### Video Transcript

Determine the absolute maximum and
minimum values of the function ๐ฆ equals ๐ฅ over two ๐ฅ plus eight on the closed
interval two to six.

Remember, to find absolute extrema
for our function ๐ of ๐ฅ over some closed interval, we follow three steps. We begin by finding any critical
points in our closed interval. We then find the values of ๐ of ๐ฅ
at these critical points. And then, we check the end points
for absolute extrema, in other words, values that are smaller than the relative
minimum or larger than the relative maximum.

Remember, the critical points are
the points on our curve where the derivative is equal to zero or possibly does not
exist. So, we need to find the derivative
of our function and set it equal to zero. But how do we differentiate ๐ฅ over
two ๐ฅ plus eight? In fact, we have a number of
methods we could use. But since itโs the quotient of two
differentiable functions, we can use the quotient rule.

This says that the derivative of
the quotient of two differentiable functions ๐ข and ๐ฃ is ๐ฃ times d๐ข by d๐ฅ minus
๐ข times d๐ฃ by d๐ฅ all over ๐ฃ squared. The numerator of our fraction is
๐ฅ, so weโre going to let ๐ข be equal to ๐ฅ and ๐ฃ be equal to two ๐ฅ plus
eight. Then, d๐ข by d๐ฅ is equal to
one. And d๐ฃ by d๐ฅ is equal to two.

So, the derivative of ๐ฆ with
respect to ๐ฅ is two ๐ฅ plus eight times d๐ข by d๐ฅ, which is one, minus ๐ข times
d๐ฃ by d๐ฅ, thatโs ๐ฅ times two, all over ๐ฃ squared, thatโs two ๐ฅ plus eight all
squared. So, we can say that d๐ฆ by d๐ฅ is
equal to eight over two ๐ฅ plus eight all squared. Next, weโre going to set this equal
to zero and solve for ๐ฅ. But look at what happens when we
do.

For an algebraic fraction to be
equal to zero, the numerator must itself be equal to zero. In this case, we end up with the
statement zero equals eight, which we know to be rubbish. This means d๐ฆ by d๐ฅ, in this
case, cannot be equal to zero. There are no turning points for us
to evaluate. And we, therefore, jump straight to
step three and evaluate the function at the end points of the interval. Thatโs ๐ of two and ๐ of six.

๐ of two is two over two times two
plus eight, thatโs two twelfths, which simplifies to one-sixth. ๐ of six is six over two times six
plus eight. Thatโs six twentieths, which
simplifies to three-tenths. Three-tenths is greater than
one-sixth, so we can say the absolute maximum value of our function is three-tenths
and the absolute minimum value is one-sixth. Now letโs backtrack a little
though.

We said that critical points occur
at places on the function where the derivative does not exist, and there is one
point on our function where the derivative doesnโt exist. Thatโs the point where two ๐ฅ plus
eight is equal to zero, or ๐ฅ is equal to negative four. Now since this was outside the
closed interval two to six, we actually didnโt need to worry about this critical
point. And we could focus solely on the
end points of the intervals ๐ of two and ๐ of six.