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For the following exercises, explain the notation in words. The volume $\text{\hspace{0.17em}}f(t)\text{\hspace{0.17em}}$ of a tank of gasoline, in gallons, $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ minutes after noon.
$f(0)=600$
$f\text{'}(30)=\mathrm{-20}$
At 12:30 p.m. , the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute.
$f(30)=0$
$f\text{'}(200)=30$
At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute.
$f(240)=500$
For the following exercises, explain the functions in words. The height, $\text{\hspace{0.17em}}s,$ of a projectile after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by $\text{\hspace{0.17em}}s(t)=-16{t}^{2}+80t.$
$s(2)=96$
The height of the projectile after 2 seconds is 96 feet.
$s\text{'}(2)=16$
$s(3)=96$
The height of the projectile at $\text{\hspace{0.17em}}t=3\text{\hspace{0.17em}}$ seconds is 96 feet.
$s\text{'}(3)=\mathrm{-16}$
$s(0)=0,s(5)=0.$
The height of the projectile is zero at $\text{\hspace{0.17em}}t=0\text{\hspace{0.17em}}$ and again at $\text{\hspace{0.17em}}t=5.\text{\hspace{0.17em}}$ In other words, the projectile starts on the ground and falls to earth again after 5 seconds.
For the following exercises, the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ of a sphere with respect to its radius $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is given by $\text{\hspace{0.17em}}V=\frac{4}{3}\pi {r}^{3}.$
Find the average rate of change of $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ changes from 1 cm to 2 cm.
Find the instantaneous rate of change of $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}r=3\text{cm}\text{.}$
$36\pi $
For the following exercises, the revenue generated by selling $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items is given by $\text{\hspace{0.17em}}R(x)=2{x}^{2}+10x.$
Find the average change of the revenue function as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ changes from $\text{\hspace{0.17em}}x=10\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}x=20.$
Find $\text{\hspace{0.17em}}R\text{'}(10)\text{\hspace{0.17em}}$ and interpret.
$50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold.
Find $\text{\hspace{0.17em}}R\text{'}(15)\text{\hspace{0.17em}}$ and interpret. Compare $\text{\hspace{0.17em}}R\text{'}(15)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}R\text{'}(10),$ and explain the difference.
For the following exercises, the cost of producing $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ cellphones is described by the function $\text{\hspace{0.17em}}C(x)={x}^{2}-4x+1000.$
Find the average rate of change in the total cost as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ changes from $\text{\hspace{0.17em}}x=10\text{to}x=15.$
$21 per unit
Find the approximate marginal cost, when 15 cellphones have been produced, of producing the 16 ^{th} cellphone.
Find the approximate marginal cost, when 20 cellphones have been produced, of producing the 21 ^{st} cellphone.
$36
For the following exercises, use the definition for the derivative at a point $\text{\hspace{0.17em}}x=a,$ $\text{\hspace{0.17em}}\underset{x\to a}{\mathrm{lim}}\frac{f(x)-f(a)}{x-a},$ to find the derivative of the functions.
$f(x)=\frac{1}{{x}^{2}}$
$f(x)=-{x}^{2}+4x+7$
$f(x)=\frac{-4}{3-{x}^{2}}$
$\frac{4}{{\left(3-x\right)}^{2}}$
For the following exercises, use [link] .
$\underset{x\to {\mathrm{-1}}^{-}}{\mathrm{lim}}f(x)$
$\underset{x\to 3}{\mathrm{lim}}f(x)$
At what values of $\text{\hspace{0.17em}}x$ is the function discontinuous? What condition of continuity is violated?
$\begin{array}{l}\text{Discontinuousat}x=-1(\underset{x\to \text{\hspace{0.17em}}a\text{\hspace{0.17em}}}{\mathrm{lim}}f(x)\text{doesnotexist}),x=3\text{}(\text{jumpdiscontinuity}),\\ \text{and}x=7\text{}(\underset{x\to \text{\hspace{0.17em}}a\text{\hspace{0.17em}}}{\mathrm{lim}}f(x)\text{doesnotexist}).\end{array}$
Using [link] , estimate $\text{\hspace{0.17em}}\underset{x\to 0}{\mathrm{lim}}f(x).$
$x$ | $F(x)$ |
−0.1 | 2.875 |
−0.01 | 2.92 |
−0.001 | 2.998 |
0 | Undefined |
0.001 | 2.9987 |
0.01 | 2.865 |
0.1 | 2.78145 |
0.15 | 2.678 |
3
For the following exercises, with the use of a graphing utility, use numerical or graphical evidence to determine the left- and right-hand limits of the function given as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ If the function has limit as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ approaches $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ state it. If not, discuss why there is no limit.
$f(x)=\{\begin{array}{lll}\left|x\right|-1,\hfill & if\hfill & x\ne 1\hfill \\ {x}^{3},\hfill & if\hfill & x=1\hfill \end{array}\text{}a=1$
$f(x)=\{\begin{array}{lll}\frac{1}{x+1},\hfill & if\hfill & x=-2\hfill \\ {(x+1)}^{2},\hfill & if\hfill & x\ne -2\hfill \end{array}\text{}a=-2$
$\begin{array}{ll}\underset{x\to -2}{\mathrm{lim}}\hfill & f(x)=1\hfill \end{array}$
$f(x)=\{\begin{array}{lll}\sqrt{x+3},\hfill & if\hfill & x<1\hfill \\ -\sqrt[3]{x},\hfill & if\hfill & x>1\hfill \end{array}\text{}a=1$
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