In Problems 5.1-5.19, find the domain and range, and draw the graph, of the function determined by the given formula.
5.1 h(x) = 4-x2.
5.2
5.3
5.4
5.5
The domain consists of all real numbers, since 4 — x2 is defined for all x. The range consists of all real numbers < 4: solving the equation y = 4 — x2 for x, we obtain x = ±\/4 — y, which is defined when and only when y < 4. The graph (Fig. 5-1) is a parabola with vertex at (0, 4) and the y-axis as its axis of symmetry.
Fig. 5-1 G(x) = -2Vx.
The domain consists of all nonnegative real numbers. The range consists of all real numbers s 0. The graph (Fig. 5-2) is the lower half of the parabola 4x = y2.
The domain is the closed interval [-2,2], since V4-x2 is defined when and only when *2s4. The graph (Fig. 5-3) is the upper half of the circle x2 + y2 - 4 with center at the origin and radius 2. The range is the closed interval [0,2].
Fig. 5-3 Fig. 5-4
omain consists of all x such that x SL 2 or x ^ —2, since we must have x2 SL 4. The graph (Fig.I The domain consists of all x such that x SL 2 or x ^ —2, since we must have x2 SL 4. The graph (Fig.
5-4) is the part of the hyperbola x2- y2 = 4 on or above the x-axis. The range consists of all nonnegative real numbers.
V(x) = \x-l\.
The domain is the set of all real numbers. The range is the set of all nonnegative real numbers. The graph (Fig. 5-5) is the graph of y = \x\ shifted one unit to the right.
23 Fig. 5-2
Fig.5.5
f(x) = [2x] = the greatest integer :£ 2x.
The domain consists of all real numbers. The range is the set of all integers. The graph (Fig. 5-6) is the graph of a step function, with each step of length | and height 1.
Fig. 5-6 Fig. 5-7
g(jc) = [x/3] (see Problem 5.6).
The domain is the set of all real numbers and the range is the set of all integers. The graph (Fig. 5-7) is the graph of a step function with each step of length 3 and height 1.
The domain is the set of all nonzero real numbers, and the range is the same set. The graph (Fig. 5-8) is the hyperbola xy = 1.
Fig. 5-8 Fig. 5-9
5.10
The domain is the set of all real numbers ^ 1. The graph (Fig. 5-9) is Fig. 5-8 shifted one unit to the right.
The range consists of all nonzero real numbers.
The domain is the set of all real numbers, and the range is the same set. See Fig. 5-10.
Fig. 5-10 Fig. 5-11
The domain and range are the set of all real numbers. The graph (Fig. 5-11) is obtained by reflecting in the jc-axis that part of the parabola y = x2 that lies to the right of the >>-axis.
CHAPTER 5 5.6
24
5.7
5.8
5.9
5.11 J(x)=-x\x\.
FUNCTIONS AND THEIR GRAPHS 5.12
5.13
The domain is the set of all real numbers. The graph (Fig. 5-12) consists of two half lines meeting at the point (1, 2). The range is the set of all real numbers a 2.
Fig. 5-12 Fig. 5-13
The domain is {1, 2, 4}. The range is {-1,3}. The graph (Fig. 5-13) consists of three points.
graph (Fig. 5-14) consists of all points on the line y = x — 2 except the point (—2, -4). The range is the set of all real numbers except -4.
5.15
Fig. 5-15
The domain is the set of all real numbers. The graph (Fig. 5-15) is made up of the left half of the line y = x for .vs2 and the right half of the line y = 4 for x>2. The range consists of all real numbers < 2, plus the number 4.
The domain is the set of all nonzero real numbers. The graph (Fig. 5-16) is the right half of the line y = 1 for x>0, plus the left half of the line y = -\ for x<Q. The range is {1,-1}.
Fig. 5-17
The domain is the set of all real numbers. The graph (Fig. 5-17) is a continuous curve consisting of three pieces: the half of the line y=\ — x to the left of jt = -l, the horizontal segment y = 2 between j t = - l and x = l, and the part of the parabola y — x2 + 1 to the right of x = l. The range consists of all real numbers == 2.
