1 FUNCTIONS AND MODELS
1.1 Four Ways to Represent a Function
In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.
1. (a) The point(−1,−2)is on the graph off, sof(−1) =−2.
(b) Whenx= 2,yis about2.8, sof(2)≈2.8.
(c)f(x) = 2is equivalent toy= 2. Wheny= 2, we havex=−3andx= 1.
(d) Reasonable estimates forxwheny= 0arex=−2.5andx= 0.3.
(e) The domain offconsists of allx-values on the graph off. For this function, the domain is−3≤x≤3, or[−3,3].
The range offconsists of ally-values on the graph off. For this function, the range is−2≤y≤3, or[−2,3].
(f ) Asxincreases from−1to3,yincreases from−2to3. Thus,fis increasing on the interval[−1,3].
3. From Figure 1 in the text, the lowest point occurs at about(t, a) = (12,−85). The highest point occurs at about(17,115).
Thus, the range of the vertical ground acceleration is−85≤a≤115. Written in interval notation, we get[−85,115].
5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails the Vertical Line Test.
7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is[−3,2]and the range is [−3,−2)∪[−1,3].
9. The person’s weight increased to about160pounds at age20and stayed fairly steady for10years. The person’s weight dropped to about120pounds for the next5years, then increased rapidly to about170pounds. The next30years saw a gradual increase to190pounds. Possible reasons for the drop in weight at30years of age: diet, exercise, health problems.
11. The water will cool down almost to freezing as the ice melts. Then, when the ice has melted, the water will slowly warm up to room temperature.
13.Of course, this graph depends strongly on the geographical location!
15. As the price increases, the amount sold decreases. 17.
9
19. (a) (b) From the graph, we estimate the number of cell-phone subscribers worldwide to be about 92 million in 1995 and 485 million in 1999.
21. f(x) = 3x2−x+ 2.
f(2) = 3(2)2−2 + 2 = 12−2 + 2 = 12.
f(−2) = 3(−2)2−(−2) + 2 = 12 + 2 + 2 = 16.
f(a) = 3a2−a+ 2.
f(−a) = 3(−a)2−(−a) + 2 = 3a2+a+ 2.
f(a+ 1) = 3(a+ 1)2−(a+ 1) + 2 = 3(a2+ 2a+ 1)−a−1 + 2 = 3a2+ 6a+ 3−a+ 1 = 3a2+ 5a+ 4.
2f(a) = 2·f(a) = 2(3a2−a+ 2) = 6a2−2a+ 4.
f(2a) = 3(2a)2−(2a) + 2 = 3(4a2)−2a+ 2 = 12a2−2a+ 2.
f(a2) = 3(a2)2−(a2) + 2 = 3(a4)−a2+ 2 = 3a4−a2+ 2.
[f(a)]2=
3a2−a+ 22
=
3a2−a+ 2
3a2−a+ 2
= 9a4−3a3+ 6a2−3a3+a2−2a+ 6a2−2a+ 4 = 9a4−6a3+ 13a2−4a+ 4.
f(a+h) = 3(a+h)2−(a+h) + 2 = 3(a2+ 2ah+h2)−a−h+ 2 = 3a2+ 6ah+ 3h2−a−h+ 2.
23. f(x) = 4 + 3x−x2, sof(3 +h) = 4 + 3(3 +h)−(3 +h)2 = 4 + 9 + 3h−(9 + 6h+h2) = 4−3h−h2, andf(3 +h)−f(3)
h = (4−3h−h2)−4
h =h(−3−h)
h =−3−h.
25. f(x)−f(a) x−a =
1 x−1
a x−a =
a−x xa
x−a = a−x
xa(x−a) =−1(x−a) xa(x−a) =− 1
ax
27. f(x) =x/(3x−1)is defined for allxexcept when0 = 3x−1 ⇔ x=13, so the domain is
x∈R|x6=13
=
−∞,13
∪1
3,∞. 29. f(t) =√
t+√3
tis defined whent≥0. These values oftgive real number results for√
t, whereas any value oftgives a real number result for√3t. The domain is[0,∞).
31. h(x) = 1√4
x2−5xis defined whenx2−5x >0 ⇔ x(x−5)>0. Note thatx2−5x6= 0since that would result in division by zero. The expressionx(x−5)is positive ifx <0orx >5. (See Appendix A for methods for solving
inequalities.) Thus, the domain is(−∞,0)∪(5,∞).
SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11 33.f(x) = 5is defined for all real numbers, so the domain isR, or(−∞,∞).
The graph offis a horizontal line withy-intercept5.
35.f(t) =t2−6tis defined for all real numbers, so the domain isR, or (−∞,∞). The graph offis a parabola opening upward since the coefficient oft2is positive. Tofind thet-intercepts, lety= 0and solve fort.
0 =t2−6t=t(t−6) ⇒ t= 0andt= 6. Thet-coordinate of the vertex is halfway between thet-intercepts, that is, att= 3. Since f(3) = 32−6·3 =−9, the vertex is(3,−9).
37.g(x) =√
x−5is defined whenx−5≥0orx≥5, so the domain is[5,∞).
Sincey=√
x−5 ⇒ y2=x−5 ⇒ x=y2+ 5, we see thatgis the top half of a parabola.
39.G(x) = 3x+|x|
x . Since|x|= x if x≥0
−x if x <0, we have
G(x) =
⎧⎪
⎨
⎪⎩ 3x+x
x if x >0 3x−x
x if x <0
=
⎧⎪
⎨
⎪⎩ 4x
x if x >0 2x
x if x <0
= 4 if x >0 2 if x <0 Note thatGis not defined forx= 0. The domain is(−∞,0)∪(0,∞).
41.f(x) = x+ 2 if x <0 1−x if x≥0 The domain isR.
43.f(x) = x+ 2 if x≤ −1 x2 if x >−1
Note that forx=−1, bothx+ 2andx2are equal to 1.
The domain isR.
45.Recall that the slopemof a line between the two points(x1, y1)and(x2, y2)ism= y2−y1
x2−x1
and an equation of the line connecting those two points isy−y1=m(x−x1). The slope of this line segment is7−(−3)
5−1 =5
2, so an equation is y−(−3) =52(x−1). The function isf(x) =52x−112,1≤x≤5.
47. We need to solve the given equation fory. x+ (y−1)2= 0 ⇔ (y−1)2=−x ⇔ y−1 =±√
−x ⇔ y= 1±√
−x. The expression with the positive radical represents the top half of the parabola, and the one with the negative radical represents the bottom half. Hence, we wantf(x) = 1−√
−x. Note that the domain isx≤0.
