### Video Transcript

Determine the absolute maximum and
minimum values of the function π of π₯ equals two π₯ to the fourth power minus
eight π₯ squared minus 13 in the closed interval negative one to two.

Remember, to find absolute extrema
for some continuous function π of π₯, we follow three steps. We start by finding all critical
points in the closed interval. We check the values of π of π₯ at
these critical points. And then, we check the end points
for any absolute extrema, values that might be smaller than any relative minima or
larger than any relative maxima. Critical points occur when the
derivative of the function is equal to zero, or does not exist. So, letβs begin by finding the
derivative of our function.

The derivative of π of π₯ is
written as π prime of π₯, and itβs four times two π₯ to the third power minus two
times eight π₯. And, of course, the derivative of
negative 13 is zero. So, we see that the derivative of
our function is eight π₯ cubed minus 16π₯. Weβll set this equal to zero and
solve for π₯. Here, we factor the expression on
the right-hand side to get eight π₯ times π₯ squared minus two.

And we see that for eight π₯ times
π₯ squared minus two to be equal to zero, either eight π₯ must be equal to zero,
which means π₯ is equal to zero. Or we can say that π₯ squared minus
two must be equal to zero. And solving this, we see that π₯ is
equal to both the positive and negative square root of two. So, we have critical points on our
function at π₯ equals zero, π₯ equals negative root two, and π₯ equals root two.

Our second step is to evaluate the
function at these critical points. Thatβs π of zero, π of root two,
and π of negative two. π of zero is two times zero to the
fourth power minus eight times zero squared minus 13, which is negative 13. π of root two is two root two to
the fourth power minus eight root two squared minus 13, which is negative 21. And, actually, π of negative root
two is also negative 21. Now this by itself doesnβt help us
much. It certainly looks like negative 21
and negative 13 might be local extrema, but we need to know whether they are
absolute extrema.

So, weβre going to evaluate our
function at the ends of our interval. Remember, this is because we know
that if we have a continuous function over some closed interval π to π, then weβre
guaranteed at some point in this interval to have both an absolute maximum and an
absolute minimum. And these extreme values are
obtained either at the place or places where local extrema occur, or at the end
points of the interval.

So, weβre going to evaluate π of
negative one, and π of two. π of negative one is two times
negative one to the fourth power minus eight times negative one squared minus 13,
which is negative 19. And π of two is two times two to
the fourth power minus eight times two squared minus 13, which is negative 13. We can quite clearly see then that
the absolute maximum value of our function in the closed interval negative one to
two is negative 13. And the absolute minimum value is
negative 21. This example involves some fairly
simple differentiation, so weβll now look at an example which involves a little more
work.