Fig. 5-16 /(!)=-!, /(2) = 3, /(4) = -l.
Fig. 5-14
I The domain is the set of all real numbers except —2. Since for the
5.14
5.16
5.17
25
for for
if if
ifif if
26 D CHAPTER 5
5.18
h(x) = x + 3 for x = 3, the graph (Fig. 5-18) is the straight line y = x + 3. The range is the set of all real numbers.
Fig. 5-18
5.20
5.21
5.22
The domain is the set of all real numbers. The graph (Fig. 5-19) is the reflection in the line y = x of the graph of y = x*. [See Problem 5.100.] The range is the set of all real numbers.
Is Fig. 5-20 the graph of a function?
Since (0,0) and (0,2) are on the graph, this cannot be the graph of a function.
Is Fig. 5-21 the graph of a function?
Since some vertical lines cut the graph in more than one point, this cannot be the graph of a function.
Is Fig. 5-22 the graph of a function?
Since each vertical line cuts the graph in at most one point, this is the graph of a function.
Fig. 5-22 Fig. 5-19
Fig. 5-20 Fig. 5-21
I The domain is the set of all real numbers. Since for and
5.19
if if
5.23 Is Fig. 5-23 the graph of a function?
Since each vertical line cuts the graph in at most one point, this is the graph of a function.
5.24
5.25
5.26
Find a formula for the function f(x) whose graph consists of all points (x, y) such that x*y — 2 = 0, and specify the domain of f(x).
f(x) = 2/x3. The domain is the set of all nonzero real numbers.
Find a formula for the function/(x) whose graph consists of all points (x, y) such that the domain of f(x).
x(\ - y) = 1 + y, x-xy = l + y, y(x + l) = x-l, the set of all real numbers different from -1.
Find a formula for the function f(x) whose graph consists of all points (x, y) such that x2 - 2xy + y2 = 0, and specify the domain of f(x).
The given equation is equivalent to (x - y)2 = 0, x - y = 0, y = x. Thus, f(x) = x, and the domain is the set of all real numbers.
In Problems 5.27-5.31, specify the domain and range of the given function.
5.31
The domain consists of all real numbers except 2 and 3. To determine the range, set v =
x is in the domain if and only if x2 <1. Thus, the domain is (-1,1). To find the range, first note that g(x)>0. Then set y = 1 /V1 - x2 and solve for A:, y2 = 1/(1 - x2), x2 = 1 - lly2 >0, l a l / y2, y22:l, y s l . Thus, the range is [1,+00).
The domain is (-1, +00). The graph consists of the open segment from (-1,0) to (1, 2), plus the half line of y = 2 with x > 1. Hence, the range is the half-open interval (0, 2].
The domain is [0,4). Inspection of the graph shows that the range is [-1,2].
G(x) = \x\-x.
The domain is the set of all real numbers. To determine the range, note that G(x) = 0 if * > 0, and G(x) = — 2x if x<0. Hence, the range consists of all nonnegative real numbers.
FUNCTIONS AND THEIR GRAPHS 27
Fig. 5-23
and solve for This has a solution when and only when This holds if and only if This holds when
and, if when Hence the range is
and specify
and the domain is So
5.27
5.28
5.29
5.30
if if
if if
28
5.32 Let
= x - 4 = /(x) if Jt^-4. Since /(-4) = -8, we must set k=-8.
/is not defined when x = 0, but g is defined when x = 0.
In Problems 5.34-5.37, define one function having set @ as its domain and set i% as its range.
2> = (0,1) and $=(0,2).
Let f(x) = 2x for 0<*<1.
2> = [0,1) and & = [-!, 4).
Look for a linear function /(*) = mx + b, taking 0 into -1 and 1 into 4. Then b = -\, and 4 = m - l , m = 5 . Hence, /(x) = 5x-l.
® = [0,+°°) and $ = {0,1}.
/(0) = 0, and /(*) = ! for *>0.
2i = (-», l)u(l,2) and $ = (!,+»).
In Problems 5.38-5.47, determine whether the function is even, odd, or neither even nor odd.