49. For0≤x≤3, the graph is the line with slope−1andy-intercept3, that is,y=−x+ 3. For3< x≤5, the graph is the line with slope2passing through(3,0); that is,y−0 = 2(x−3), ory= 2x−6. So the function is
f(x) = −x+ 3 if 0≤x≤3 2x−6 if 3< x≤5
51. Let the length and width of the rectangle beLandW. Then the perimeter is2L+ 2W = 20and the area isA=LW.
Solving thefirst equation forWin terms ofLgivesW = 20−2L
2 = 10−L. Thus,A(L) =L(10−L) = 10L−L2. Since lengths are positive, the domain ofAis0< L <10. If we further restrictLto be larger thanW, then5< L <10would be the domain.
53. Let the length of a side of the equilateral triangle bex. Then by the Pythagorean Theorem, the heightyof the triangle satisfies y2+ 12x 2=x2, so thaty2 =x2−14x2=34x2andy=√23x. Using the formula for the areaAof a triangle,
A=12(base)(height), we obtainA(x) = 12(x) √23x =√43x2, with domainx >0.
55. Let each side of the base of the box have lengthx, and let the height of the box beh. Since the volume is2, we know that 2 =hx2, so thath= 2/x2, and the surface area isS=x2+ 4xh. Thus,S(x) =x2+ 4x(2/x2) =x2+ (8/x), with domainx >0.
57. The height of the box isxand the length and width areL= 20−2x,W= 12−2x. ThenV =LW xand so V(x) = (20−2x)(12−2x)(x) = 4(10−x)(6−x)(x) = 4x(60−16x+x2) = 4x3−64x2+ 240x.
The sidesL,W, andxmust be positive. Thus,L >0 ⇔ 20−2x >0 ⇔ x <10;
W >0 ⇔ 12−2x >0 ⇔ x <6; andx >0. Combining these restrictions gives us the domain0< x <6.
59. (a) (b) On$14,000, tax is assessed on$4000, and10%($4000) = $400.
On$26,000, tax is assessed on$16,000, and
10%($10,000) + 15%($6000) = $1000 + $900 = $1900.
(c) As in part (b), there is $1000 tax assessed on $20,000 of income, so the graph ofT is a line segment from(10,000,0)to(20,000,1000).
The tax on $30,000 is $2500, so the graph ofT forx >20,000is the ray with initial point(20,000,1000)that passes through (30,000,2500).
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 13 61.fis an odd function because its graph is symmetric about the origin.gis an even function because its graph is symmetric with
respect to they-axis.
63. (a) Because an even function is symmetric with respect to they-axis, and the point(5,3)is on the graph of this even function, the point(−5,3)must also be on its graph.
(b) Because an odd function is symmetric with respect to the origin, and the point(5,3)is on the graph of this odd function, the point(−5,−3)must also be on its graph.
65.f(x) = x x2+ 1. f(−x) = −x
(−x)2+ 1= −x
x2+ 1=− x
x2+ 1 =−f(x).
Sofis an odd function.
67.f(x) = x
x+ 1, sof(−x) = −x
−x+ 1= x x−1. Since this is neitherf(x)nor−f(x), the functionfis neither even nor odd.
69.f(x) = 1 + 3x2−x4.
f(−x) = 1 + 3(−x)2−(−x)4= 1 + 3x2−x4=f(x).
Sofis an even function.
1.2 Mathematical Models: A Catalog of Essential Functions
1. (a)f(x) =√5xis a root function withn= 5.
(b)g(x) =√
1−x2is an algebraic function because it is a root of a polynomial.
(c)h(x) =x9+x4is a polynomial of degree9.
(d)r(x) = x2+ 1
x3+xis a rational function because it is a ratio of polynomials.
(e)s(x) = tan 2xis a trigonometric function.
(f )t(x) = log10xis a logarithmic function.
3.We notice from thefigure thatgandhare even functions (symmetric with respect to they-axis) and thatfis an odd function (symmetric with respect to the origin). So (b) y=x5 must bef. Sincegisflatter thanhnear the origin, we must have (c) y=x8 matched withgand (a) y=x2 matched withh.
5. (a) An equation for the family of linear functions with slope2 isy=f(x) = 2x+b, wherebis they-intercept.
(b)f(2) = 1means that the point(2,1)is on the graph off. We can use the point-slope form of a line to obtain an equation for the family of linear functions through the point(2,1). y−1 =m(x−2), which is equivalent toy=mx+ (1−2m)in slope-intercept form.
(c) To belong to both families, an equation must have slopem= 2, so the equation in part (b),y=mx+ (1−2m), becomesy= 2x−3. It is theonlyfunction that belongs to both families.
7. All members of the family of linear functionsf(x) =c−xhave graphs that are lines with slope−1. They-intercept isc.
9. Sincef(−1) =f(0) =f(2) = 0,fhas zeros of−1,0, and2, so an equation forfisf(x) =a[x−(−1)](x−0)(x−2), orf(x) =ax(x+ 1)(x−2). Becausef(1) = 6, we’ll substitute1forxand6forf(x).
6 =a(1)(2)(−1) ⇒ −2a= 6 ⇒ a=−3, so an equation forfisf(x) =−3x(x+ 1)(x−2).
11. (a)D= 200, soc= 0.0417D(a+ 1) = 0.0417(200)(a+ 1) = 8.34a+ 8.34. The slope is8.34, which represents the change in mg of the dosage for a child for each change of 1 year in age.
(b) For a newborn,a= 0, soc= 8.34mg.
13. (a) (b) The slope of95 means thatFincreases 95degrees for each increase of1◦C. (Equivalently,Fincreases by 9whenCincreases by5 andFdecreases by9whenCdecreases by5.) TheF-intercept of 32is the Fahrenheit temperature corresponding to a Celsius temperature of0.
SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 15 15. (a) UsingNin place ofxandT in place ofy, wefind the slope to be T2−T1
N2−N1
= 80−70 173−113= 10
60= 1
6. So a linear equation isT−80 = 16(N−173) ⇔ T−80 = 16N−1736 ⇔ T =16N+3076 3076 = 51.16 .
(b) The slope of 16 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket chirps per minute. Said differently, each increase of6cricket chirps per minute corresponds to an increase of1◦F.
(c) WhenN= 150, the temperature is given approximately byT =16(150) +3076 = 76.16◦F≈76◦F.
17. (a) We are given change in pressure
10feet change in depth = 4.34
10 = 0.434. UsingPfor pressure anddfor depth with the point (d, P) = (0,15), we have the slope-intercept form of the line,P = 0.434d+ 15.
(b) WhenP = 100, then100 = 0.434d+ 15 ⇔ 0.434d= 85 ⇔ d=0.43485 ≈195.85feet. Thus, the pressure is 100 lb/in2at a depth of approximately196feet.
19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form f(x) =acos(bx) +cseems appropriate.
(b) The data appear to be decreasing in a linear fashion. A model of the formf(x) =mx+bseems appropriate.
Some values are given to many decimal places. These are the results given by several computer algebra systems—rounding is left to the reader.
21. (a)
A linear model does seem appropriate.
(b) Using the points(4000,14.1)and(60,000,8.2), we obtain y−14.1 = 8.2−14.1
60,000−4000(x−4000)or, equivalently, y≈ −0.000105357x+ 14.521429.
