Chapter 1

$f(1)=3$ and $f(a+h)=a^2+2ah+h^2-3a-3h+5$

Domain = $\left\{x|x\le 2\right\},$ range = $\left\{y\right|y\ge 5\}$

$x=0,2,3$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{x^2+3}{2x-5}.$ The domain is $\left\{x|x\ne \frac{5}{2}\right\}.$

$\left(f\circ g\right)\left(x\right)=2-5\sqrt{x}.$

$(g\circ f)(x)=0.63x$

$f(x)$ is odd.

Domain = $(\text{−}\infty ,\infty ),$ range = $\left\{y\right|y\ge \mathrm{-4}\}.$

$m=1\text{/}2.$ The point-slope form is

$y-4=\frac{1}{2}\left(x-1\right).$

The slope-intercept form is

$y=\frac{1}{2}x+\frac{7}{2}.$

The zeros are $x=1\pm \sqrt{3}\text{/}3.$ The parabola opens upward.

The domain is the set of real numbers $x$ such that $x\ne 1\text{/}2.$ The range is the set $\left\{y\right|y\ne 5\text{/}2\}.$

The domain of $f$ is $\text{(−∞, ∞).}$ The domain of $g$ is $\left\{x\right|x\ge 1\text{/}5\}.$

Algebraic

$C\left(x\right)=\{\begin{array}{c}49,0<x\le 1\\ 70,1<x\le 2\\ 91,2<x\le 3\end{array}$

Shift the graph $y=x^2$ to the left 1 unit, reflect about the $x$-axis, then shift down 4 units.

$7\pi \text{/}6;$ 330°

$\text{cos}(3\pi \text{/}4)=\text{−}\sqrt{2}\text{/}2;\text{sin}(\text{−}\pi \text{/}6)=\mathrm{-1}\text{/}2$

$10$ ft

$\theta =\frac{3\pi }{2}+2n\pi ,\frac{\pi }{6}+2n\pi ,\frac{5\pi }{6}+2n\pi$ for $n=0,\pm 1,\pm 2\text{,…}$

$\begin{array}{rcl}1+\theta & =& 1+\frac{\theta }{\theta }\\ & =& \frac{\theta }{\theta }+\frac{\theta }{\theta }\\ & =& \frac{\theta +\theta }{\theta }\\ & =& \frac{1}{\theta }\\ & =& \theta \end{array}$

To graph $f\left(x\right)=3\text{sin}\left(4x\right)-5,$ the graph of $y=\text{sin}(x)$ needs to be compressed horizontally by a factor of 4, then stretched vertically by a factor of 3, then shifted down 5 units. The function $f$ will have a period of $\pi \text{/}2$ and an amplitude of 3.

No.

$f^{\mathrm{-1}}\left(x\right)=\frac{2x}{x-3}.$ The domain of $f^{\mathrm{-1}}$ is $\left\{x\right|x\ne 3\}.$ The range of $f^{\mathrm{-1}}$ is $\left\{y\right|y\ne 2\}.$

The domain of $f^{\mathrm{-1}}$ is $\left(0,\infty \right).$ The range of $f^{\mathrm{-1}}$ is $(\text{−}\infty ,0).$ The inverse function is given by the formula $f^{\mathrm{-1}}\left(x\right)=\mathrm{-1}\text{/}\sqrt{x}.$

$f\left(4\right)=900;f\left(10\right)=24,300.$

$x\text{/}\left(2y^3\right)$

$A\left(t\right)=750e^{0.04t}.$ After $30$ years, there will be approximately $\text{\$}2,490.09.$

$x=\frac{\text{ln}3}{2}$

$x=\frac{1}{e}$

$1.29248$

The magnitude $8.4$ earthquake is roughly $10$ times as severe as the magnitude $7.4$ earthquake.

