Chapter 3

$\frac{1}{4}$

$6$

$f^{\prime }\left(1\right)=5$

$\mathrm{-32}$ ft/s

$P^{\prime }\left(3.25\right)=20>0;$ raise prices

$f^{\prime }\left(x\right)=2x$

$\left(0,\text{+}\infty \right)$

$a=6$ and $b=\mathrm{-9}$

$f^{″}\left(x\right)=2$

$a\left(t\right)=6t$

0

$4x^3$

$f^{\prime }\left(x\right)=7x^6$

$f^{\prime }\left(x\right)=6x^2-12x.$

$y=12x-23$

$j^{\prime }\left(x\right)=10x^4\left(4x^2+x\right)+\left(8x+1\right)\left(2x^5\right)=56x^6+12x^5.$

$k^{\prime }\left(x\right)=-\frac{13}{{(4x-3)}^2}.$

$g^{\prime }\left(x\right)=\mathrm{-7}{x}^{\mathrm{-8}}.$

$3f^{\prime }\left(x\right)-2g^{\prime }\left(x\right).$

$\frac{5}{8}$

$\mathrm{-4.4}$

left to right

3,300

$2

$f^{\prime }\left(x\right)={\text{cos}}^{2}x-{\text{sin}}^{2}x$

$\frac{\text{cos}x+x\text{sin}x}{{\text{cos}}^{2}x}$

$t=\frac{\pi }{3},t=\frac{2\pi }{3}$

$f^{\prime }\left(x\right)=\text{−}{\text{csc}}^{2}x$

$f^{\prime }\left(x\right)=2{\text{sec}}^{2}x+3{\text{csc}}^{2}x$

$\frac{4}{3}$

$\text{cos}x$

$\text{−}\text{cos}x$

$v\left(\frac{5\pi }{6}\right)=\text{−}\sqrt{3}<0$ and $a\left(\frac{5\pi }{6}\right)=\mathrm{-1}<0.$ The block is speeding up.

$h^{\prime }\left(x\right)=4{\left(2x^3+2x-1\right)}^3\left(6x^2+2\right)=8\left(3x^2+1\right){\left(2x^3+2x-1\right)}^3$

$y=\mathrm{-48}x-88$

$h^{\prime }\left(x\right)=7\text{cos}\left(7x+2\right)$

$h^{\prime }\left(x\right)=\frac{3-4x}{{\left(2x+3\right)}^4}$

$h^{\prime }\left(x\right)=18x^2{\text{sin}}^5\left(x^3\right)\text{cos}\left(x^3\right)$

$a\left(t\right)=\mathrm{-16}\text{sin}(4t)$

$28$

$\frac{dy}{dx}=\mathrm{-3}{x}^2\text{sin}\left(x^3\right)$

$g^{\prime }\left(x\right)=-\frac{1}{{(x+2)}^2}$

$g\left(x\right)=\frac{1}{5}{x}^{\text{−}4\text{/}5}$

$s^{\prime }\left(t\right)={\left(2t+1\right)}^{\text{−}1\text{/}2}$

$g^{\prime }\left(x\right)=\frac{1}{1+x^2}$

$h^{\prime }\left(x\right)=\frac{\mathrm{-3}}{\sqrt{6x-9x^2}}$

$y=x$

$\frac{dy}{dx}=\frac{5-20x^4}{{\text{sec}}^{2}y-2y}$

$y=\frac{5}{3}x-\frac{16}{3}$

$h^{\prime }\left(x\right)=e^{2x}+2x{e}^{2x}$

996

$f^{\prime }\left(x\right)=\frac{15}{3x+2}$

$9\text{ln}\left(3\right)$

$\frac{dy}{dx}=x^x\left(1+\text{ln}x\right)$

$y^{\prime }=\pi {\left(\text{tan}x\right)}^{\pi -1}{\text{sec}}^{2}x$

$4$

$8.5$

$-\frac{3}{4}$

$0.2$

$0.25$

a. $\mathrm{-4}$ b. $y=3-4x$

a. $3$ b. $y=3x-1$

a. $\frac{\mathrm{-7}}{9}$ b. $y=\frac{\mathrm{-7}}{9}x+\frac{14}{3}$

a. $12$ b. $y=12x+14$

a. $\mathrm{-2}$ b. $y=\mathrm{-2}x-10$

$5$

$13$

$\frac{1}{4}$

$-\frac{1}{4}$

$\mathrm{-3}$

a. $\text{(i)}5.100000,$ $\text{(ii)}5.010000,$ $\text{(iii)}5.001000,$ $\text{(iv)}5.000100,$ $\text{(v)}5.000010,$ $\text{(vi)}5.000001,$ $\text{(vii)}4.900000,$ $\text{(viii)}4.990000,$ $\text{(ix)}4.999000,$ $\text{(x)}4.999900,$ $\text{(xi)}4.999990,$ $\text{(x)}4.999999$ b. $m_{\text{tan}}=5$ c. $y=5x+3$