(c) Using a computing device, we obtain the least squares regression liney=−0.0000997855x+ 13.950764.
The following commands and screens illustrate how tofind the least squares regression line on a TI-83 Plus.
Enter the data into list one (L1) and list two (L2). Press to enter the editor.
Find the regession line and store it in Y1. Press .
Note from the lastfigure that the regression line has been stored in Y1and that Plot1 has been turned on (Plot1 is highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing or by
pressing .
Now press to produce a graph of the data and the regression line. Note that choice 9 of the ZOOM menu automatically selects a window that displays all of the data.
(d) Whenx= 25,000,y≈11.456; or about11.5per100population.
(e) Whenx= 80,000,y≈5.968; or about a6%chance.
(f ) Whenx= 200,000,yis negative, so the model does not apply.
23. (a)
A linear model does seem appropriate.
(b)
Using a computing device, we obtain the least squares regression liney= 0.089119747x−158.2403249, wherexis the year andyis the height in feet.
(c) Whenx= 2000, the model givesy≈20.00ft. Note that the actual winning height for the 2000 Olympics isless thanthe winning height for 1996—so much for that prediction.
(d) Whenx= 2100,y≈28.91ft. This would be an increase of9.49ft from 1996 to 2100. Even though there was an increase of8.59ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next100years.
25. Using a computing device, we obtain the cubic
functiony=ax3+bx2+cx+dwith a= 0.0012937,b=−7.06142,c= 12,823, andd=−7,743,770. Whenx= 1925, y≈1914(million).
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 17
1.3 New Functions from Old Functions
1. (a) If the graph offis shifted3units upward, its equation becomesy=f(x) + 3.
(b) If the graph offis shifted3units downward, its equation becomesy=f(x)−3.
(c) If the graph offis shifted3units to the right, its equation becomesy=f(x−3).
(d) If the graph offis shifted3units to the left, its equation becomesy=f(x+ 3).
(e) If the graph offis reflected about thex-axis, its equation becomesy=−f(x).
(f ) If the graph offis reflected about they-axis, its equation becomesy=f(−x).
(g) If the graph offis stretched vertically by a factor of3, its equation becomesy= 3f(x).
(h) If the graph offis shrunk vertically by a factor of3, its equation becomesy= 13f(x).
3. (a) (graph 3) The graph offis shifted4units to the right and has equationy=f(x−4).
(b) (graph 1) The graph offis shifted3units upward and has equationy=f(x) + 3.
(c) (graph 4) The graph offis shrunk vertically by a factor of3and has equationy=13f(x).
(d) (graph 5) The graph offis shifted4units to the left and reflected about thex-axis. Its equation isy=−f(x+ 4).
(e) (graph 2) The graph offis shifted6units to the left and stretched vertically by a factor of2. Its equation is y= 2f(x+ 6).
5. (a) To graphy=f(2x)we shrink the graph off horizontally by a factor of2.
The point(4,−1)on the graph offcorresponds to the point 12·4,−1 = (2,−1).
(b) To graphy=f 12x we stretch the graph off horizontally by a factor of2.
The point(4,−1)on the graph offcorresponds to the point(2·4,−1) = (8,−1).
(c) To graphy=f(−x)we reflect the graph offabout they-axis.
The point(4,−1)on the graph offcorresponds to the point(−1·4,−1) = (−4,−1).
(d) To graphy=−f(−x)we reflect the graph offabout they-axis, then about thex-axis.
The point(4,−1)on the graph offcorresponds to the point(−1·4,−1·−1) = (−4,1).
7. The graph ofy=f(x) =√
3x−x2has been shifted4units to the left, reflected about thex-axis, and shifted downward 1unit. Thus, a function describing the graph is
y= −1· reflect aboutx-axis
f (x+ 4) shift 4 units left
− 1 shift 1 unit left This function can be written as
y=−f(x+ 4)−1 =− 3(x+ 4)−(x+ 4)2−1 =− 3x+ 12−(x2+ 8x+ 16)−1 =−√
−x2−5x−4−1
9. y=−x3: Start with the graph ofy=x3and reflect about thex-axis. Note: Reflecting about they-axis gives the same result since substituting−xforxgives usy= (−x)3=−x3.
11. y= (x+ 1)2: Start with the graph ofy=x2 and shift1unit to the left.
13. y= 1 + 2 cosx: Start with the graph ofy= cosx, stretch vertically by a factor of2, and then shift1unit upward.
15. y= sin(x/2): Start with the graph ofy= sinxand stretch horizontally by a factor of2.
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 19 17.y=√
x+ 3: Start with the graph of y=√
xand shift3units to the left.
19.y=12(x2+ 8x) = 12(x2+ 8x+ 16)−8 = 12(x+ 4)2−8: Start with the graph ofy=x2, compress vertically by a factor of2, shift4units to the left, and then shift8units downward.
0 0 0 0
21.y= 2/(x+ 1): Start with the graph ofy= 1/x, shift1unit to the left, and then stretch vertically by a factor of2.
23.y=|sinx|: Start with the graph ofy= sinxand reflect all the parts of the graph below thex-axis about thex-axis.
25.This is just like the solution to Example 4 except the amplitude of the curve (the 30◦N curve in Figure 9 on June 21) is 14−12 = 2. So the function isL(t) = 12 + 2 sin 3652π(t−80) . March 31 is the90th day of the year, so the model gives L(90)≈12.34h. The daylight time (5:51AMto 6:18PM) is12hours and27minutes, or12.45h. The model value differs from the actual value by 12.4512.45−12.34 ≈0.009, less than1%.
27. (a) To obtainy=f(|x|), the portion of the graph ofy=f(x)to the right of they-axis is reflected about they-axis.
(b)y= sin|x| (c)y= |x|
29. f(x) =x3+ 2x2; g(x) = 3x2−1. D=Rfor bothfandg.
(f+g)(x) = (x3+ 2x2) + (3x2−1) =x3+ 5x2−1, D=R. (f−g)(x) = (x3+ 2x2)−(3x2−1) =x3−x2+ 1, D=R. (f g)(x) = (x3+ 2x2)(3x2−1) = 3x5+ 6x4−x3−2x2, D=R.
f
g (x) = x3+ 2x2
3x2−1, D= x|x6=± 1
√3 since3x2−16= 0.
31. f(x) =x2−1, D=R; g(x) = 2x+ 1, D=R.
(a)(f◦g)(x) =f(g(x)) =f(2x+ 1) = (2x+ 1)2−1 = (4x2+ 4x+ 1)−1 = 4x2+ 4x, D=R. (b)(g◦f)(x) =g(f(x)) =g(x2−1) = 2(x2−1) + 1 = (2x2−2) + 1 = 2x2−1, D=R. (c)(f◦f)(x) =f(f(x)) =f(x2−1) = (x2−1)2−1 = (x4−2x2+ 1)−1 =x4−2x2, D=R. (d)(g◦g)(x) =g(g(x)) =g(2x+ 1) = 2(2x+ 1) + 1 = (4x+ 2) + 1 = 4x+ 3, D=R.
