negative four 𝑢 to the negative two
power. Putting all of this together, we
have that d𝑧 by d𝑥 is equal to d𝑧 by d𝑢, which is negative four 𝑢 to the
negative two power, multiplied by d𝑢 by d𝑥, which is just one. So d𝑧 by d𝑥 is just negative four
𝑢 to the negative two power. But remember that we let 𝑢 be
equal to 𝑥 add two. So let’s replace 𝑢 with 𝑥 add
two.
So we’ve now found d𝑦 by d𝑥 for
our original equation to be negative one minus negative four multiplied by 𝑥 add
two to the negative two power. As we have a double negative here,
we can write this with an add instead.
Remember, we said that the points
where we have a local maximum and a local minimum, we have that d𝑦 by d𝑥 is equal
to zero. So, in order to find the local
maximum and local minimum points, we need to set what we found for d𝑦 by d𝑥 to be
equal to zero. One thing that might make it a
little bit easier to solve this is to write four multiplied by 𝑥 add two to the
negative two power back in its fractional form, that is, four over 𝑥 add two
squared.
Let’s start to solve this by adding
one to both sides and then multiplying both sides by 𝑥 add two squared. From here, we can square root both
sides of the equation. We’ve got to be careful when we do
this because the square root of four could either take the value of two or negative
two. If it takes the value of two, then
the solution is that 𝑥 is equal to zero. But if it takes the solution of
negative two, then 𝑥 takes the value of negative four. So these are two possible
solutions.
Let’s find the corresponding values
of 𝑦. We can do this by substituting 𝑥
is equal to zero and 𝑥 is equal to negative four into our original equation for
𝑦. Firstly, when 𝑥 is equal to zero,
we have that 𝑦 equals negative zero minus four over zero add two, which just gives
us negative four over two, which is just negative two. So, when 𝑥 is equal to zero, 𝑦 is
equal to negative two.
When 𝑥 is equal to negative four,
𝑦 is equal to minus negative four minus four over negative four add two, which is
just four minus four over negative two, which is the same as four add two. And that just gives us six. So, when 𝑥 is equal to negative
four, 𝑦 is equal to six.
So these are the two points that
we’ve identified as maximum or minimum points. It could be that they’re both local
maximums, it could be that they’re both local minimums, or we might have that one of
them is a local maximum and one of them is a local minimum.
We can determine this using the
second derivative. If at our value of 𝑥 we have that
the second derivative is greater than zero, we have a local minimum. If the second derivative is less
than zero, we have a local maximum. So we need to go ahead and find the
second derivative, that is, the derivative of the first derivative.
Well, negative one is a
constant. So this is going to differentiate
to zero. But four multiplied by 𝑥 add two
to the negative two power, we need to differentiate this using the chain rule
because this is a function of a function. So I’ll clear some space so that we
can use the chain rule again.
Remember, 𝑢 is our inner function,
which again is going to be 𝑥 add two. Then, we can write our outer
function 𝑧 as four 𝑢 to the negative two power. So let’s go ahead and find d𝑧 by
d𝑢 and d𝑢 by d𝑥. d𝑢 by d𝑥 is the derivative of 𝑥 add two with respect to 𝑥,
which is just going to be one. d𝑧 by d𝑢 is a derivative of four 𝑢 to the negative
two power with respect to 𝑢.
Using the power rule for
differentiation, which tells us to multiply by the power and then subtract one from
the power, gives us that d𝑧 by d𝑢 is equal to negative eight 𝑢 to the negative
three power. So then d𝑧 by d𝑥, which we’re
trying to find, is d𝑧 by d𝑢, which is negative eight 𝑢 to the negative three
power, multiplied by d𝑢 by d𝑥, which is one. So we have that d𝑧 by d𝑥 is equal
to negative eight 𝑢 to the negative three power. But we need to remember that we let
𝑢 equal 𝑥 add two. So d𝑧 by d𝑥 is equal to negative
eight multiplied by 𝑥 add two to the negative three power.
So we found that our second
derivative is equal to negative eight multiplied by 𝑥 add two to the negative three
power. Let’s rewrite this in its
fractional form as negative eight over 𝑥 add two cubed. So we’re going to use this to
determine whether the coordinates that we found are maximum or minimum points.
Firstly, when 𝑥 is equal to zero,
the second derivative is equal to negative eight over zero add two cubed. This gives us negative eight over
two cubed, which is negative eight over eight, which is just negative one. So, when 𝑥 is equal to zero, the
second derivative is negative one, which is less than zero. Therefore, at the point zero,
negative two, we have a local maximum.
When 𝑥 is equal to negative four,
we have that the second derivative is equal to negative eight over negative four add
two cubed. This is negative eight over
negative two cubed, which is negative eight over negative eight. And this just gives us one. So, when 𝑥 is equal to negative
four, the second derivative is equal to one. Because this is greater than zero,
we have that the point negative four, six is a local minimum.
So what we can conclude is that
zero, negative two is a local maximum point and negative four, six is a local
minimum point.
