- b. = = =
– 39 ((xx++ 11 ))
2
= –3(x+ 1) = –g(x)- d.Since the function 2g(x)h(x) = is a
rational function, its domain is the set of all
those x-values for which the denominator is
not equal to zero. There is no real number x
that satisfies the equation x^2 + 1 = 0. There-
fore, the domain is the set of all real numbers.- b.
3 f(x) – 2xg(x) –= 3[– (2x– (–1 – x^2 ))] – 2x[3(x+ 1)] –= –3(x^2 + 2x+ 1) – 6x(x+ 1) – (x^2 + 1)
= –3x^2 – 6x– 3 – 6x^2 – 6x– x^2 – 1
= – 10x^2 – 12x– 4
= – 2(5x^2 + 6x+ 2)Set 44 (Page 105)
- b.The graph of the equation in diagram Ais
not a function. A function is an equation in
which each unique input yields no more than
one output. The equation in diagram Afails
the vertical line test for all x-values where
–2x 2. For each of these x-values
(inputs), there are two y-values (outputs). - d.The range of a function is the set of possible
outputs of the function. In each of the five
equations, the set of possible y-values that can
be generated for the equation is the range of
the equation. Find the coordinate planes that
show a graph that extends below the x-axis.
These equations have negative y-values, which
means that the range of the equation contains
negative values. The graphs of the equations in
diagrams A,B, andDextend below the x-axis.
However, the graph of the equation in diagram
Ais not a function. It fails the vertical line test
for all x-values where –2 x 2. The equa-
tions graphed in diagrams BandDare func-
tions whose ranges contain negative values.
691. e.The equation of the graph in diagram Bis
y= |x| – 3. Any real number can be substituted
into this equation. There are no x-values that
will generate an undefined or imaginary y-value.
The equation of the graph in diagram Eis y=
(x– 3)^2 + 1.With this equation as well, any real
number can be substituted for x—there are no
x-values that will generate an undefined or
imaginary y-value. The equation of the graph
in diagramDis y= ^1 x.Ifx= 0, this function
will be undefined. Therefore, the domain of
this function is all real numbers excluding 0.
Only the functions in diagrams Band Ehave a
domain of all real numbers with no exclusions.- b.The equation of the graph in diagram Cis y
= x. Since the square root of a negative
number is imaginary, the domain of this equa-
tion is all real numbers greater than or equal
to 0. The square roots of real numbers greater
than or equal to 0 are also real numbers that
are greater than or equal to 0. Therefore, the
range of the equation y= xis all real num-
bers greater than 0, and the domain and range
of the equation are the same. The equation of
the graph in diagramDis y= ^1 x.Ifx= 0, this
function will be undefined. Therefore, the
domain of this function is all real numbers
excluding 0. Dividing 1 by a real number
(excluding 0) will yield real numbers, exclud-
ing 0. Therefore, the range of the equation
y= ^1 xis all real numbers excluding 0, and the
domain and range of the equation are the
same. The equation of the graph in diagram B
is y= |x| – 3. Any real number can be substituted
^11
x^2 + 1^1
h(x)6(x+ 1)
x^2 + 1–9(x^2 + 2x+ 1)
3(x+ 1)
9[ –(2x– (–1 – x^2 ))]
3(1 + x)
^9 f(x)
g(x)ANSWERS & EXPLANATIONS–