33. f(x) = 1−3x; g(x) = cosx. D=Rfor bothfandg, and hence for their composites.
(a)(f◦g)(x) =f(g(x)) =f(cosx) = 1−3 cosx.
(b)(g◦f)(x) =g(f(x)) =g(1−3x) = cos(1−3x).
(c)(f◦f)(x) =f(f(x)) =f(1−3x) = 1−3(1−3x) = 1−3 + 9x= 9x−2.
(d)(g◦g)(x) =g(g(x)) =g(cosx) = cos(cosx) [Note that this isnotcosx·cosx.]
35. f(x) =x+1
x, D={x|x6= 0}; g(x) = x+ 1
x+ 2, D={x|x6=−2} (a)(f◦g)(x) =f(g(x)) =f x+ 1
x+ 2 = x+ 1 x+ 2+ 1
x+ 1 x+ 2
=x+ 1
x+ 2+x+ 2 x+ 1
=(x+ 1)(x+ 1) + (x+ 2)(x+ 2)
(x+ 2)(x+ 1) = x2+ 2x+ 1 + x2+ 4x+ 4
(x+ 2)(x+ 1) = 2x2+ 6x+ 5 (x+ 2)(x+ 1) Sinceg(x)is not defined forx=−2andf(g(x))is not defined forx=−2andx=−1,
the domain of (f◦g)(x)isD={x|x6=−2,−1}.
(b)(g◦f)(x) =g(f(x)) =g x+1
x =
x+1 x + 1 x+1
x + 2
=
x2+ 1 +x x x2+ 1 + 2x
x
= x2+x+ 1
x2+ 2x+ 1= x2+x+ 1 (x+ 1)2
Sincef(x)is not defined forx= 0andg(f(x))is not defined forx=−1, the domain of(g◦f)(x)isD={x|x6=−1,0}.
(c)(f◦f)(x) =f(f(x)) =f x+1
x = x+1
x + 1
x+1x =x+1 x+ 1
x2+1 x
=x+1 x+ x
x2+ 1
= x(x) x2+ 1 + 1 x2+ 1 +x(x)
x(x2+ 1) =x4+x2+x2+ 1 +x2 x(x2+ 1)
= x4+ 3x2+ 1
x(x2+ 1) , D={x|x6= 0}
SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 21
(d)(g◦g)(x) =g(g(x)) =g x+ 1 x+ 2 =
x+ 1 x+ 2+ 1 x+ 1 x+ 2+ 2
=
x+ 1 + 1(x+ 2) x+ 2 x+ 1 + 2(x+ 2)
x+ 2
= x+ 1 +x+ 2
x+ 1 + 2x+ 4 =2x+ 3 3x+ 5 Sinceg(x)is not defined forx=−2andg(g(x))is not defined forx=−53,
the domain of(g◦g)(x)isD= x|x6=−2,−53 .
37.(f◦g◦h)(x) =f(g(h(x))) =f(g(x−1)) =f(2(x−1)) = 2(x−1) + 1 = 2x−1 39.(f◦g◦h)(x) =f(g(h(x))) =f(g(x3+ 2)) =f[(x3+ 2)2]
=f(x6+ 4x3+ 4) = (x6+ 4x3+ 4)−3 =√
x6+ 4x3+ 1
41.Letg(x) =x2+ 1andf(x) =x10. Then(f◦g)(x) =f(g(x)) =f(x2+ 1) = (x2+ 1)10=F(x).
43.Letg(x) =√3xandf(x) = x
1 +x. Then(f◦g)(x) =f(g(x)) =f(√3x) =
√3
x 1 +√3
x =F(x).
45.Letg(t) = costandf(t) =√t. Then(f◦g)(t) =f(g(t)) =f(cost) =√
cost=u(t).
47.Leth(x) =x2,g(x) = 3x, andf(x) = 1−x. Then
(f◦g◦h)(x) =f(g(h(x))) =f(g(x2)) =f 3x2 = 1−3x2 =H(x).
49.Leth(x) =√x,g(x) = secx, andf(x) = x4. Then
(f◦g◦h)(x) =f(g(h(x))) =f(g(√x)) =f(sec√x) = (sec√x)4= sec4(√x) =H(x).
51. (a)g(2) = 5, because the point(2,5)is on the graph ofg. Thus,f(g(2)) =f(5) = 4, because the point(5,4)is on the graph off.
(b)g(f(0)) =g(0) = 3
(c)(f◦g)(0) =f(g(0)) =f(3) = 0
(d)(g◦f)(6) =g(f(6)) =g(6). This value is not defined, because there is no point on the graph ofgthat has x-coordinate6.
(e)(g◦g)(−2) =g(g(−2)) =g(1) = 4 (f )(f◦f)(4) =f(f(4)) =f(2) =−2
53. (a) Using the relationshipdistance=rate·timewith the radiusras the distance, we haver(t) = 60t.
(b)A=πr2 ⇒ (A◦r)(t) =A(r(t)) =π(60t)2= 3600πt2. This formula gives us the extent of the rippled area (in cm2) at any timet.
55. (a) From thefigure, we have a right triangle with legs6andd, and hypotenuses.
By the Pythagorean Theorem,d2+ 62=s2 ⇒ s=f(d) =√ d2+ 36.
(b) Usingd=rt, we getd= (30km/hr)(thr) = 30t(in km). Thus, d=g(t) = 30t.
(c)(f◦g)(t) =f(g(t)) =f(30t) = (30t)2+ 36 =√
900t2+ 36. This function represents the distance between the lighthouse and the ship as a function of the time elapsed since noon.
57. (a)
H(t) = 0 if t <0 1 if t≥0
(b)
V(t) = 0 if t <0
120 if t≥0 soV(t) = 120H(t).
(c) Starting with the formula in part (b), we replace120with240to reflect the different voltage. Also, because we are starting5units to the right oft= 0, we replacetwitht−5. Thus, the formula isV(t) = 240H(t−5).
59. Iff(x) =m1x+b1andg(x) =m2x+b2, then
(f◦g)(x) =f(g(x)) =f(m2x+b2) =m1(m2x+b2) +b1=m1m2x+m1b2+b1. Sof◦gis a linear function with slopem1m2.
61. (a) By examining the variable terms ingandh, we deduce that we must squaregto get the terms4x2and4xinh. If we let f(x) =x2+c, then(f◦g)(x) =f(g(x)) =f(2x+ 1) = (2x+ 1)2+c= 4x2+ 4x+ (1 +c). Since
h(x) = 4x2+ 4x+ 7, we must have1 +c= 7. Soc= 6andf(x) =x2+ 6.
