### Video Transcript

The function π of π₯ can be
represented by the graph shown below. Which of the following is the
piecewise representation of π of π₯? Is it option (A) π of π₯ is equal
to π₯ squared minus one when π₯ is greater than or equal to negative two and π₯ is
less than four and π of π₯ is equal to 25 when π₯ is greater than four and π₯ is
less than or equal to 10 and π of π₯ is equal to two π₯ plus eight when π₯ is
greater than 10? Or is it option (B) π of π₯ is
equal to π₯ squared minus one when π₯ is greater than or equal to negative two and
π₯ is less than four and π of π₯ is equal to 25 when π₯ is greater than or equal to
four and π₯ is less than or equal to 10 and π of π₯ is equal to two π₯ plus eight
when π₯ is greater than 10? Or is it option (C) π of π₯ is
equal to π₯ squared minus one when π₯ is greater than or equal to negative two and
π₯ is less than four and π of π₯ is equal to 25π₯ when π₯ is greater than four and
π₯ is less than or equal to 10 and π of π₯ is equal to two π₯ plus eight when π₯ is
greater than 10? Is it option (D) π of π₯ is equal
to π₯ squared plus one when π₯ is greater than or equal to negative two and π₯ is
less than four and π of π₯ is equal to 25 when π₯ is greater than four and π₯ is
less than or equal to 10 and π of π₯ is equal to two π₯ plus eight when π₯ is
greater than 10. Or is it option (E) π of π₯ is
equal to π₯ squared plus one when π₯ is greater than or equal to negative two and π₯
is less than four, π of π₯ is equal to 25 when π₯ is greater than four and less
than 10, and π of π₯ is equal to two π₯ plus eight when π₯ is greater than or equal
to 10. Find the value of π evaluated at
three.

There are two parts of this
question. In the first part of this question,
we need to determine which of the five given piecewise representations of π of π₯
is the correct piecewise representation of π of π₯. In the second part of this
question, we need to determine the value of π at three. And there are many different ways
of answering the first part of this question. The easiest way is to start by
considering how would we construct a piecewise representation of π of π₯ from its
graph.

And there are three sections to
this graph, so we would expect three subfunctions for our piecewise-defined
function. And the easiest of these
subfunctions to find is the second subfunction because itβs a constant value of
25. We want to find the subdomain for
the second subfunction. Thatβs the values of π₯ we input
into our piecewise-defined function to output the second part of our function.

On our graph, this will be the
π₯-coordinate of any point on the second part of our function. We can see these lie between four
and 10. And we need to know on the diagram
we can see on the left side when π₯ is equal to four, we have a hollow circle, and
on the right side when π₯ is equal to 10, we have a solid circle. When itβs hollow, we donβt include
the endpoint, and when itβs solid, we do include the endpoint. So, our subdomain will be the
values of π₯ greater than four and the values of π₯ less than or equal to 10.

This gives us our first
subfunction. Itβs a constant value of 25 and the
values of π₯ must be greater than four and less than or equal to 10. Although itβs not necessary, we can
use this to eliminate some of our options. First, we can see in option (B) our
second subdomain includes the value of four. However, in the diagram, when π₯ is
equal to four, our second subfunction has a hollow dot. So, four is not in the subdomain of
our second subfunction. So, option (B) is not correct. In option (C), we can see the
second subfunction is 25π₯. However, we know itβs a constant
value of 25. So, (C) is not the correct option
either. Finally, in option (E), we can see
that 10 is not included in the second subdomain. However, in the diagram, when π₯ is
equal to 10, we can see we have a solid dot in the second subfunction. So, 10 is in the second subdomain
of our function. So, our answer canβt be option
(E).

We can do the same to find
expressions for the first and third subfunctions and subdomains. Since the third subfunction is a
linear function, this one will be easier. So, letβs do this one next. Weβll start by finding the
subdomain of this subfunction. We can see that the graph of this
subfunction starts when π₯ is equal to 10, and it continues indefinitely. And since the graph of this
subfunction has a hollow dot at its endpoint, we donβt include π₯ is equal to 10 in
the subdomain. The subdomain of this function is
just all values of π₯ greater than 10.

Next, we need to find an expression
for this linear function. Thereβs a few different ways of
doing this. One way is to find the slope and
the coordinates of a point on the line. We can find the slope of this line
directly from the graph. For every one unit we move across,
we move two units upwards. The slope of this line is two. We can then find an equation from
this line, either by extending the line and finding its π¦-intercept or by using the
coordinates of a point on the line. We see that our line passes through
the point 13, 34. This allows us to find the equation
of this line by using the pointβslope form of the equation of a line.

A line with slope π passing
through the point with coordinates π₯ one, π¦ one will have the equation π¦ minus π¦
one is equal to two times π₯ minus π₯ one. In this case, thatβs π¦ minus 34 is
equal to two multiplied by π₯ minus 13. And if we distribute, rearrange,
and simplify this equation, we get the equation of this line is π¦ is equal to two
π₯ plus eight. Therefore, the third subfunction of
our piecewise-defined function is two π₯ plus eight with the subdomain π₯ being
greater than 10.

However, we can just notice both of
our remaining options already have this as the third subfunction and third
subdomain. But it is useful to be able to
determine subfunctions and subdomains from a graph because we wonβt always be given
the options. Instead, letβs clear some space and
determine the first subfunction and subdomain. Once again, weβll start by
determining the subdomain of this subfunction. We see the values of π₯ for this
part of the graph range from negative two to four. And thereβs a solid circle when π₯
is equal to negative two and a hollow circle when π₯ is equal to four. This means the subdomain will
include the value of negative two, and it wonβt include the value of four. So, our subdomain is π₯ is greater
than or equal to negative two and π₯ is less than four.

All thatβs left to do is to
determine the first subfunction. Thereβs a few different ways of
doing this. The easiest way is to notice that
this has the same shape as a quadratic parabola. So, if we sketch π¦ is equal to π₯
squared on the closed interval from negative two to four onto our diagram, we can
see that our subfunction is translated one unit down from this parabola. And to translate a function down,
we subtract one from its output. So, our first subfunction is π₯
squared minus one.

This gives us the full piecewise
representation of our function π of π₯. And we can see this is given by
option (A) π of π₯ is equal to π₯ squared minus one when π₯ is greater than or
equal to negative two and less than four, π of π₯ is equal to 25 when π₯ is greater
than four and less than or equal to 10, and π of π₯ is equal to two π₯ plus eight
when π₯ is greater than 10.

But weβre not done yet. The second part of our question
wants us to determine π evaluated at three. And thereβs two different ways we
can do this. Letβs start by clearing some space
and do this from the diagram. In a graph, the π₯-coordinates of
any point on our curve tell us the input values of the function and the
corresponding π¦-coordinates tell us the output of our function. So, we can determine π evaluated
at three by reading off the π¦-coordinate of the point on the curve with
π₯-coordinate three. And from the graph, we can see this
is eight. Therefore, π evaluated at three is
equal to eight.

However, this isnβt the only way we
can answer this question. We can also do this directly from
the piecewise definition of π of π₯ we determined in the first part. To input the value of three into
this function, we need to determine which subdomain the value of three lies in. And three is bigger than negative
two and less than four, so three lies in the first subdomain of our function. Therefore, we evaluate π at three
by substituting π₯ is equal to three into our first subfunction. π of three is three squared minus
one, which is also equal to eight. Therefore, π evaluated at three is
equal to eight.