Domain
Page 269. # 53.
Solution: Since
this equation is in the form 
,
the graphing calculator will probably be a good way to find domain and range.
Just enter
this in the calculator. In the standard window, and the graph should look
like this. In the two sketches below, it is clear that there are points at
(-1,0) and (6,0) that will be critical to finding the domain and range.
Notice that the graph actually touches the x axis at these two points, and
from these points the graph extends upward from y=0. The graph extends to
the left from x=-1, and to the right from x=6.
From this
graph, it should be clear that the Domain is
.
Likewise, the range is all values that are on or above the x-axis, or
,
or in interval notation 
.
Page 270: #70. 
Solution:
In order to
find the domain, you must solve for y in terms of x, in order to see if
there are any restrictions on x. To do this, notice that there is only one
y term already on the left side. You should start by adding
to
both sides to get all the non-y terms on the right side
.
In order to
solve for y, you must divide both sides by
,
To find the
domain, you are looking for restrictions on the denominator. Denominators
can NOT be zero. In this case, the denominator is
must never be zero, so
.
Therefore,
the domain is all values of x, such that
.
Page 271:
#75. 
Solution:
In order to find the domain, you must solve for y
in terms of x, in order to see if there are any restrictions on x. To do
this, notice that all the y terms are already on the left side, and you can
start by factoring out the y.
In order to solve for y, you must divide both
sides by 
,
To find the domain, you are looking for
restrictions on the denominator. Denominators can NOT be zero. But in this
case, the denominator is
, which can never be
zero anyway, so there are no restrictions on x.
Therefore, the domain is all real x,
or 
Range
Page 274.
# 2.
Solution: If
you recognize that this is a straight line, then you may already know that
the domain and range are both all real values. However, the general
practice in finding the range is to solve for
in terms of 
in order to determine what restrictions there might be
for To find the range, solve for
.
Subtract
from each side of the equation.
Divide both sides by
or 
or 
Either way you look at it, there are no denominators and no radicals, and
therefore there are no restrictions on
. Range is therefore all real
, or in interval notation
.
Page 275.
# 18.
Solution: If
you recognize that this is a parabola that opens upward, then you may
already know that the domain is all real values, and the range is
. However, as always, the general practice in finding the
range is to solve for in terms of in order to determine what restrictions there might be
for To find the range, solve for
.
Take the square root of both sides.
The
restriction here is that, because of the square root, the radicand
must be greater than or equal to zero. That is, the
range is
, or in interval notation
.
Page 276. # 25.
Solution: You
may recognize that this is a parabola that opens downward with vertex at
(0, 16).
Window:
x=[10,10]
y=[-10,20]
If so, then
you already know that the domain is all real values, and the range is all
values BELOW the vertex, which would be
or
.
However, as always, the general practice in finding the range is to
solve for
in
terms of
in
order to determine what restrictions there might be for To
find the range, solve for
.
Start by
adding
to both sides.
Next
add
to each side.
Take the
square root of both sides.
The
restriction here is that, because of the square root, the radicand
must
be greater than or equal to zero. That is, the range is
Divide by -1:
Range:
Page 276. # 26.
Solution: You
may recognize that this is a parabola that opens downward with vertex at
(0, −9).
Window:
x=[10,10]
y=[-20,10]
If so, then
you already know that the domain is all real values, and the range is all
values BELOW the vertex, which would be
or
.
However, as always, the general practice in finding the range is to
solve for
in
terms of
in
order to determine what restrictions there might be for To
find the range, solve for
.
Start by
adding
to both sides.
Next
add
to each side.
Take the
square root of both sides.
The
restriction here is that, because of the square root, the radicand
must
be greater than or equal to zero. That is, the range is
Divide by -1:
Range:
Page 278. # 35.
Solution: Since
this equation is in the form ,
the graphing calculator will probably be a good way to find domain and range. Just
enter this in the calculator, and it should look like this:
From this
graph (or from squaring both sides of the equation!), you may recognize this
as the upper half of a parabola that extends upward and to the right to
infinity. The range is all values that are on or above the x-axis, or
,
or in interval notation .
The domain
(it’s easy, and no extra charge!) just from looking at the graph is
Page 278.
# 38.
Solution: The
best bet here is to graph the function, since it is in the form
.
Just enter this in the calculator,
and it should look like this:
From this graph (or from squaring both sides of
the equation!), you may recognize this as the lower half of a parabola that
extends downward and to the right to infinity. The range is all values
below the x-axis, or 
, or in interval notation
.
The domain (it’s easy, and no extra charge!) just
from looking at the graph is 
Page 280.
# 44.
Solution: The
best bet here is to graph the function, since it is in the form
.
Just enter this in the calculator,
and it should look like this:
From this graph (or from squaring both sides of
the equation!), you may recognize this as the lower half of a circle that
extends 3 units down, 3 units to the right, and 3 units to the left. The
range is all values from -3 up to 0, or in interval notation
.
The domain (again, it’s easy, and no extra
charge!) just from looking at the graph is
.
Page 280.
# 46.
Solution: It
will be best is to graph the function, since it is in the form
.
Just enter this in the calculator,
and it should look like this:
From this graph you probably won’t recognize
this graph, but it doesn’t matter, does it? It should be clear that the
values of y extend from negative infinity up to zero. The range is
therefore in interval notation .
The domain (again, it’s easy, and no extra
charge!) just from looking at the graph is
.
.
Functions, Domain, and Range
Page
283.
#3.
