### Video Transcript

Find where, if at all, the function
π¦ equals negative π₯ minus one over π₯ add eight has its local maxima and
minima.

Letβs first of all remind ourselves
what local maxima and local minima are. The local maxima and minima are the
largest and smallest values of the function within a given range. At these points, we can see that
the slope, the derivative, is equal to zero. So we need to differentiate our
function to find the slope function. And then we work out where the
slope function is equal to zero. So we need to look at
differentiating this function. We can differentiate this term by
term. So letβs start with negative
π₯. We can differentiate this, knowing
that this is just negative π₯ to the power of one, and apply the power rule. This gives us that the derivative
of negative π₯ is negative one.

Now we move on to the second term,
one over π₯ add eight. We start by writing it as a
negative exponent, π₯ add eight to the power of negative one. And we have a really useful tool
for differentiating a function raised to a power. Itβs the chain rule. This tells us that the derivative
with respect to π₯ of a function π of π₯ raised to the power of π is π multiplied
by the function of π₯ raised to the power of π minus one multiplied by π prime of
π₯. And so, the derivative with respect
to π₯ of π₯ add eight to the power of negative one is π the power, which is
negative one, multiplied by our function, π₯ add eight, raised to the power of π
minus one. So negative one minus one, which is
negative two, multiplied by π dash of π₯, which is the derivative of π₯ add eight
with respect to π₯.

So whatβs the derivative of π₯ add
eight? Well, differentiating term by term,
π₯ differentiates to one. And eight is a constant. So it differentiates to zero. So the derivative of π₯ add eight
is just one. So this is just negative π₯ add
eight to the power of negative two. We can rewrite this as negative one
over π₯ add eight squared. So dπ¦ by dπ₯ equals negative one
minus, because it was a minus in the question, negative one over π₯ add eight
squared. Because weβre subtracting a
negative term here, we can write this as an add. Remember, we said that local maxima
and minima occur where dπ¦ by dπ₯ equals zero. So we set our derivative equal to
zero and solve for π₯.

We can start by adding one to both
sides and then multiplying both sides by π₯ add eight squared. From here, we can square root both
sides. And we must be careful here as the
square root of one can take the value positive one or negative one. So when π₯ add eight equals
positive one, π₯ equals negative seven. And when π₯ add eight equals
negative one, π₯ equals negative nine. So now, we have the π₯-values of
possible maxima or minima. But we donβt know which is maxima
or minima. So letβs find out. Iβm going to clear some space to do
this.

Letβs start by finding the
corresponding π¦-values by substituting our π₯-values into our original equation for
π¦. So for our points negative seven,
six and negative nine, 10, letβs check for maxima or minima. To check this, we evaluate the
values of π₯. We found at the second
derivative. If the second derivative is
positive, π₯ is a minima. And if the second derivative is
negative, π₯ is a maxima. To find the second derivative, we
differentiate the first derivative. Negative one is a constant. So this differentiates to zero. For one over π₯ add eight squared,
we can proceed in a similar way to how we found the first derivative. We rewrite this as a negative
exponent and use the chain rule formula.

So the derivative of π₯ add eight
to the power of negative two is negative two multiplied by π₯ add eight to the power
of negative three multiplied by the derivative of π₯ add eight, which is one. And so π₯ add eight to the power of
negative two differentiates to negative two multiplied by π₯ add eight to the power
of negative three, which we can write as negative two over π₯ add eight cubed. And so the second derivative is
negative two over π₯ add eight cubed.

Iβm just going to clear some space
here. So at π₯ equals negative seven, the
second derivative is equal to negative two over negative seven add eight cubed,
which is negative two over one cubed, which is just negative two. This is less than zero. So at π₯ equals negative seven, we
have a maximum. At π₯ equals negative nine, the
second derivative is equal to negative two over negative nine add eight cubed, which
is equal to negative two over negative one cubed. And remember that negative one
cubed is negative one. So this gives us two. This is greater than zero. So at π₯ equals negative nine, we
have a minimum. So we found our local maximum to be
negative seven, six and our local minimum to be negative nine, 10.

And hence the local maximum value
is six. And the local minimum value is
10.