Fig. 5-24 /« = 9-x2.
ce /(-*) = 9 - (-*)2 = 9 - *2 = /(*), /W is even.I Since /(-*) = 9 - (-*)2 = 9 - *2 = /(*), /W is even.
A*) = V3t.
For a function/(x) to be considered even or odd, it must be defined at -x whenever it is defined at x. Since /(I) is defined but/(-I) is not defined, f(x) is neither even nor odd.
/(*) = 4-*2.
This function is even, since /(-*)=/(*).
Since |-x| = |;e|, this function is even.
Determine k so that f(x) = g(*) for all x.
5.33
5.34
5.35
5.36
5.37
5.38
5.39
5.40
5.41 /W =
W-Let and g(x) = x — 1. Why is it wrong to assert that / and g are the same function?
Let for and for See Fig. 5-24.
CHAPTER 5
f(x)=x-4 if
if
5.43
[j] = 0 and [-§] = —1. Hence, this function is neither even nor odd.
f(x)=
f(x)=
/« =
|*-1|-/(1) = 0 and /(-1) = |-1-1| = 2. So, fix) is neither even nor odd.
The function J(x) of Problem 5.11.
J(—x) = —(—x) \—x\ = x \x\ = —J(x), so this is an odd function.
fix) = 2x + l.
/(I) = 3 and /(-!)=-!. So, f(x) is neither even nor odd.
Show that a function f(x) is even if and only if its graph is symmetric with respect to the y-axis.
Assume that fix) is even. Let (jc, y) be on the graph of/. We must show that (—x, y) is also on the graph of /. Since (x, y) is on the graph of /, f(x) = y. Hence, since/is even, f(~x)=f(x) = y, and, thus, (—x, y) is
on the graph of/. Conversely, assume that the graph of/is symmetric with respect to the _y-axis. Assume that fix) is defined and f(x) = y. Then (x, y) is on the graph of/. By assumption, (-x, y) also is on the graph of/.
Hence, f ( ~ x ) = y. Then, /(-*)=/(*), and fix) is even.
Show that/(*) is odd if and only if the graph of/is symmetric with respect to the origin.
Assume that fix) is odd. Let (x, y) be on the graph of/. Then fix) = y. Since fix) is odd, fi—x) =
—fix) = —y, and, therefore, (—x, -y) is on the graph of /. But, (x, y) and (~x, -y) are symmetric with respect to the origin. Conversely, assume the graph of / is symmetric with respect to the origin. Assume fix) = y. Then, (x, y) is on the graph of/. Hence, by assumption, (—x, -y) is on the graph of/. Thus, fi-x) = — y = -fix), and, therefore, fix) is odd.
Show that, if a graph is symmetric with respect to both the x-axis and the y-axis, then it is symmetric with respect to the origin.
Assume (x, y) is on the graph. Since the graph is symmetric with respect to the jt-axis, (x, -y) is also on the graph, and, therefore, since the graph is symmetric with respect to the y-axis, (—x, —y) is also on the graph.
Thus, the graph is symmetric with respect to the origin.
Show that the converse of Problem 5.50 is false.
The graph of the odd function fix) = x is symmetric with respect to the origin. However, (1,1) is on the graph but (—1,1) is not; therefore, the graph is not symmetric with respect to the y-axis. It is also not symmetric with respect to the x-axis.
If / is an odd function and /(O) is denned, must /(O) = 0?
Yes. /(0)=/(-0) = -/(0). Hence, /(0) = 0.
If fix) = x2 + kx + 1 for all x and / is an even function, find k.
il) = 2 + k and /(-l) = 2-fc. By the evenness of/, /(-!) = /(!). Hence, 2+k = 2-k, k = -k, k = 0.
FUNCTIONS AND THEIR GRAPHS 29 5.42
5.44
5.45
5.46
5.47
5.48
5.49
5.50
5.51
5.52
5.53
fix) = [x]
Hence, this function is odd.
So, fix) is odd.
Show that any function F(x) that is defined for all x may be expressed as the sum of an even function and an odd function: F(x) = E(x) + O(x).
Take E(x)= |[F(*) + F(-x)] and O(x) = \[F(x) - F(-x)].