(b) We need a functiongso thatf(g(x)) = 3(g(x)) + 5 =h(x). But
h(x) = 3x2+ 3x+ 2 = 3(x2+x) + 2 = 3(x2+x−1) + 5, so we see thatg(x) =x2+x−1.
63. (a) Iffandgare even functions, thenf(−x) =f(x)andg(−x) =g(x).
(i)(f+g)(−x) =f(−x) +g(−x) =f(x) +g(x) = (f+g)(x), sof+gis anevenfunction.
(ii)(f g)(−x) =f(−x)·g(−x) =f(x)·g(x) = (f g)(x), sof gis anevenfunction.
(b) Iffandgare odd functions, thenf(−x) =−f(x)andg(−x) =−g(x).
(i)(f+g)(−x) =f(−x) +g(−x) =−f(x) + [−g(x)] =−[f(x) +g(x)] =−(f+g)(x), sof+gis anoddfunction.
(ii)(f g)(−x) =f(−x)·g(−x) =−f(x)·[−g(x)] =f(x)·g(x) = (f g)(x), sof gis anevenfunction.
65. We need to examineh(−x).
h(−x) = (f◦g)(−x) =f(g(−x)) =f(g(x)) [becausegis even] =h(x) Becauseh(−x) =h(x),his an even function.
SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 23
1.4 Graphing Calculators and Computers
1.f(x) =√ x3−5x2 (a)[−5,5]by[−5,5]
(There is no graph shown.)
(b)[0,10]by[0,2] (c)[0,10]by[0,10]
The most appropriate graph is produced in viewing rectangle (c).
3.Since the graph off(x) = 5 + 20x−x2is a parabola opening downward, an appropriate viewing rectangle should include the maximum point.
5.f(x) =√4
81−x4is defined when81−x4≥0 ⇔ x4≤81 ⇔ |x|≤3, so the domain offis[−3,3]. Also 0≤√4
81−x4≤√481 = 3, so the range is[0,3].
7.The graph off(x) =x3−225xis symmetric with respect to the origin.
Sincef(x) =x3−225x=x(x2−225) =x(x+ 15)(x−15), there arex-intercepts at0,−15, and15.f(20) = 3500.
9.The period ofg(x) = sin(1000x)is10002π ≈0.0063and its range is [−1,1]. Sincef(x) = sin2(1000x)is the square ofg, its range is [0,1]and a viewing rectangle of[−0.01,0.01]by[0,1.1]seems appropriate.
11.The domain ofy=√xisx≥0, so the domain off(x) = sin√xis[0,∞) and the range is[−1,1]. With a little trial-and-error experimentation, wefind that an Xmax of 100 illustrates the general shape off, so an appropriate viewing rectangle is[0,100]by[−1.5,1.5].
13. Thefirst term,10 sinx, has period2πand range[−10,10]. It will be the dominant term in any “large” graph of
y= 10 sinx+ sin 100x, as shown in thefirstfigure. The second term,sin 100x, has period1002π =50π and range[−1,1].
It causes the bumps in thefirstfigure and will be the dominant term in any “small” graph, as shown in the view near the origin in the secondfigure.
15. We must solve the given equation foryto obtain equations for the upper and lower halves of the ellipse.
4x2+ 2y2= 1 ⇔ 2y2= 1−4x2 ⇔ y2=1−4x2
2 ⇔
y=± 1−4x2 2
17. From the graph ofy= 3x2−6x+ 1 andy= 0.23x−2.25in the viewing rectangle[−1,3]by[−2.5,1.5], it is difficult to see if the graphs intersect.
If we zoom in on the fourth quadrant, we see the graphs do not intersect.
19. From the graph off(x) =x3−9x2−4, we see that there is one solution of the equationf(x) = 0and it is slightly larger than 9. By zooming in or using a root or zero feature, we obtainx≈9.05.
21. We see that the graphs off(x) =x2andg(x) = sinxintersect twice. One solution isx= 0.The other solution off=gis thex-coordinate of the point of intersection in thefirst quadrant. Using an intersect feature or zooming in, wefind this value to be approximately 0.88. Alternatively, we couldfind that value byfinding the positive zero ofh(x) =x2−sinx.
Note: After producing the graph on a TI-83 Plus, we canfind the approximate value 0.88 by using the following keystrokes:
. The “1” is just a guess for0.88.
SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 25 23.g(x) =x3/10is larger thanf(x) = 10x2wheneverx >100.
25. We see from the graphs ofy=|sinx−x|andy= 0.1that there are
two solutions to the equation|sinx−x|= 0.1:x≈ −0.85and x≈0.85. The condition|sinx−x|<0.1holds for anyxlying between these two values, that is,−0.85< x <0.85.
27. (a) The root functionsy=√x, y=√4
xandy=√6 x
(b) The root functionsy=x, y=√3
xandy=√5 x
(c) The root functionsy=√x,y=√3x, y=√4
xandy=√5 x
(d) •For anyn, thenth root of0is0and thenth root of1is1; that is, allnth root functions pass through the points(0,0) and(1,1).
•For oddn, the domain of thenth root function isR, while for evenn, it is{x∈R|x≥0}.
•Graphs of even root functions look similar to that of√x, while those of odd root functions resemble that of√3x.
•Asnincreases, the graph of √nxbecomes steeper near0andflatter forx >1.
29.f(x) =x4+cx2+x. Ifc <−1.5, there are three humps: two minimum points and a maximum point. These humps getflatter ascincreases, until atc=−1.5 two of the humps disappear and there is only one minimum point. This single hump then moves to the right and approaches the origin ascincreases.
31.y=xn2−x. Asnincreases, the maximum of the function moves further from the origin, and gets larger. Note, however, that regardless ofn, the function approaches0asx→ ∞.
33. y2=cx3+x2. Ifc <0, the loop is to the right of the origin, and ifcis positive, it is to the left. In both cases, the closercis to0, the larger the loop is. (In the limiting case,c= 0, the loop is “infinite,” that is, it doesn’t close.) Also, the larger|c|is, the steeper the slope is on the loopless side of the origin.
35. The graphing window is 95 pixels wide and we want to start withx= 0and end withx= 2π. Since there are 94 “gaps”
between pixels, the distance between pixels is2π94−0. Thus, thex-values that the calculator actually plots arex= 0 +2π94 ·n, wheren= 0,1,2,. . .,93,94. Fory= sin 2x, the actual points plotted by the calculator are 2π94·n,sin 2·2π94·n for n= 0,1,. . .,94.Fory= sin 96x, the points plotted are 2π94·n,sin 96·2π94 ·n forn= 0,1,. . .,94.But
sin 96·2π94 ·n = sin 94·2π94 ·n+ 2·2π94 ·n = sin 2πn+ 2·2π94·n
= sin 2·2π94 ·n [by periodicity of sine], n= 0,1,. . .,94 So they-values, and hence the points, plotted fory= sin 96xare identical to those plotted fory= sin 2x.
