### Video Transcript

In the closed interval negative one
to two, determine the absolute maximum and minimum values of the function π of π₯
equals four π₯ squared plus three π₯ minus seven if π₯ is less than or equal to one,
six π₯ minus five if π₯ is greater than one. And round them to the nearest
hundredth.

In order to find absolute maximum
and minimum values of a function, we need to consider a couple of things. We need to look for any critical
points of the function. Remember, these are points where
the first derivative is equal to zero. A critical point can indicate a
relative maximum and a relative minimum. And these of course could also be
absolute extrema. We also need to consider the end
points of the function.

So what weβre going to do is begin
by finding the first derivative of our function. Now, our function is piecewise. So weβre going to differentiate
each expression with respect to π₯. Letβs begin with the expression
four π₯ squared plus three π₯ minus seven. The derivative of four π₯ squared
with respect to π₯ is two times four π₯, which is eight π₯. The derivative of three π₯ is
three. And the derivative of negative
seven is zero. And thatβs for values of π₯ less
than or equal to one.

Next, we differentiate six π₯ minus
five with respect to π₯. Well, the derivative of six π₯
minus five is simply six. So we see that the first derivative
is either equal to π₯ plus three if π₯ is less than or equal to one or six if π₯ is
greater than one. And actually, if we look carefully,
we see that the expression six cannot be equal to zero for any values of π₯. This indicates to us there are no
critical points in this part of the function. But we will set eight π₯ plus three
equal to zero and solve for π₯. By subtracting three from both
sides and then dividing through by eight, we find that π₯ equals negative
three-eighths. So we have a critical point when π₯
is equal to negative three-eighths.

So weβre going to do a number of
things. Weβre going to evaluate the
function at this critical point. Thatβs π of negative
three-eighths. Weβll evaluate it at the lower and
upper end of our interval. Thatβs π of negative one and π of
two. And weβll evaluate it here when π₯
is equal to one. This is the end point of each bit
of the function. Now, when π₯ is equal to negative
three-eighths, it is indeed less than or equal to one. So we use the first bit of our
function to evaluate π of negative three-eighths. Itβs four times negative three
eighths squared plus three times negative three-eighths minus seven. Thatβs negative 7.5625.

When π₯ is equal to negative one,
we use the same bit of the function. So we get four times negative one
squared plus three times negative one minus seven, which is negative six. Now, negative six is greater than
negative 7.5625. So when π₯ is equal to negative
one, we absolutely canβt have an absolute minimum. But we could still have an absolute
maximum. Weβre now going to test π of
two. When π₯ is equal to two, this is
obviously greater than one. So weβre going to use the second
part of our function. Thatβs six times two minus five,
which is equal to seven.

Finally, we just need to check when
π₯ is equal to one. We go back to the first part of our
function because this is defined for values of π₯ less than or equal to one. We get four times one squared plus
three times one minus seven, which is equal to zero. Well, if we look carefully, we see
that the smallest value we have over our interval is negative 7.5625. And the largest value we get is
seven. We round negative 7.5625 correct to
the nearest hundredth.

And we find that the absolute
maximum value of our function is seven. And the absolute minimum is
negative 7.56.