Prove that the representation of F(x) in Problem 5.55 is unique.
If F(x) = E(x) + O(x) and F(x) = E*(x) + O*(x), then, by subtraction,
0 = e(x) + o(x) (1)
where e(x) = E*(x) - E(x) is even and o(x) = O*(x) - O(x) is odd. Replace x by -x in (1) to obtain 0=e(x)-o(x) (2) But (1) and (2) together imply e(x) = o(x) = 0; that is, £*(*) = E(x) and O*(x) = O(x).
In Problems 5.57-5.63, determine whether the given function is one-one.
f(x) = mx + b for all x, where m^O.
Assume f(u)=f(v). Then, mu + b = mv + b, mu = mv, u = v. Thus, /is one-one.
/(*) = Vx for all nonnegative x.
Assume f ( u ) = f ( v ) . Then, Vu = Vv. Square both sides; u = v. Thus,/is one-one.
f(x) = x2 for all x.
/(-I) = 1 =/(!). Hence, /is not one-one.
f(x) = - for all nonzero x.
f(x) = \x\ for all x.
/(—I) = 1 = /(I). Hence, /is not one-one.
f(x) = [x] for all x.
/(O) = 0 = /(|). Hence, /is not one-one.
f(x) = x3 for all x.
Assume /(M)=/(U). Then u3 = v3. Taking cube roots (see Problem 5.84), we obtain u = v. Hence,/
is one-one.
In Problems 5.64-5.68, evaluate the expression f(x) = x2-2x.
f(x + h) = (x + h)2 - 2(x + h) = (x2 +2xh + h2)-2x- 2h. So, f(x + h)- f(x) = [(x2 + 2xh + h2)
f(x) = x + 4.
f(x + h) = x + h + 4. So, f(x + h)-f(x) = (x + h+4)-(x + 4) = h. Hence,
30 CHAPTER 5
5.54 If f(x) = x3-kx2 + 2x for all x and if / is an odd function, find k.
f ( l ) = 3-k and /(-l)=-3-Jfc. Since / is odd, -3 - k = -(3- k) = -3 + k. Hence, -k = k, t = n
5.55
5.56
5.57
5.58
5.59
5.60
5.61
5.62
5.63
5.64
5.65
for the given function /.
Hence,
I Assume f(u)=f(v). Then Hence, u = v. Thus,/is one-one.
5.72
Does a self-inverse function exist? Is there more than one?
See Problems 5.69 and 5.74.
In Problems 5.76-5.82, find all real roots of the given polynomial.
x4 - 10x2 + 9.
x4- Wx2 + 9=(x2- 9)(x2 -!) = (*- 3)(* + 3)(x - l)(x + 1). Hence, the roots are 3, -3,1, -1.
x3 + 2x2 - 16x - 32.
Inspection of the divisors of the constant term 32 reveals that -2 is a root. Division by x + 2 yields the
factorization (x + 2)(x2 - 16) = (x + 2)(x2 + 4)(x2 - 4) = (x + 2)(x2 + 4)(* - 2)(x + 2). So, the roots are 2 andfactorization (x + 2)(x2 - 16) = (x + 2)(x2 + 4)(x2 - 4) = (x + 2)(x2 + 4)(* - 2)(x + 2). So, the roots are 2 and _2
FUNCTIONS AND THEIR GRAPHS 31 5.66
5.67
f(x) =f(x) = X3 + l.
f(x + h) = (x + h)3 + l = x3 + 3x2h + 3xh2 + h3 + l. So, f(x + h) f(x) = (x3 + 3x2h + 3xh2 + h3 + 1) -(x3 + l) = 3x2h + 3xh2 + h3 = h(3x2 + 3xh + h2).
f(x) = Vx.
5.68
Hence.
In Problems 5.69-5.74, for each of the given one-one functions f(x), find a formula for the inverse function
r\y).
f(x) = x.
Let >>=/(*) = *. So, x = y. Thus, f~\y) = y.
f(x) = 2x + l.
Let y = 2x + l and solve for x. x=\(y-\). Thus, f~\y) = \(y - 1).
f(x) = x3.