To find the domain, you must solve for
in terms of
. This means you
must first get all the terms on one side, and the non-y terms on the other side.
To do this, subtract 
from each side.
Factor out the
:
Divide both sides by
Because the denominator must never equal zero,
this means that 
.
Domain is all.
Also, this
IS a function, since it can be
expressed in the form = _____.
To find the range, you must solve for
in terms of
by subtracting
from each side.
Next, divide both sides by
in order to solve for
.
From this equation, you can see that the
denominator cannot equal zero. That is, the
range is all
.
If you
choose to look at the graph of with
a graphing calculator, it looks like this:
Notice from
this graph, that
domain is all values of x except the asymptote at x=1
(the value
of f(x) is undefined at x=1), Likewise, the
range is
all values of y except .
D: all
.
R:
all
.
#4. 
To find the domain, you must solve for
in terms of
. In this case
there is only one term, so subtract
from each side to isolate the y term.
Now to solve for
, you must divide both sides by
.
From this equation, you can see that the
denominator cannot equal zero. That is, the
domain is all
.
Also, this
IS a function, since it can be
expressed in the form = _____.
To find the range, you must solve for
in terms of
. To do this, you must first get all the x terms on one
side by subtracting
from each side.
Next, factor out the
Next, divide both sides by
in order to solve
for
.
Because the denominator must never equal zero,
this means that 
.
Range is all
.
If you
choose to look at the graph of with
a graphing calculator, it looks like this:
.
The first graph above is the standard window, illustrating that if x=0, the
value of y is undefined. In the second graph, the x window is extended from
[-100,100] to illustrate that the value of y is approaching but never quite
reaches y=4. Notice
from these graphs, that the
domain is all values of x except the asymptote at x=0
(the value
of f(x) is undefined at x=0), Likewise, the
range is
all values of y except .
D: all
.
R:
all .
Page 285. # 18.
Solution: Since
this equation is in the form of a circle, you already know what the graph
looks like. You don’t even need your graphing calculator for this one.
NOTE: If you used a graphing calculator to graph two semicircles,
the semicircles on the calculator may not appear to connect. However,
in reality, the semicircles actually DO connect to form a circle!!
From this
graph of a circle with center at the origin and of radius 4, you can see
that the values of x extend from -4 to 4 inclusive, and the values of y also
extend from -4 to 4 inclusive. It is NOT a function.
Domain: 
Range:
Function: NO!
Page 286. # 19.
Solution: While this is NOT in the form
,
you can always solve for y in terms of x, and then use the graphing
calculator as a way to find domain and range. Begin by adding
to
each side:
Next,
divide both sides by 
,
which gives you
Take the
square root of each side:
Of course
to graph this you must use two graphs:
and 
Just enter
these in the calculator, and the graph should look like this:
Although
you entered the equation as an upper half and lower half graph, the actual
graph looks like a left half and a right half graph. The domain consists
of all values to left of and including -4 and to the right of and including
4 
.
The range extends all the way down to negative infinity and all the way up
to positive infinity. This would be all real values or
.
This is NOT a function.
Domain:
Range:
Function: NO!
P. 287 #25. Preliminary Note: This is NOT a
good calculator problem, unless you enlarge the window!! The standard
window will be VERY misleading!! Please see the calculator explanation
that follows this algebraic solution.
To find the
domain,
you must solve for in
terms of .
In this case there is only one term,
so add 
to
each side to isolate the y term.
Now to
solve for ,
you must divide both sides by .
From this
equation, you can see that the denominator cannot equal zero. That is, the
domain is all
.
Also,
this IS a function,
since it can be expressed in the form
=
_____.
To find the
range,
you must solve for 
in
terms of 
.
To do this, you must first get all the
terms
on one side of the equation. However, surprise!! All the
terms
are already on the left side!!
Now, just
factor out the
Next,
divide both sides by 
in
order to solve for .
Because the
denominator must never equal zero, this means that
.
Range is
all.
CALCULATOR SOLUTION FOR
#25.
If you
choose to solve this with a graphing calculator, the standard window will be
absolutely NO help to you at all. In fact, the standard window,
illustrated on the left below, is VERY misleading!!
The first
graph (on the left above) is the standard window, which does not accurately
reveal the behavior of this graph!! In the second graph, the y window is
extended from [-10, 20]. From this graph, you can see that the
domain is
all values of x except the asymptote at x=0
(the value of f(x) is undefined at x=0). To find the range, you
may suspect that there is a horizontal asymptote—that is, a forbidden value
of y that might be determined by looking at points on the right and/or left
edge of the graph. The next two graphs show the x window from [-20,20] and
y window [-10,20]. To see this, do [TRACE] at x=-20 and x= 20, the extreme
left and right sides of the window.
Can you see
from looking at the right and left sides of this window that the value of y
is approaching but never quite reaches y=10? Therefore, the
range is
all values of y except
.
FINAL
ANSWER: D: all
.
R:
all
,
and this IS a function!!
Extra
Problem:
Solution:
There are
actually three restrictions here:
1.
The first denominator is also a radical, so
.
2.
The second numerator is a radical, so
3.
The second denominator cannot equal zero, so
The domain of this
function is the intersection of all three of these restrictions. However,
in this case, since x must be greater than -5 and greater than or equal to
4, it will automatically not equal -6. The third restriction does not
affect the solution.
and
and
“And”
means “intersection”,
so choose only the regions that are common to all three graphs:
Interval notation:
[4, ∞)
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