$(x^2+x^{\mathrm{-2}})\text{/}2$

$\frac{1}{2}\text{ln}\left(3\right)\approx 0.5493.$

a. Domain = $\left\{\mathrm{-3},\mathrm{-2},\mathrm{-1},0,1,2,3\right\},$ range = $\left\{0,1,4,9\right\}$ b. Yes, a function

a. Domain = $\left\{0,1,2,3\right\},$ range = $\left\{\mathrm{-3},\mathrm{-2},\mathrm{-1},0,1,2,3\right\}$ b. No, not a function

a. Domain = $\left\{3,5,8,10,15,21,33\right\},$ range = $\left\{0,1,2,3\right\}$ b. Yes, a function

a. $\mathrm{-2}$ b. 3 c. 13 d. $\mathrm{-5}x-2$ e. $5a-2$ f. $5a+5h-2$

a. Undefined b. 2 c. $\frac{2}{3}$ d. $-\frac{2}{x}$ e $\frac{2}{a}$ f. $\frac{2}{a+h}$

a. $\sqrt{5}$ b. $\sqrt{11}$ c. $\sqrt{23}$ d. $\sqrt{\mathrm{-6}x+5}$ e. $\sqrt{6a+5}$ f. $\sqrt{6a+6h+5}$

a. 9 b. 9 c. 9 d. 9 e. 9 f. 9

$x\ge \frac{1}{8};y\ge 0;x=\frac{1}{8};$ no y-intercept

$x\ge \mathrm{-2};y\ge \mathrm{-1};x=\mathrm{-1};y=\mathrm{-1}+\sqrt{2}$

$x\ne 4;y\ne 0;$ no x-intercept; $y=-\frac{3}{4}$

$x>5;y>0;$ no intercepts

Function; a. Domain: all real numbers, range: $y\ge 0$ b. $x=\pm 1$ c. $y=1$ d. $\mathrm{-1}<x<0$ and $1<x<\infty$ e. $\text{−}\infty <x<-1$ and $0<x<1$ f. Not constant g. y-axis h. Even

Function; a. Domain: all real numbers, range: $\mathrm{-1.5}\le y\le 1.5$ b. $x=0$ c. $y=0$ d. $\text{all real numbers}$ e. None f. Not constant g. Origin h. Odd

Function; a. Domain: $\text{−}\infty <x<\infty ,$ range: $\mathrm{-2}\le y\le 2$ b. $x=0$ c. $y=0$ d. $\mathrm{-2}<x<2$ e. Not decreasing f. $\text{−}\infty <x<-2$ and $2<x<\infty$ g. Origin h. Odd

Function; a. Domain: $\mathrm{-4}\le x\le 4,$ range: $\mathrm{-4}\le y\le 4$ b. $x=1.2$ c. $y=4$ d. Not increasing e. $0<x<4$ f. $\mathrm{-4}<x<0$ g. No Symmetry h. Neither

a. $5x^2+x-8;$ all real numbers b. $\mathrm{-5}{x}^2+x-8;$ all real numbers c. $5x^3-40x^2;$ all real numbers d. $\frac{x-8}{5x^2};x\ne 0$

a. $\mathrm{-2}x+6;$ all real numbers b. $\mathrm{-2}{x}^2+2x+12;$ all real numbers c. $\text{−}{x}^4+2x^3+12x^2-18x-27;$ all real numbers d. $-\frac{x+3}{x+1};x\ne \text{−}1,3$

a. $6+\frac{2}{x};x\ne 0$ b. 6; $x\ne 0$ c. $\frac{6}{x}+\frac{1}{x^2};x\ne 0$ d. $6x+1;x\ne 0$

a. $4x+3;$ all real numbers b. $4x+15;$ all real numbers

a. $x^4-6x^2+16;$ all real numbers b. $x^4+14x^2+46;$ all real numbers

a. $\frac{3x}{4+x};x\ne 0,\mathrm{-4}$ b. $\frac{4x+2}{3};x\ne -\frac{1}{2}$

a. Yes, because there is only one winner for each year. b. No, because there are three teams that won more than once during the years 2001 to 2012.