Note:Try graphingy= sin 94x. Can you see why all they-values are zero?
1.5 Exponential Functions
1. (a)f(x) =ax, a >0 (b)R (c)(0,∞) (d) See Figures 4(c), 4(b), and 4(a), respectively.
3. All of these graphs approach0asx→ −∞, all of them pass through the point (0,1), and all of them are increasing and approach∞asx→ ∞. The larger the base, the faster the function increases forx >0, and the faster it approaches0as x→ −∞.
Note:The notation “x→ ∞” can be thought of as “xbecomes large” at this point.
More details on this notation are given in Chapter 2.
5. The functions with bases greater than1(3xand10x) are increasing, while those with bases less than1 13 x and 101 x are decreasing. The graph of 13 xis the reflection of that of3xabout they-axis, and the graph of 101 xis the reflection of that of10xabout they-axis. The graph of10xincreases more quickly than that of 3xforx >0, and approaches0faster asx→ −∞.
7. We start with the graph ofy= 4x(Figure 3) and then shift 3 units downward. This shift doesn’t affect the domain, but the range ofy= 4x−3is(−3,∞). There is a horizontal asymptote ofy=−3.
y= 4x y= 4x−3
SECTION 1.5 EXPONENTIAL FUNCTIONS ¤ 27 9.We start with the graph ofy= 2x(Figure 3),
reflect it about they-axis, and then about the x-axis (or just rotate180◦to handle both reflections) to obtain the graph ofy=−2−x. In each graph,y= 0is the horizontal asymptote.
y= 2x y= 2−x y=−2−x
11.We start with the graph ofy=ex(Figure 13) and reflect about they-axis to get the graph ofy=e−x. Then we compress the graph vertically by a factor of2to obtain the graph ofy= 12e−xand then reflect about thex-axis to get the graph of y=−12e−x. Finally, we shift the graph upward one unit to get the graph ofy= 1−12e−x.
13. (a) Tofind the equation of the graph that results from shifting the graph ofy=ex2units downward, we subtract2from the original function to gety=ex−2.
(b) Tofind the equation of the graph that results from shifting the graph ofy=ex2units to the right, we replacexwithx−2 in the original function to gety=e(x−2).
(c) Tofind the equation of the graph that results from reflecting the graph ofy=exabout thex-axis, we multiply the original function by−1to gety=−ex.
(d) Tofind the equation of the graph that results from reflecting the graph ofy=exabout they-axis, we replacexwith−xin the original function to gety=e−x.
(e) Tofind the equation of the graph that results from reflecting the graph ofy=exabout thex-axis and then about the y-axis, wefirst multiply the original function by−1(to gety=−ex) and then replacexwith−xin this equation to gety=−e−x.
15. (a) The denominator1 +exis never equal to zero becauseex>0, so the domain off(x) = 1/(1 +ex)isR. (b)1−ex= 0 ⇔ ex= 1 ⇔ x= 0, so the domain off(x) = 1/(1−ex)is(−∞,0)∪(0,∞).
17.Usey=Caxwith the points(1,6)and(3,24). 6 =Ca1 C= 6a and24 =Ca3 ⇒ 24 = 6
a a3 ⇒ 4 =a2 ⇒ a= 2 [sincea >0] andC=62 = 3. The function isf(x) = 3·2x.
19.Iff(x) = 5x, thenf(x+h)−f(x)
h =5x+h−5x
h = 5x5h−5x
h = 5x 5h−1
h = 5x 5h−1
h .
21. 2ft= 24in,f(24) = 242in= 576in= 48ft. g(24) = 224in= 224/(12·5280)mi≈265mi 23. The graph ofgfinally surpasses that offatx≈35.8.
25. (a) Fifteen hours represents5doubling periods (one doubling period is three hours).100·25= 3200 (b) Inthours, there will bet/3doubling periods. The initial population is100,
so the populationyat timetisy= 100·2t/3. (c)t= 20 ⇒ y= 100·220/3≈10,159
(d) We graphy1= 100·2x/3andy2= 50,000. The two curves intersect at x≈26.9, so the population reaches50,000in about26.9hours.
27. An exponential model isy=abt, wherea= 3.154832569×10−12 andb= 1.017764706. This model givesy(1993)≈5498million and y(2010)≈7417million.
29. From the graph, it appears thatfis an odd function (fis undefined forx= 0).
To prove this, we must show thatf(−x) =−f(x).
f(−x) =1−e1/(−x)
1 +e1/(−x) = 1−e(−1/x) 1 +e(−1/x) =
1− 1 e1/x 1 + 1
e1/x
·e1/x
e1/x = e1/x−1 e1/x+ 1
=−1−e1/x
1 +e1/x =−f(x) sofis an odd function.
1.6 Inverse Functions and Logarithms
1. (a) See Definition 1.
(b) It must pass the Horizontal Line Test.
3. fis not one-to-one because26= 6, butf(2) = 2.0 =f(6).
5. No horizontal line intersects the graph offmore than once. Thus, by the Horizontal Line Test,fis one-to-one.
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 29 7.The horizontal liney= 0(thex-axis) intersects the graph offin more than one point. Thus, by the Horizontal Line Test,
fis not one-to-one.
9.The graph off(x) =x2−2xis a parabola with axis of symmetryx=−b
2a =−−2
2(1) = 1. Pick anyx-values equidistant from 1 tofind two equal function values. For example,f(0) = 0andf(2) = 0, sofis not one-to-one.
11.g(x) = 1/x. x16=x2 ⇒ 1/x16= 1/x2 ⇒ g(x1)6=g(x2), sogis one-to-one.
Geometric solution: The graph ofgis the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test, sogis one-to-one.
13.A football will attain every heighthup to its maximum height twice: once on the way up, and again on the way down. Thus, even ift1does not equalt2,f(t1)may equalf(t2), sofis not1-1.
15.Sincef(2) = 9andfis1-1, we know thatf−1(9) = 2. Remember, if the point(2,9)is on the graph off, then the point (9,2)is on the graph off−1.
17.First, we must determinexsuch thatg(x) = 4. By inspection, we see that ifx= 0, theng(x) = 4. Sincegis1-1(gis an increasing function), it has an inverse, andg−1(4) = 0.
19.We solveC=59(F−32)forF:95C=F−32 ⇒ F =95C+ 32. This gives us a formula for the inverse function, that is, the Fahrenheit temperatureFas a function of the Celsius temperatureC.F≥ −459.67 ⇒ 95C+ 32≥ −459.67 ⇒
9
5C≥ −491.67 ⇒ C≥ −273.15, the domain of the inverse function.
21.f(x) =√
10−3x ⇒ y=√
10−3x (y≥0) ⇒ y2 = 10−3x ⇒ 3x= 10−y2 ⇒ x=−13y2+103. Interchangexandy: y=−13x2+103. Sof−1(x) =−13x2+103. Note that the domain off−1isx≥0.
