### Video Transcript

Consider the function π of π₯ is
equal to π₯ minus three all squared if π₯ is less than or equal to five and π of π₯
is equal to four π₯ minus 16 if π₯ is greater than five over the closed interval
from zero to seven. Determine the absolute minimum of
π of π₯ over the given interval. Determine the absolute maximum of
π of π₯ over the given interval.

The question gives us a
piecewise-defined function π of π₯ defined on a closed interval. And weβre asked to find the
absolute minimum and absolute maximum of our function π of π₯ over this
interval. We know how to find the absolute
maximum and absolute minimum of a continuous function over a closed interval.

To find the absolute extrema of a
continuous function on a closed interval, we follow the following three steps. First, we find all the critical
points of our function. Then we evaluate our function at
all of the critical points. And finally, we evaluate our
function at the endpoints of our closed interval. But in this case, weβre given a
piecewise function. We donβt know if this is
necessarily continuous or not.

We could check if our function π
of π₯ is continuous and then carry on using this method, and this would work. However, thereβs a more simple
method for finding absolute extrema of piecewise-defined functions. We know that π₯ minus three all
squared is continuous and four π₯ minus 16 is continuous. So we could find the absolute
extrema on each of these pieces separately. Then weβll just check the value of
π of π₯ around π₯ is equal to five separately. If this function has a jump
discontinuity, it will have a different value as π₯ approaches five from the left
and five from the right.

So letβs start with the first part
of our piecewise-defined function π of π₯. We want to find all of the extrema
of π₯ minus three all squared on the closed interval from zero to five. And this is a continuous function
on a closed interval. So we can do this by following our
three steps.

First, we need to find the critical
points of this function. This is where the derivative of our
function is equal to zero or where the derivative does not exist. However, this is a polynomial. So its derivative exists for all
real values of π₯. To find the derivative of π₯ minus
three all squared, weβll start by distributing the square over our parentheses. We get π₯ squared minus six π₯ plus
nine.

Then we can differentiate this term
by term by using the power rule for differentiation. We get two π₯ minus six. And we can then solve this equal to
zero. We get that π₯ is equal to
three. So we found one critical point of
this function. Itβs when π₯ is equal to three.

Next, we need to evaluate π₯ minus
three all squared, thatβs our critical point, and the endpoints of our interval. Evaluating this at π₯ is equal to
three, we get three minus three all squared, which is equal to zero. Now, we evaluated our first
endpoint π₯ is equal to zero. We get zero minus three all
squared, which is equal to nine. Finally, we evaluate to the other
end of our closed interval. This is π₯ is equal to five. So we get five minus three all
squared, which is equal to four.

Letβs now find any extrema for the
second part of our piecewise-defined function π of π₯. Weβll look for extrema of four π₯
minus 16 on the closed interval from five to seven. However, itβs worth reiterating
weβve not shown that our function π of π₯ is equal to four π₯ minus 16 when π₯ is
equal to five. So if we get an extrema at this
point, weβll need to consider what this means graphically. But for now, weβll just carry on
and find the extrema of this continuous function by using our three steps.

First, we need to find the
derivative of this linear function. Since itβs a linear function, the
derivative will be the coefficient of π₯, which is four. And so our derivative is not equal
to zero for any value of π₯. So we only need to evaluate this at
the endpoints of our interval. This time, our closed interval is
from five to seven.

Weβll start with π₯ is equal to
five. We get four times five minus 16,
which is equal to four. And itβs worth pointing out this is
the same as what we got when we evaluated our other endpoint. In other words, we couldβve shown
that our function π of π₯ was continuous by using direct substitution on the left-
and right-hand limit. But our current method will still
work. Weβll now check the other endpoint
of our closed interval π₯ is equal to seven. We get four times seven minus 16,
which is equal to 12.

Weβve now evaluated our function π
of π₯ at all of the possible values where there could be an absolute minimum or an
absolute maximum on this closed interval. We can see our lowest value was
when π₯ is equal to three. It gave us zero. And the largest value is when π₯
was equal to seven. It gave us an output of 12.

So the absolute minimum of our
piecewise-defined function π of π₯ over the closed interval from zero to seven must
be zero. And the absolute maximum of our
piecewise-defined function π of π₯ on the closed interval must be equal to 12.