Let y = x3. Then x = \/y. So, f~\y)=^/J.
5.69
5.70
5.71
Thus, rl(y) = (y-l)/(y + l).
5.73
5.74
Then So
5.75
5.76
5.77 So,
Hence, = 3x2 + 3xh + h2.
Let y = l/x. Then x = \ly. So, f~\y) = l/y.
I Let
I Let Then y(\ - x) = 1 + x, y-yx = l + x, y - 1 = x(\ + y), x = (y -l)/(y + 1).
for
x* - jc3 - lOx2 + 4x + 24.
ng the divisors of 24 yields the root 2. Division by x 2 yields the factorization (x 2)(*3 + x2 I Testing the divisors of 24 yields the root 2. Division by x 2 yields the factorization (x 2)(*3 + x2 -8x - 12). It turns out that-2 is another root; division of x3 + x2 - 8x - 12 by x + 2 gives (x-2)(x + 2)(x2 -x-6) = (x- 2)(x + 2)(x -3)(x + 2). So, the roots are 2, -2, and 3.
x3 - 2x2 + x - 2.
x3 - 2x2 + x - 2 = x\x -2) + x-2 = (x- 2)(x2 + 1). Thus, the only real root is 2.
x3 + 9x2 + 26x + 24.
Testing the divisors of 24, reveals the root -2. Dividing by x + 2 yields the factorization (x + 2)(x2 + Ix + 12) = (x + 2)(x + 3)(;t + 4). Thus, the roots are -2, -3, and -4.
Ar3-5jc-2.
-2 is a root. Dividing by x + 2 yields the factorization (x + 2)(x2 — 2x — 1). The quadratic formula applied to x2 - 2x — I gives the additional roots 1 ± V2.
x3 - 4x2 -2x + 8.
x3-4x2-2x + 8 = x2(x -4) - 2(x -4) = (x- 4)(x2 - 2). Thus, the roots are 4 and ±V2.
Establish the factorization
w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l)w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l) for n = 2, 3,
Simply multiply out the right-hand side. The cross-product terms will cancel in pairs (u times the kth term of the second factor will cancel with -v times the (k - l)st term).
Prove algebraically that a real number x has a unique cube root.
Suppose there were two cube roots, u and i>, so that u3 = i>3 = x, or u3 - v3 = 0. Then, by Problem 5.83,
Unless both u and v are zero, the factor in brackets is positive (being a sum of squares); hence the other factor must vanish, giving u = v. If both u and v are zero, then again u = v.
If f(x) — (x + 3)(x + k), and the remainder is 16 when f(x) is divided by x ~ 1, find k.
f(x) = (x~l)q(x) + \6. Hence, /(I) = 16. But, /(I) = (1 + 3)(1 + k) =4(1 + k). So, l + k = 4, k = 3.
If f(x) = (x + 5)(x — k) and the remainder is 28 when f(x) is divided by x — 2, find k.
f(x) = (x-2)q(x) + 2&. Hence, /(2) = 28. But, f(2) = (2 + 5)(2 - k) = 7(2 - k). So, 2 - it = 4, k= -2.
If the zeros of a function f(x) are 3 and -4, what are the zeros of the function g(x) =/(jt/3)?
/(jt/3) = 0 if and only if x/3 = 3 or */3=-4, that is, if and only if * = 9 or x=-12.
Describe the function f(x) = |jt| + j* — 1| and draw its graph.
Case 1. jcsl. Then /(*) = * + *-1 =2x - 1. Case 2. Os * < 1. Then f(x) = x - (x ~ l)= \.
Case 3. Jt<0. Then /(*)= -x - (x - 1) = -2x + 1. So, the graph (Fig. 5-25) consists of a horizontal line segment and two half lines.
CHAPTER 5 32
5.78
5.79
5.80
5.81
5.82
5.83
5.84
5.85
5.86
5.87
5.88
FUNCTIONS AND THEIR GRAPHS
Fig. 5-25 Find the domain and range of f(x) = V5 — 4x - x2.
ompleting the square, x2 + 4x - 5 = (x + 2)2 - 9. So, 5 - 4x - x2 = 9 - (x + 2)2. For the functionI By completing the square, x2 + 4x - 5 = (x + 2)2 - 9. So, 5 - 4x - x2 = 9 - (x + 2)2. For the function to be defined we must have (x + 2)2s9, -3==* +2s3, -5<*sl. Thus, the domain is [-5,1]. For*
in the domain, 9 > 9 - (x + 2)2 > 0, and, therefore, the range will be [0,3].
