### Video Transcript

Determine the absolute maximum and
minimum values of the function π of π₯ is equal to six π₯ minus three squared, if
π₯ is less than or equal to two and two minus nine π₯, if π₯ is greater than two in
the closed interval one to six.

Remember, to find absolute extrema
for continuous functions, we follow three steps. We begin by finding any critical
points in the closed interval weβre looking at. We find the values of our function
π of π₯ at these critical points. And then, we check the end points
for absolute extrema, values that are smaller than the relative minimum or larger
than the relative maximum.

Now we have a slight problem
here. This is a piecewise function. And we donβt yet know if this
piecewise function is itself continuous. We do have a note that the function
six π₯ minus three all squared is continuous, and the function two minus nine π₯ is
continuous. So, what weβll do is consider that
π₯ equals two might be a critical point. And to test this, weβre going to
evaluate the right-hand and left-hand derivative of our function at π₯ equals
two.

Weβll begin by evaluating the
right-hand derivative of π at π₯ equals two. This is given by the limit as β
approaches zero from the right of π of two plus β minus π of two all over β. Weβre looking at the right-hand
derivative, so weβre interested in the function that applies when π₯ is greater than
two. Thatβs π of π₯ equals two minus
nine π₯.

So, weβre looking for the limit as
β approaches zero from the right of two minus nine times two plus β minus two minus
nine times two over β. Thatβs two minus 18 minus nine β
minus two plus 18 all over β. This quite quickly simplifies to
negative nine β over β. And then, we simplify further, and
we see that weβre looking for the limit as β tends to zero from the right of
negative nine. But this is independent of β, so we
know that this is just going to be equal to negative nine. And so, our right-hand derivative
is negative nine.

Weβll now repeat this process for
the left-hand derivative. This time, weβre evaluating π of
two plus β minus π of two over β as β approaches zero from the left. And so, weβre interested in the
part of π of π₯ where π₯ is less than or equal to two. So, we have the limit as β
approaches zero from the left of six times two plus β minus three all squared minus
six times two minus three all squared all over β. This simplifies to nine plus six β
all squared minus nine squared over β.

And then, if we distribute the
parentheses, we see that weβre left with the limit as β approaches zero from the
left of 108β plus 36β squared. We can do a little bit of
simplification, and this becomes 108 plus 36β. And then, we see that as β
approaches zero from the left, weβre left with 108.

We can see that our left-hand and
right-hand derivatives are not equal. And so, π prime of π₯, our
derivative, doesnβt actually exist at π₯ equals two. And so, we know that we have a
critical point at π₯ equals two. And we know weβre going to need to
find π of π₯ at this point. We should also though check for any
critical points on each part of our piecewise function. So, weβll differentiate each part
with respect to π₯ and set that equal to zero.

We can use the general power rule
to differentiate six π₯ minus three all squared with respect to π₯. Itβs two times six π₯ minus
three. And then, we reduce the power by
one. And then, we multiply that by the
derivative of six π₯ minus three, which is just six. So, the derivative of this bit is
12 times six π₯ minus three. And the derivative of two minus
nine π₯ is negative nine.

It should be quite clear that
thereβs no way for negative nine to be equal to zero. The derivative of this part of our
function is always negative nine. But we can set 12 times six π₯
minus three equal to zero. And when we solve for π₯, we get π₯
equals 0.5. So, we have one more critical point
at π₯ equals 0.5. Weβre going to evaluate our
function at the points π₯ equals two and π₯ equals 0.5 then. Letβs clear some space.

0.5 is less than two, so we
evaluate the function at this point by using six π₯ minus three all squared. And when we substitute 0.5 in, we
get zero. So, π of 0.5 is zero. We use the same part of our
piecewise function to evaluate π of two. And when we do, we get 81. So, weβve found π of π₯ at the
critical points on our function. Next, we need to check the end
points.

These are π of one and π of
six. We use six π₯ minus three all
squared once again to evaluate π of one. And that gives us nine. But six is greater than two. So, to evaluate π of six, we use
two minus nine π₯ and we get negative 52. We can see that the absolute
maximum value of our function is 81 and the absolute minimum value is negative
52. The key point to remember here is
that if weβre dealing with a piecewise function, we must check the behaviour at the
end of each piece of our function. In our final example, weβll
consider how we can apply ideas about finding absolute extrema to exponential
functions.