### Video Transcript

Determine the absolute maximum and
minimum values of the function π¦ equals π₯ over two π₯ plus eight on the closed
interval two to six.

Remember, to find absolute extrema
for our function π of π₯ 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 π of π₯
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 π₯ over
two π₯ plus eight? In fact, we have a number of
methods we could use. But since itβs the quotient of two
differentiable functions, we can use the quotient rule.

This says that the derivative of
the quotient of two differentiable functions π’ and π£ is π£ times dπ’ by dπ₯ minus
π’ times dπ£ by dπ₯ all over π£ squared. The numerator of our fraction is
π₯, so weβre going to let π’ be equal to π₯ and π£ be equal to two π₯ plus
eight. Then, dπ’ by dπ₯ is equal to
one. And dπ£ by dπ₯ is equal to two.

So, the derivative of π¦ with
respect to π₯ is two π₯ plus eight times dπ’ by dπ₯, which is one, minus π’ times
dπ£ by dπ₯, thatβs π₯ times two, all over π£ squared, thatβs two π₯ plus eight all
squared. So, we can say that dπ¦ by dπ₯ is
equal to eight over two π₯ plus eight all squared. Next, weβre going to set this equal
to zero and solve for π₯. 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 dπ¦ by dπ₯, 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. Thatβs π of two and π of six.

π of two is two over two times two
plus eight, thatβs two twelfths, which simplifies to one-sixth. π of six is six over two times six
plus eight. Thatβs 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 letβs 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 doesnβt exist. Thatβs the point where two π₯ plus
eight is equal to zero, or π₯ is equal to negative four. Now since this was outside the
closed interval two to six, we actually didnβt need to worry about this critical
point. And we could focus solely on the
end points of the intervals π of two and π of six.