Show that the product of two even functions and the product of two odd functions are even functions.
If / and g are even, then f(~x)-g(-x) = f(x)-g(x). On the other hand, if / and g are odd, then
/(-*) • g(-x) = [-/(*)] • [-«<*)] = /W •
gM-Show that the product of an even function and an odd function is an odd function.
Let /be even and g odd. Then f(-x)-g(-x) =/(*)• [-g«] = -f(x)-g(x).
Prove that if an odd function f(x) is one-one, the inverse function g(y) is also odd.
Write y=f(*)', then * = g(}') and, by oddness, f(~x)=—y, or —x — g(—y). Thus, g(—y) = -g(y), and g is odd.
5.93
5.94
5.95
5.%
5.97
5.98
What can be said about the inverse of an even, one-one function?
Anything you wish, since no even function is one-one [/(-*) =/(*)]•
Find an equation of the new curve C* when the graph of the curve C with the equation x2 - xy + y2 = 1 is reflected in the x-axis.
(x, y) is on C* if and only if (x, —y) is on C, that is, if and only if x2 — x(—y) + (—y)2 — 1, which reduces to x2 + xy + y2 = 1.
Find the equation of the new curve C* when the graph of the curve C with the equation y3 — xy2 + x3 = 8 is reflected in the y-axis.
is on C* if and only if (-x, y) is on C, that is, if and only if y3 - (~x)y2 + (-x)3 = 8, which reducesI (x, y) is on C* if and only if (-x, y) is on C, that is, if and only if y3 - (~x)y2 + (-x)3 = 8, which reduces to y3 + xy2 - x3 = 8.
Find the equation of the new curve C* obtained when the graph of the curve C with the equation x2 - 12x + 3y = 1 is reflected in the origin.
(x, y) is on C* if and only if (-x, -y) is on C, that is, if and only if (-x)2 - 12(-x) + 3(-y) = 1, which reduces to x2 + 12x — 3y = 1.
Find the reflection of the line y = mx + b in the y-axis.
We replace x by —AC, obtaining y = — mx + b. Thus, the y-intercept remains the same and the slope changes to its negative.
Find the reflection of the line y = mx + b in the x-axis.
We replace y by -y , obtaining -y = mx + b, that is, y = - mx ~ b. Thus, both the y-intercept and the slope change to their negatives.
5.89
5.90
5.91
5.92
33
Find the reflection of the line y = mx + b in the origin.
e replace x by -x and y by -y, obtaining -y = -mx+b, that is, y = mx-b. Thus, the y-intercept changes to its negative and the slope remains unchanged.
Show geometrically that when the graph of a one-one function is reflected in the 45° line y = x, the result is the graph of the inverse function.
It is evident from Fig. 5-26 that right triangles ORP and OR'P' are congruent. Hence,
Thus the locus of P' is the graph of x as a function of y; i.e., of x = f~i(y). Note that because y = f(x) meets the horizontal-line test (/being one-one), x = f ~ ( y ) meets the vertical-line test.
Fig. 5-26 5.101 Graph the function f(x) = V|*-l|-l.
The complement of the domain is given by |jt-l|<l, or -Kx-Kl, or 0<*<2. Hence, the domain consists of all x such that x < 0 or x a 2. Case 1. x a 2. Then y = Vx^2, y2 = * - 2.
So, we have the top half of a parabola with its vertex at (2,0) and the jt-axis as axis of symmetry. Case 2.
x < 0. Then, y = V^x, y2 = -x. So, we have the top half of a parabola with vertex at the origin and with as axis of symmetry. The graph is shown in Fig. 5-27.
Fig. 5-27 34
5.99
5.100
CHAPTER 5
and
R'P'=RP=y OR'=OR=x