a. $V(s)=s^3$ b. $V\left(11.8\right)\approx 1643;$ a cube of side length 11.8 each has a volume of approximately 1643 cubic units.

a. $N(x)=15x$ b. i. $N\left(20\right)=15\left(20\right)=300;$ therefore, the vehicle can travel 300 mi on a full tank of gas. Ii. $N\left(15\right)=225;$ therefore, the vehicle can travel 225 mi on 3/4 of a tank of gas. c. Domain: $0\le x\le 20;$ range: $[0,300]$ d. The driver had to stop at least once, given that it takes approximately 39 gal of gas to drive a total of 578 mi.

a. $A\left(t\right)=A\left(r\left(t\right)\right)=\pi ·{\left(6-\frac{5}{t^2+1}\right)}^2$ b. Exact: $\frac{121\pi }{4};$ approximately 95 cm² c. $C\left(t\right)=C\left(r\left(t\right)\right)=2\pi \left(6-\frac{5}{t^2+1}\right)$ d. Exact: $11\pi ;$ approximately 35 cm

a. $S\left(x\right)=8.5x+750$ b. $962.50, $1090, $1217.50 c. 77 skateboards

a. −1 b. Decreasing

a. 3/4 b. Increasing

a. 4/3 b. Increasing

a. 0 b. Horizontal

$y=\mathrm{-6}x+9$

$y=\frac{1}{3}x+4$

$y=\frac{1}{2}x$

$y=\frac{3}{5}x-3$

a. $\left(m=2,b=\mathrm{-3}\right)$ b.

a. $\left(m=\mathrm{-6},b=0\right)$ b.

a. $\left(m=0,b=\mathrm{-6}\right)$ b.

a. $\left(m=-\frac{2}{3},b=2\right)$ b.

a. 2 b. $\frac{5}{2},\mathrm{-1};$ c. −5 d. As $x\to \pm \infty ,y\to \infty$ e. Neither

a. 2 b. $\pm \sqrt{2}$ c. −1 d. As $x\to \pm \infty ,y\to \infty$ e. Even

a. 3 b. 0, $\pm \sqrt{3}$ c. 0 d. As $x\to -\infty ,y\to \infty$ and $x\to \infty ,y\to -\infty$ e. Odd

a. $13,\mathrm{-3},5$ b.

a. $\frac{\mathrm{-3}}{2},\frac{\mathrm{-1}}{2},4$ b.

True; $n=3$

False; $f\left(x\right)=x^b,$ where $b$ is a real-valued constant, is a power function

a. $V\left(t\right)=\mathrm{-2733}t+20500$ b. $\left(0,20,500\right)$ means that the initial purchase price of the equipment is $20,500; $\left(7.5,0\right)$ means that in 7.5 years the computer equipment has no value. c. $6835 d. In approximately 6.4 years

a. $C=0.75x+125$ b. $245 c. 167 cupcakes

a. $V\left(t\right)=\mathrm{-1500}t+\mathrm{26,000}$ b. In 4 years, the value of the car is $20,000.

$30,337.50

96% of the total capacity

$\frac{4\pi }{3}\text{rad}$

$\frac{\text{−}\pi }{3}$

$\frac{11\pi }{6}\text{rad}$

$210\text{°}$

$\mathrm{-540}\text{°}$

$\mathrm{-0.5}$

$-\frac{\sqrt{2}}{2}$

$\frac{\sqrt{3}-1}{2\sqrt{2}}$

a. $b=5.7$ b. $\text{sin}A=\frac{4}{7},\text{cos}A=\frac{5.7}{7},\text{tan}A=\frac{4}{5.7},\text{csc}A=\frac{7}{4},\text{sec}A=\frac{7}{5.7},\text{cot}A=\frac{5.7}{4}$