23.y=f(x) =ex3 ⇒ lny=x3 ⇒ x=√3
lny. Interchangexandy: y=√3
lnx. Sof−1(x) =√3 lnx.
25.y=f(x) = ln (x+ 3) ⇒ x+ 3 =ey ⇒ x=ey−3. Interchangexandy:y=ex−3. Sof−1(x) =ex−3.
27.y=f(x) =x4+ 1 ⇒ y−1 =x4 ⇒ x=√4
y−1 (not ± since x≥0). Interchangex andy:y=√4
x−1. Sof−1(x) =√4
x−1. The graph ofy=√4
x−1is just the graph ofy=√4
xshifted right one unit.
From the graph, we see thatfandf−1are reflections about the liney=x.
29.Reflect the graph offabout the liney=x. The points(−1,−2),(1,−1), (2,2), and(3,3)onfare reflected to(−2,−1),(−1,1),(2,2), and(3,3) onf−1.
31. (a) It is defined as the inverse of the exponential function with basea, that is,logax=y ⇔ ay=x.
(b)(0,∞) (c)R (d) See Figure 11.
33. (a)log5125 = 3since 53 = 125. (b)log3 1
27 =−3since3−3= 1 33 = 1
27. 35. (a)log26−log215 + log220 = log2(156) + log220 [by Law 2]
= log2(156 ·20) [by Law 1]
= log28, and log28 = 3since23= 8.
(b)log3100−log318−log350 = log3 10018 −log350 = log3 18100·50
= log3(19), and log3 1
9 =−2since3−2=19. 37. ln 5 + 5 ln 3 = ln 5 + ln 35 [by Law 3]
= ln(5·35) [by Law 1]
= ln 1215
39. ln(1 +x2) +12lnx−ln sinx= ln(1 +x2) + lnx1/2−ln sinx= ln[(1 +x2)√x]−ln sinx= ln(1 +x2)√ x sinx 41. To graph these functions, we uselog1.5x= lnx
ln 1.5andlog50x= lnx ln 50. These graphs all approach−∞asx→0+, and they all pass through the point(1,0). Also, they are all increasing, and all approach∞asx→ ∞. The functions with larger bases increase extremely slowly, and the ones with smaller bases do so somewhat more quickly. The functions with large bases approach they-axis more closely asx→0+.
43. 3ft= 36in, so we needxsuch thatlog2x= 36 ⇔ x= 236= 68,719,476,736. In miles, this is 68,719,476,736in· 1ft
12in· 1mi
5280ft ≈1,084,587.7mi.
45. (a) Shift the graph ofy= log10xfive units to the left to obtain the graph ofy= log10(x+ 5). Note the vertical asymptote ofx=−5.
y= log10x y= log10(x+ 5)
(b) Reflect the graph ofy= lnxabout thex-axis to obtain the graph ofy=−lnx.
y= lnx y=−lnx 47. (a)2 lnx= 1 ⇒ lnx= 12 ⇒ x=e1/2=√
e (b)e−x= 5 ⇒ −x= ln 5 ⇒ x=−ln 5
49. (a)2x−5= 3 ⇔ log23 =x−5 ⇔ x= 5 + log23.
Or:2x−5= 3 ⇔ ln 2x−5 = ln 3 ⇔ (x−5) ln 2 = ln 3 ⇔ x−5 =ln 3
ln 2 ⇔ x= 5 +ln 3 ln 2
SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 31
(b)lnx+ ln(x−1) = ln(x(x−1)) = 1 ⇔ x(x−1) =e1 ⇔ x2−x−e= 0. The quadratic formula (witha= 1, b=−1, andc=−e) givesx= 12 1±√
1 + 4e , but we reject the negative root since the natural logarithm is not defined forx <0. Sox=12 1 +√
1 + 4e .
51. (a)ex<10 ⇒ lnex<ln 10 ⇒ x <ln 10 ⇒ x∈(−∞,ln 10) (b)lnx >−1 ⇒ elnx> e−1 ⇒ x > e−1 ⇒ x∈(1/e,∞) 53. (a) Forf(x) =√
3−e2x, we must have3−e2x≥0 ⇒ e2x≤3 ⇒ 2x≤ln 3 ⇒ x≤ 12ln 3. Thus, the domain offis(−∞,12ln 3].
(b)y=f(x) =√
3−e2x [note thaty≥0] ⇒ y2= 3−e2x ⇒ e2x= 3−y2 ⇒ 2x= ln(3−y2) ⇒ x= 12ln(3−y2). Interchangexandy: y=12ln(3−x2). Sof−1(x) =12ln(3−x2). For the domain off−1, we must have3−x2>0 ⇒ x2<3 ⇒ |x|<√
3 ⇒ −√
3< x <√
3 ⇒ 0≤x <√
3sincex≥0. Note that the domain off−1,[0,√
3 ), equals the range off.
55.We see that the graph ofy=f(x) =√
x3+x2+x+ 1is increasing, sofis1-1.
Enterx= y3+y2+y+ 1and use your CAS to solve the equation fory.
Using Derive, we get two (irrelevant) solutions involving imaginary expressions, as well as one which can be simplified to the following:
y=f−1(x) =−√364
√3
D−27x2+ 20−√3
D+ 27x2−20 +√3 2 whereD= 3√
3√
27x4−40x2+ 16.
Maple and Mathematica each give two complex expressions and one real expression, and the real expression is equivalent to that given by Derive. For example, Maple’s expression simplifies to1
6
M2/3−8−2M1/3
2M1/3 , where
M = 108x2+ 12√
48−120x2+ 81x4−80.
57. (a)n= 100·2t/3 ⇒ n
100= 2t/3 ⇒ log2
n 100 = t
3 ⇒ t= 3 log2
n
100 . Using formula (10), we can write this ast=f−1(n) = 3·ln(n/100)
ln 2 . This function tells us how long it will take to obtainnbacteria (given the numbern).
(b)n= 50,000 ⇒ t=f−1(50,000) = 3·ln 50,000100
ln 2 = 3 ln 500
ln 2 ≈26.9 hours 59. (a)sin−1 √23 =π3 sincesinπ3 =√23 andπ3 is in −π2,π2 .
(b)cos−1(−1) =πsincecosπ=−1andπis in[0, π].
61. (a)arctan 1 =π4 sincetanπ4 = 1andπ4 is in −π2,π2 . (b)sin−1√12 =π4 sincesinπ4 =√12 andπ4 is in −π2,π2 .
63. (a) In general,tan(arctanx) =xfor any real numberx. Thus,tan(arctan 10) = 10.
(b)sin−1 sin7π3 = sin−1 sinπ3 = sin−1 √23 =π3 sincesinπ3 =√23 andπ3 is in −π2,π2 . [Recall that7π3 = π3 + 2πand the sine function is periodic with period2π.]