a. $\text{(i)}4.8771,$ $\text{(ii)}4.9875\text{(iii)}4.9988,$ $\text{(iv)}4.9999,$ $\text{(v)}4.9999,$ $\text{(vi)}4.9999$ b. $m_{\text{tan}}=5$ c. $y=5x+10$

a. $\frac{1}{3};$ b. $\text{(i)}0.\bar{3}$ m/s, $\text{(ii)}0.\bar{3}$ m/s, $\text{(iii)}0.\bar{3}$ m/s, $\text{(iv)}0.\bar{3}$ m/s; c. $0.\bar{3}=\frac{1}{3}$ m/s

a. $2\left(h^2+6h+12\right);$ b. $\text{(i)}25.22$ m/s, $\text{(ii)}24.12$ m/s, $\text{(iii)}24.01$ m/s, $\text{(iv)}24$ m/s; c. $24$ m/s

a. $1.25;$ b. $0.5$

$\underset{x\to 0^{-}}{\text{lim}}\frac{x^{1\text{/}3}-0}{x-0}=\underset{x\to 0^{-}}{\text{lim}}\frac{1}{x^{2\text{/}3}}=\infty$

$\underset{x\to 1^{-}}{\text{lim}}\frac{1-1}{x-1}=0\ne 1=\underset{x\to 1^{+}}{\text{lim}}\frac{x-1}{x-1}$

a. $\text{(i)}61.7244$ ft/s, $\text{(ii)}61.0725$ ft/s $\text{(iii)}61.0072$ ft/s $\text{(iv)}61.0007$ ft/s b. At $4$ seconds the race car is traveling at a rate/velocity of $61$ ft/s.

a. The vehicle represented by $f\left(t\right),$ because it has traveled $2$ feet, whereas $g\left(t\right)$ has traveled $1$ foot. b. The velocity of $f\left(t\right)$ is constant at $1$ ft/s, while the velocity of $g\left(t\right)$ is approximately $2$ ft/s. c. The vehicle represented by $g\left(t\right),$ with a velocity of approximately $4$ ft/s. d. Both have traveled $4$ feet in $4$ seconds.

a.

b. $a\approx -1.361,2.694$

a. $N\left(x\right)=\frac{x}{30}$ b. $\sim 3.3$ gallons. When the vehicle travels $100$ miles, it has used $3.3$ gallons of gas. c. $\frac{1}{30}.$ The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled $100$ miles.

a.

b. $\mathrm{-0.028},\mathrm{-0.16},0.16,0.028$

$\mathrm{-3}$

$8x$

$\frac{1}{\sqrt{2x}}$

$\frac{\mathrm{-9}}{x^2}$

$\frac{\mathrm{-1}}{2x^{3\text{/}2}}$

$f\left(x\right)=3x^2+2,a=2$

$f\left(x\right)=x^4,a=2$

$f\left(x\right)=e^x,a=0$

a.

b. $\underset{h\to 1^{-}}{\text{lim}}\frac{3-3}{h}\ne \underset{h\to 1^{+}}{\text{lim}}\frac{3h}{h}$

a.

b. $\underset{h\to 1^{-}}{\text{lim}}\frac{2h}{h}\ne \underset{h\to 1^{+}}{\text{lim}}\frac{\frac{2}{x+h}-\frac{2}{x}}{h}.$

a. $x=1,$ b. $x=2$

$0$

$\frac{2}{x^3}$

$f^{\prime }\left(x\right)=6x+2$

$f^{\prime }\left(x\right)=-\frac{1}{{\left(2x\right)}^{3\text{/}2}}$

$f^{\prime }\left(x\right)=3x^2$

a. Average rate at which customers spent on concessions in thousands per customer. b. Rate (in thousands per customer) at which $x$ customers spent money on concessions in thousands per customer.

a. Average grade received on the test with an average study time between two values. b. Rate (in percentage points per hour) at which the grade on the test increased or decreased for a given average study time of $x$ hours.

a. Average change of atmospheric pressure between two different altitudes. b. Rate (torr per foot) at which atmospheric pressure is increasing or decreasing at $x$ feet.

a. The rate (in degrees per foot) at which temperature is increasing or decreasing for a given height $x.$ b. The rate of change of temperature as altitude changes at $1000$ feet is $\mathrm{-0.1}$ degrees per foot.

a. The rate at which the number of people who have come down with the flu is changing $t$ weeks after the initial outbreak. b. The rate is increasing sharply up to the third week, at which point it slows down and then becomes constant.