a. $c=151.7$ b. $\text{sin}A=0.5623,\text{cos}A=0.8273,\text{tan}A=0.6797,\text{csc}A=1.778,\text{sec}A=1.209,\text{cot}A=1.471$

a. $c=85$ b. $\text{sin}A=\frac{84}{85},\text{cos}A=\frac{13}{85},\text{tan}A=\frac{84}{13},\text{csc}A=\frac{85}{84},\text{sec}A=\frac{85}{13},\text{cot}A=\frac{13}{84}$

a. $y=\frac{24}{25}$ b. $\text{sin}\theta =\frac{24}{25},\text{cos}\theta =\frac{7}{25},\text{tan}\theta =\frac{24}{7},\text{csc}\theta =\frac{25}{24},\text{sec}\theta =\frac{25}{7},\text{cot}\theta =\frac{7}{24}$

a. $x=\frac{\text{−}\sqrt{2}}{3}$ b. $\text{sin}\theta =\frac{\sqrt{7}}{3},\text{cos}\theta =\frac{\text{−}\sqrt{2}}{3},\text{tan}\theta =\frac{\text{−}\sqrt{14}}{2},\text{csc}\theta =\frac{3\sqrt{7}}{7},\text{sec}\theta =\frac{\mathrm{-3}\sqrt{2}}{2},\text{cot}\theta =\frac{\text{−}\sqrt{14}}{7}$

${\text{sec}}^{2}x$

${\text{sin}}^{2}x$

${\text{sec}}^2\theta$

$\frac{1}{\text{sin}t}(=\text{csc}t)$

$\left\{\frac{\pi }{6},\frac{5\pi }{6}\right\}$

$\left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\right\}$

$\left\{\frac{2\pi }{3},\frac{5\pi }{3}\right\}$

$\left\{0,\pi ,\frac{\pi }{3},\frac{5\pi }{3}\right\}$

$y=4\text{sin}\left(\frac{\pi }{4}x\right)$

$y=\text{cos}\left(2\pi x\right)$

a. 1 b. $2\pi$ c. $\frac{\pi }{4}$ units to the right

a. $\frac{1}{2}$ b. $8\pi$ c. No phase shift

a. 3 b. $2$ c. $\frac{2}{\pi }$ units to the left

Approximately 42 in.

a. 0.550 rad/sec b. 0.236 rad/sec c. 0.698 rad/min d. 1.697 rad/min

$\approx 30.9{\text{in}}^2$

a. π/184; the voltage repeats every π/184 sec b. Approximately 59 periods

a. Amplitude = $10;\text{period}=24$ b. $47.4\text{°}F$ c. 14 hours later, or 2 p.m. d.

Not one-to-one

Not one-to-one

One-to-one

a. $f^{\mathrm{-1}}\left(x\right)=\sqrt{x+4}$ b. Domain $\text{:}x\ge \mathrm{-4},\text{range}\text{:}y\ge 0$

a. $f^{\mathrm{-1}}\left(x\right)=\sqrt[3]{x-1}$ b. Domain: all real numbers, range: all real numbers

a. $f^{\mathrm{-1}}\left(x\right)=x^2+1,$ b. Domain: $x\ge 0,$ range: $y\ge 1$

These are inverses.

These are not inverses.

These are inverses.

These are inverses.

$\frac{\pi }{6}$

$\frac{\pi }{4}$

$\frac{\pi }{6}$

$\frac{\sqrt{2}}{2}$

$-\frac{\pi }{6}$

a. $x=f^{\mathrm{-1}}\left(V\right)=\sqrt{0.04-\frac{V}{500}}$ b. The inverse function determines the distance from the center of the artery at which blood is flowing with velocity V. c. 0.1 cm; 0.14 cm; 0.17 cm

