- b.The domain of the function is the set of all
real numbers, so any real number can be sub-
stituted for x. The range of a function is the set
of all possible outputs of the function. Since
the xterm is squared, then made negative, the
largest value that this term can equal is 0 (when
x= 0). Every other xvalue will result in a neg-
ative value for f(x). As such, the range off(x)
is the set of all real numbers less than or equal
to 0.
- c.You must identify all possible y-values that
are attained within the graph off. The graph
offis comprised of three distinct components,
each of which contributes an interval of values
to the range off. The set ofy-values correspond-
ing to the bottommost segment is (–2,–1];
note that –2 is excluded due to the open hole
at (5,–2) on the graph, and there is no other x-
value in [–5,5] whose functional value is –2.
Next, the portion of the range corresponding
to the middle segment is [0,2); note that 2 is
excluded from the range for the same reason
–2 is excluded. Finally, the horizontal segment
contributes the singleton {3} to the range; even
though there is a hole in the graph at (0,3),
there are infinitely many other x-values in
[–5,5] whose functional value is 3, thereby
requiring that it be included in the range.
Thus, the range is (–2, –1]∪[0, 2)∪{3}.
- c.The graph ofgis steadily decreasing from
left to right, beginning at the point (–5,–4)
and ending at (5,–4), with the only gap occur-
ring in the form of a hole at (0,1). Since there
is no x-value in [–5,5] whose functional value
is 1, this value must be excluded from the
range. All other values in the interval [–4,4]
do belong to the range. Thus, the range is
[–4,1)∪(1, 4].
680. d.Using the graphs yields f(0) = 0,f(2) = –1,
and g(4) = –4. Substituting these values into
the given expression yields
2 f(0) + [f(2)g(4)]^2 = 2(0) + [(–1)( –4)]^2 =
0 + 4^2 = 16
- b.The zeros of a polynomial are precisely its x-
intercepts, which are –3, 1, and 3.
- c.The lowest point on the graph ofy= p(x)
occurs at (2, –1), so the smallest possible y-
value attained is –1. Further, every real num-
ber greater than –1 is also attained at some
x-value. Hence, the range is [–1,∞).
- b.The domain of any polynomial function is
because any real number can be substituted
in for xin p(x) and yield another real number.
- d.We must identify the x-values of the por-
tion of the graph ofy= p(x) that lies between
the horizontal lines y= –1 and y= 0 (i.e.,
the x-axis). Once this is done, we exclude
the x-values of the points where the graph
ofy= p(x) intersects the line y= –1 (because
of the strict inequality), and we include those
x-values of the points where the graph ofy=
p(x) intersects the line y= 0. This yields the set
[1, 2)∪(2, 3]∪{–3}.
- b.The graph offhas a vertical asymptote at
x= 1 and a horizontal asymptote at y= –2.
Since the graph follows the vertical asymptote
up to positive infinity as xapproaches x= 1
from the left and down to negative infinity as x
approaches x= 1 from the right, and it does
not cross the horizontal asymptote, we con-
clude that the graph attains all y-values except
–2. Hence, the range is (– ∞, – 2)∪(–2,∞).
ANSWERS & EXPLANATIONS–