$G^{\prime }\left(t\right)=2.858t+0.0857$

$H^{″}\left(t\right)=0,G^{″}\left(t\right)=2.858\text{and}{f}^{″}\left(t\right)=1.222t+5.912$ represent the acceleration of the rocket, with units of meters per second squared $({\text{m/s}}^2).$

$f^{\prime }\left(x\right)=15x^2-1$

$f^{\prime }\left(x\right)=32x^3+18x$

$f^{\prime }\left(x\right)=270x^4+\frac{39}{{\left(x+1\right)}^2}$

$f^{\prime }\left(x\right)=\frac{\mathrm{-5}}{x^2}$

$f^{\prime }\left(x\right)=\frac{4x^4+2x^2-2x}{x^4}$

$f^{\prime }\left(x\right)=\frac{\text{−}{x}^2-18x+64}{{\left(x^2-7x+1\right)}^2}$

$T\left(x\right)=–4x+7$

$T\left(x\right)=4x-5$

$h^{\prime }\left(x\right)=3x^{2}f\left(x\right)+x^3{f}^{\prime }\left(x\right)$

$h^{\prime }\left(x\right)=\frac{3f^{\prime }\left(x\right)\left(g\left(x\right)+2\right)-3f\left(x\right)g^{\prime }\left(x\right)}{{\left(g\left(x\right)+2\right)}^2}$

$\frac{16}{9}$

Undefined

a. $2,$ b. does not exist, c. $2.5$

a. 23, b. $y=23x-28$

a. 3, b. $y=3x+2$

$y=\mathrm{-7}x-3$

$y=\mathrm{-5}x+7$

$y=-\frac{3}{2}x+\frac{15}{2}$

$y=\mathrm{-3}{x}^2+9x-1$

$\frac{12}{121}$ or 0.0992 ft/s

a. $\frac{\mathrm{-2}{t}^4-2t^3+200t+50}{{\left(t^3+50\right)}^2}$ b. $\mathrm{-0.02395}$ mg/L-hr, −0.01344 mg/L-hr, −0.003566 mg/L-hr, −0.001579 mg/L-hr c. The rate at which the concentration of drug in the bloodstream decreases is slowing to 0 as time increases.

a. $F\prime (d)=\frac{\mathrm{-2}G{m}_1{m}_2}{d^3}$ b. $\mathrm{-1.33}\times {10}^{\mathrm{-7}}$ N/m

a. $v\left(t\right)=6t^2-30t+36,a\left(t\right)=12t-30;$ b. speeds up $\left(2,2.5\right){\displaystyle \cup }\left(3,\infty \right),$ slows down $\left(0,2\right){\displaystyle \cup }\left(2.5,3\right)$

a. $464{\text{ft/s}}^2$ b. $\mathrm{-32}{\text{ft/s}}^2$

a. 5 ft/s b. 9 ft/s

a. 84 ft/s, −84 ft/s b. 84 ft/s c. $\frac{25}{8}\text{s}$ d. $\mathrm{-32}{\text{ft/s}}^2$ in both cases e. $\frac{1}{8}\left(25+\sqrt{965}\right)\text{s}$ f. $\mathrm{-4}\sqrt{965}\text{ft/s}$

a. Velocity is positive on $\left(0,1.5\right){\displaystyle \cup }\left(6,7\right),$ negative on $\left(1.5,2\right){\displaystyle \cup }\left(5,6\right),$ and zero on $\left(2,5\right).$ b.

c. Acceleration is positive on $\left(5,7\right),$ negative on $\left(0,2\right),$ and zero on $\left(2,5\right).$ d. The object is speeding up on $\left(6,7\right){\displaystyle \cup }\left(1.5,2\right)$ and slowing down on $\left(0,1.5\right){\displaystyle \cup }\left(5,6\right).$

a. $R\left(x\right)=10x-0.001x^2$ b. $R^{\prime }\left(x\right)=10-0.002x$ c. $6 per item, $0 per item

a. $C^{\prime }\left(x\right)=65$ b. $R\left(x\right)=143x-0.03x^2,R^{\prime }\left(x\right)=143-0.06x$ c. $83,\mathrm{-97}.$ At a production level of 1000 cordless drills, revenue is increasing at a rate of $83 per drill; at a production level of 4000 cordless drills, revenue is decreasing at a rate of $97 per drill. d. $P\left(x\right)=\mathrm{-0.03}{x}^2+78x-75000,P^{\prime }\left(x\right)=\mathrm{-0.06}x+78$ e. $18,\mathrm{-162}.$ At a production level of 1000 cordless drills, profit is increasing at a rate of $18 per drill; at a production level of 4000 cordless drills, profit is decreasing at a rate of $162 per drill.