a. $31,250, $66,667, $107,143 b. $\left(p=\frac{85C}{C+75}\right)$ c. 34 ppb

a. $~92\text{°}$ b. $~42\text{°}$ c. $~27\text{°}$

$x\approx 6.69,8.51;$ so, the temperature occurs on June 21 and August 15

$~1.5\text{sec}$

${\text{tan}}^{\mathrm{-1}}\left(\text{tan}(2.1)\right)\approx -1.0416;$ the expression does not equal 2.1 since $2.1>1.57=\frac{\pi }{2}$—in other words, it is not in the restricted domain of $\text{tan}x.{\text{cos}}^{\mathrm{-1}}\left(\text{cos}(2.1)\right)=2.1,$ since 2.1 is in the restricted domain of $\text{cos}x.$

a. 125 b. 2.24 c. 9.74

a. 0.01 b. 10,000 c. 46.42

d

b

e

Domain: all real numbers, range: $\left(2,\infty \right),y=2$

Domain: all real numbers, range: $\left(0,\infty \right),y=0$

Domain: all real numbers, range: $\left(\text{−}\infty ,1\right),y=1$

Domain: all real numbers, range: $\left(\mathrm{-1},\infty \right),y=\mathrm{-1}$

$8^{1\text{/}3}=2$

$5^2=25$

$e^{\mathrm{-3}}=\frac{1}{e^3}$

$e^0=1$

${\text{log}}_4\left(\frac{1}{16}\right)=\mathrm{-2}$

${\text{log}}_{9}1=0$

${\text{log}}_{64}4=\frac{1}{3}$

${\text{log}}_{9}150=y$

${\text{log}}_{4}0.125=-\frac{3}{2}$

Domain: $(1,\infty ),$ range: $\left(\text{−}\infty ,\infty \right),x=1$

Domain: $\left(0,\infty \right),$ range: $\left(\text{−}\infty ,\infty \right),x=0$

Domain: $\left(\mathrm{-1},\infty \right),$ range: $\left(\text{−}\infty ,\infty \right),x=\mathrm{-1}$

$2+3{\text{log}}_{3}a-{\text{log}}_{3}b$

$\frac{3}{2}+\frac{1}{2}{\text{log}}_{5}x+\frac{3}{2}{\text{log}}_{5}y$

$-\frac{3}{2}+\text{ln}6$

$\frac{\text{ln}15}{3}$

$\frac{3}{2}$

$\text{log}7.21$

$\frac{2}{3}+\frac{\text{log}11}{3\text{log}7}$

$x=\frac{1}{25}$

$x=4$

$x=3$

$1+\sqrt{5}$

$\left(\frac{\text{log}82}{\text{log}7}\approx 2.2646\right)$

$\left(\frac{\text{log}211}{\text{log}0.5}\approx -7.7211\right)$

$\left(\frac{\text{log}0.452}{\text{log}0.2}\approx 0.4934\right)$

$~17,491$

Approximately $131,653 is accumulated in 5 years.

i. a. pH = 8 b. Base ii. a. pH = 3 b. Acid iii. a. pH = 4 b. Acid

a. $~333$ million b. 94 years from 2013, or in 2107

a. $k\approx 0.0578$ b. $\approx 92$ hours

The San Francisco earthquake was ${10}^{3.4}\text{or}\approx 2512$ times more intense than the Japanese earthquake.

False

False

Domain: $x>5,$ range: all real numbers

Domain: $x>2$ and $x<-4,$ range: all real numbers

Degree of 3, $y$-intercept: 0, zeros: 0, $\sqrt{3}-1,\mathrm{-1}-\sqrt{3}$

$\begin{array}{l}{\cos }^{2}x-{\sin }^{2}x=\cos 2x\\ =1-2{\sin }^{2}x\\ =2{\cos }^{2}x-1\end{array}$

$0,\pm 2\pi$

4

One-to-one; yes, the function has an inverse; inverse: $f^{\mathrm{-1}}(x)=\frac{1}{y}$

$x\ge -\frac{3}{2},f^{\mathrm{-1}}(x)=-\frac{3}{2}+\frac{1}{2}\sqrt{4y-7}$