65. Lety= sin−1x. Then−π2 ≤y≤π2 ⇒ cosy≥0, socos(sin−1x) = cosy= 1−sin2y=√ 1−x2. 67. Lety= tan−1x. Thentany=x, so from the triangle we see that
sin(tan−1x) = siny= x
√1 +x2.
69. The graph ofsin−1xis the reflection of the graph of
sinxabout the liney=x.
71. g(x) = sin−1(3x+ 1).
Domain(g) ={x|−1≤3x+ 1≤1}={x|−2≤3x≤0}= x|−23 ≤x≤0 = −23,0 . Range(g) = y|−π2 ≤y≤ π2 = −π2,π2 .
73. (a) If the point(x, y)is on the graph ofy=f(x), then the point(x−c, y)is that point shiftedcunits to the left. Sincefis 1-1, the point(y, x)is on the graph ofy=f−1(x)and the point corresponding to(x−c, y)on the graph offis (y, x−c)on the graph off−1. Thus, the curve’s reflection is shifteddownthe same number of units as the curve itself is shifted to the left. So an expression for the inverse function isg−1(x) =f−1(x)−c.
(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the liney=xis compressed (or stretched) verticallyby the same factor. Using this geometric principle, we see that the inverse ofh(x) =f(cx)can be expressed as h−1(x) = (1/c)f−1(x).
CHAPTER 1 REVIEW ¤ 33
1 Review
1. (a) Afunctionfis a rule that assigns to each elementxin a setAexactly one element, calledf(x), in a setB. The setAis called thedomainof the function. Therangeoffis the set of all possible values off(x)asxvaries throughout the domain.
(b) Iffis a function with domainA, then itsgraphis the set of ordered pairs{(x, f(x))|x∈A}. (c) Use the Vertical Line Test on page16.
2.The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given below.
Verbally:An assignment of students to chairs in a classroom (a description in words) Numerically:A tax table that assigns an amount of tax to an income (a table of values) Visually:A graphical history of the Dow Jones average (a graph)
Algebraically:A relationship between distance, rate, and time:d=rt(an explicit formula)
3. (a) Aneven functionfsatisfiesf(−x) =f(x)for every numberxin its domain. It is symmetric with respect to they-axis.
(b) Anodd functiongsatisfiesg(−x) =−g(x)for every numberxin its domain. It is symmetric with respect to the origin.
4.A functionfis calledincreasingon an intervalIiff(x1)< f(x2)wheneverx1< x2inI.
5.Amathematical modelis a mathematical description (often by means of a function or an equation) of a real-world phenomenon.
6. (a) Linear function:f(x) = 2x+ 1,f(x) =ax+b 7.
(b) Power function:f(x) =x2,f(x) =xa (c) Exponential function:f(x) = 2x,f(x) =ax
(d) Quadratic function:f(x) =x2+x+ 1,f(x) =ax2+bx+c (e) Polynomial of degree5:f(x) =x5+ 2
(f ) Rational function:f(x) = x
x+ 2,f(x) =P(x)
Q(x) whereP(x)and Q(x)are polynomials
8. (a) (b)
(c) (d) (e)
(f ) (g) (h)
9. (a) The domain off+gis the intersection of the domain offand the domain ofg; that is,A∩B.
(b) The domain off gis alsoA∩B.
(c) The domain off /gmust exclude values ofxthat makegequal to0; that is,{x∈A∩B|g(x)6= 0}.
10. Given two functionsfandg, thecompositefunctionf◦gis defined by(f◦g) (x) =f(g(x)). The domain off◦gis the set of allxin the domain ofgsuch thatg(x)is in the domain off.
11. (a) If the graph offis shifted2units upward, its equation becomesy=f(x) + 2.
(b) If the graph offis shifted2units downward, its equation becomesy=f(x)−2.
(c) If the graph offis shifted2units to the right, its equation becomesy=f(x−2).
(d) If the graph offis shifted2units to the left, its equation becomesy=f(x+ 2).
(e) If the graph offis reflected about thex-axis, its equation becomesy=−f(x).
(f ) If the graph offis reflected about they-axis, its equation becomesy=f(−x).
(g) If the graph offis stretched vertically by a factor of2, its equation becomesy= 2f(x).
(h) If the graph offis shrunk vertically by a factor of2, its equation becomesy=12f(x).
(i) If the graph offis stretched horizontally by a factor of2, its equation becomesy=f1
2x . (j) If the graph offis shrunk horizontally by a factor of2, its equation becomesy=f(2x).
12. (a) A functionfis called aone-to-one functionif it never takes on the same value twice; that is, iff(x1)6=f(x2)whenever x16=x2. (Or,fis1-1if each output corresponds to only one input.)
Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than once.
(b) Iffis a one-to-one function with domainAand rangeB, then itsinverse functionf−1has domainBand rangeAand is defined by
f−1(y) =x ⇔ f(x) =y
for anyyinB. The graph off−1is obtained by reflecting the graph offabout the liney=x.
CHAPTER 1 REVIEW ¤ 35
13. (a) The inverse sine functionf(x) = sin−1xis defined as follows:
sin−1x=y ⇔ siny=x and −π
2 ≤y≤π 2 Its domain is−1≤x≤1and its range is−π
2 ≤y≤π 2. (b) The inverse cosine functionf(x) = cos−1xis defined as follows:
cos−1x=y ⇔ cosy=x and 0≤y≤π Its domain is−1≤x≤1and its range is0≤y≤π.
(c) The inverse tangent functionf(x) = tan−1xis defined as follows:
tan−1x=y ⇔ tany=x and −π
2 < y <π 2 Its domain isRand its range is−π
2 < y < π 2.
1.False. Letf(x) =x2,s=−1, andt= 1. Thenf(s+t) = (−1 + 1)2= 02= 0, but f(s) +f(t) = (−1)2+ 12= 26= 0 =f(s+t).
3.False. Letf(x) =x2. Thenf(3x) = (3x)2= 9x2and3f(x) = 3x2. Sof(3x)6= 3f(x).
5.True. See the Vertical Line Test.
7.False. Letf(x) =x3. Thenfis one-to-one andf−1(x) =√3x. But1/f(x) = 1/x3, which is not equal tof−1(x).
9.True. The functionlnxis an increasing function on(0,∞).
11.False. Letx=e2anda=e. Thenlnx lna = lne2
lne =2 lne
lne = 2andlnx a = lne2
e = lne= 1, so in general the statement is false. Whatistrue, however, is thatlnx
a = lnx−lna.
13.False. For example,tan−120is defined;sin−120andcos−120are not.
1. (a) Whenx= 2,y≈2.7. Thus,f(2)≈2.7.
(b)f(x) = 3 ⇒ x≈2.3,5.6
(c) The domain offis−6≤x≤6, or[−6,6].
(d) The range offis−4≤y≤4, or[−4,4].
(e)fis increasing on[−4,4], that is, on−4≤x≤4.
(f )fis not one-to-one since it fails the Horizontal Line Test.
(g)fis odd since its graph is symmetric about the origin.