a. $N^{\prime }\left(t\right)=3000\left(\frac{\mathrm{-4}{t}^2+400}{{\left(t^2+100\right)}^2}\right)$ b. $120,0,\mathrm{-14.4},\mathrm{-9.6}$ c. The bacteria population increases from time 0 to 10 hours; afterwards, the bacteria population decreases. d. $0,\mathrm{-6},0.384,0.432.$ The rate at which the bacteria is increasing is decreasing during the first 10 hours. Afterwards, the bacteria population is decreasing at a decreasing rate.

a. $P\left(t\right)=0.03983+0.4280$ b. $P^{\prime }\left(t\right)=0.03983.$ The population is increasing. c. $P^{″}\left(t\right)=0.$ The rate at which the population is increasing is constant.

a. $p(t)=\mathrm{-0.6071}{x}^2+0.4357x-0.3571$ b. $p^{\prime }\left(t\right)=\mathrm{-1.214}x+0.4357.$ This is the velocity of the sensor. c. $p^{″}\left(t\right)=\mathrm{-1.214}.$ This is the acceleration of the sensor; it is a constant acceleration downward.

a.

b. $f^{\prime }\left(x\right)=a.$ The more increase in prey, the more growth for predators. c. $f^{″}\left(x\right)=0.$ As the amount of prey increases, the rate at which the predator population growth increases is constant. d. This equation assumes that if there is more prey, the predator is able to increase consumption linearly. This assumption is unphysical because we would expect there to be some saturation point at which there is too much prey for the predator to consume adequately.

a.

b. $f^{\prime }\left(x\right)=\frac{2ax{n}^2}{{\left(n^2+x^2\right)}^2}.$ When the amount of prey increases, the predator growth increases. c. $f^{″}\left(x\right)=\frac{2a{n}^2\left(n^2-3x^2\right)}{{\left(n^2+x^2\right)}^3}.$ When the amount of prey is extremely small, the rate at which predator growth is increasing is increasing, but when the amount of prey reaches above a certain threshold, the rate at which predator growth is increasing begins to decrease. d. At lower levels of prey, the prey is more easily able to avoid detection by the predator, so fewer prey individuals are consumed, resulting in less predator growth.

$\frac{dy}{dx}=2x-\text{sec}x\text{tan}x$

$\frac{dy}{dx}=2x\text{cot}x-x^2{\text{csc}}^{2}x$

$\frac{dy}{dx}=\frac{x\text{sec}x\text{tan}x-\text{sec}x}{x^2}$

$\frac{dy}{dx}=\left(1-\text{sin}x\right)\left(1-\text{sin}x\right)-\text{cos}x\left(x+\text{cos}x\right)$

$\frac{dy}{dx}=\frac{2{\text{csc}}^{2}x}{{\left(1+\text{cot}x\right)}^2}$

$y=\text{−}x$

$y=x+\frac{2-3\pi }{2}$

$y=\text{−}x$

$3\text{cos}x-x\text{sin}x$

$\frac{1}{2}\text{sin}x$

$2\text{csc}x\left({\text{csc}}^{2}x+{\text{cot}}^{2}x\right)$

$\frac{(2n+1)\pi }{4},\text{where}n\text{is an integer}$

$(\frac{\pi }{4}, 1), (\frac{3\pi }{4}, \mathrm{-1}), (\frac{5\mathrm{\pi }}{4}, 1), (\frac{7\mathrm{\pi }}{4}, -1)$

$a=0,b=3$

$y^{\prime }=5\text{cos}\left(x\right),$ increasing on $\left(0,\frac{\pi }{2}\right),\left(\frac{3\pi }{2},\frac{5\pi }{2}\right),$ and $\left(\frac{7\pi }{2},12\right)$

$3\text{sin}x$

$5\text{cos}x$

$720x^7-5\text{tan}\left(x\right){\text{sec}}^3\left(x\right)-{\text{tan}}^3\left(x\right)\text{sec}\left(x\right)$

$18u^2·7=18{\left(7x-4\right)}^2·7$