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Your answer to 50vs51 should get you thinking. Suppose a bookstore estimates it will sell 50 copies of the book. How many
books can, in fact, be ordered for the same price as those 50 copies? (Round your answer to a whole number of
books.)
56 books.
An on-line comic book retailer charges shipping costs according to the following formula
\[ S(n) = \left \{ \begin{array}{rcl} 1.5 n + 2.5 & \mbox { if } & 1 \leq n \leq 14 \\ 0 & \mbox { if } & n \geq 15 \end{array} \right . \]
where \(n\) is the number of comic
books purchased and \(S(n)\) is the shipping cost in dollars.
What is the cost to ship 10 comic books?
\(\$\answer {17.50}\)
What is the significance of the formula \(S(n) = 0\) for \(n \geq 15\)?
There is free shipping on orders of \(15\) or more comic books.
The cost in dollars \(C(m)\) to talk \(m\) minutes a month on a mobile phone plan is modeled by
\[ C(m) = \left \{ \begin{array}{rcl} 25 & \mbox { if } & 0 \leq m \leq 1000 \\ 25+0.1(m-1000) & \mbox { if } & m > 1000 \end{array} \right . \]
How much does it cost to talk \(750\) minutes per month with this plan?
\(\$\answer {25}\)
How much does it cost to talk \(20\) hours a month with this plan?
\(\$\answer {45}\)
Explain the terms of the plan verbally.
It costs \(\$25\) for up to \(1000\) minutes and \(10\) cents per minute for each minute over \(1000\) minutes.
Jeff can walk comfortably at \(3\) miles per hour. Find an expression for a linear function \(d(t)\) that represents the total distance Jeff
can walk in \(t\) hours, assuming he doesn’t take any breaks.
\(d(t) = 3t\), \(t \geq 0\).
Carl can stuff \(6\) envelopes per minute. Find an expression for a linear function \(E(t)\) that represents the total number of envelopes
Carl can stuff after \(t\)hours, assuming he doesn’t take any breaks.
\(E(t) = 360t\), \(t \geq 0\).
A landscaping company charges \(\$45\) per cubic yard of mulch plus a delivery charge of \(\$20\). Find an expression for a linear function \(C(x)\)
which computes the total cost in dollars to deliver \(x\) cubic yards of mulch.
\(C(x) = 45x+20\), \(x \geq 0\).
A plumber charges \(\$50\) for a service call plus \(\$80\) per hour. If she spends no longer than 8 hours a day at any one site, find an
expression for a linear function \(C(t)\) that computes her total daily charges in dollars as a function of the amount of time spent in
hours, \(t\) at any one given location.
\(C(t) = 80t + 50\), \(0 \leq t \leq 8\).
A salesperson is paid $200 per week plus 5% commission on her weekly sales of \(x\) dollars. Find an expression for a linear
function \(W(x)\) which computes her total weekly pay in dollars as a function of \(x\). What must her weekly sales be in order for her to
earn $475.00 for the week?
\(W(x) = 200 + .05x,\, x \geq 0\;\;\) She must make $5500 in weekly sales.
An on-demand publisher charges \(\$22.50\) to print a 600 page book and \(\$15.50\) to print a 400 page book. Find an expression for a linear
function which models the cost of a book in dollars \(C(p)\) as a function of the number of pages \(p\). Find and interpret both the slope of
the linear function and \(C(0)\).
\(C(p) = 0.035p + 1.5 \;\) The slope \(0.035\) means it costs \(3.5\)¢ per page. \(C(0) = 1.5\) means there is a fixed, or start-up, cost of \(\$1.50\) to make each book.
The Topology Taxi Company charges \(\$2.50\) for the first fifth of a mile and \(\$0.45\) for each additional fifth of a mile. Find an expression for
a linear function which models the taxi fare \(F(m)\) as a function of the number of miles driven, \(m\). Find and interpret both the slope of
the linear function and \(F(0)\).
\(F(m) = 2.25m + 2.05 \;\) The slope \(2.25\) means it costs an additional \(\$2.25\) for each mile beyond the first 0.2 miles. \(F(0) = 2.05\), so according to the model, it would cost \(\$2.05\)
for a trip of \(0\) miles. Would this ever really happen? Depends on the driver and the passenger, we suppose.
Water freezes at \(0^{\circ }\) Celsius and \(32^{\circ }\) Fahrenheit and it boils at \(100^{\circ }\)C and \(212^{\circ }\)F.
Find an expression for a linear function \(F(T)\) that computes temperature in the Fahrenheit scale as a function of the
temperature \(T\) given in degrees Celsius. Use this function to convert \(20^{\circ }\)C into Fahrenheit.
\(F(T) = \frac {9}{5}T + 32\)
Find an expression for a linear function \(C(T)\) that computes temperature in the Celsius scale as a function of the
temperature \(T\) given in degrees Fahrenheit. Use this function to convert \(110^{\circ }\)F into Celsius.
Is there a temperature \(T\) such that \(F(T) = C(T)\)?
\(\answer {-40}\)
Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is \(40^{\circ }F\) outside and only 5 times per
hour if it’s \(70^{\circ }F\). Assuming that the number of howls per hour, \(N\), can be represented by a linear function of temperature Fahrenheit,
find the number of howls per hour he’ll make when it’s only \(20^{\circ }F\) outside. What troubles do you encounter when trying to determine
a reasonable applied domain?
\(N(T) = -\frac {2}{15}T + \frac {43}{3}\) and \(N(20) = \frac {35}{3} \approx 12\) howls per hour.
Having a negative number of howls makes no sense and since \(N(107.5) = 0\) we can put an upper bound of \(107.5^{\circ }F\) on the domain. The lower
bound is trickier because there’s nothing other than common sense to go on. As it gets colder, he howls more often. At some
point it will either be so cold that he freezes to death or he’s howling non-stop. So we’re going to say that he can withstand
temperatures no lower than \(-42^{\circ }F\) so that the applied domain is \([-42, 107.5]\).
Economic forces have changed the cost function for PortaBoys to \(C(x) = 105x + 175\). Rework Example PortaBoyCost with this new cost
function.
\(C(0) = 175\), so our start-up costs are \(\$ 175\). \(C(5) = 700\), so to produce \(5\) systems, it costs \(\$ 700\).
Since we can’t make a negative number of game systems, \(x \geq 0\).
The slope is \(m = 105\) so for each additional system produced, it costs an additional \(\$105\).
Solving \(C(x) = 15000\) gives \(x \approx 141.19\) so \(141\) can be produced for \(\$ 15, \! 000\).
In response to the economic forces in Exercise exerredoportaboy above, the local retailer sets the selling price of a PortaBoy at $250.
Remarkably, 30 units were sold each week. When the systems went on sale for $220, 40 units per week were sold. Rework
Example PortaBoyDemand with this new data.
\(p(x) = -3x+340\), \(0 \leq x \leq 113\).
The slope is \(m = -3\) so for each \(\$3\) drop in price, we sell one additional game system.
Since \(x = 150\) is not in the domain of \(p\), \(p(150)\) is not defined. (In other words, under these conditions, it is impossible to sell 150 game
systems.)
Solving \(p(x) = 150\) gives \(x \approx 63.33\) so if the price \(\$150\) per system, we would sell \(63\) systems.
A local pizza store offers medium two-topping pizzas delivered for \(\$6.00\) per pizza plus a \(\$1.50\) delivery charge per order. On
weekends, the store runs a ‘game day’ special: if six or more medium two-topping pizzas are ordered, they are \(\$5.50\) each with no
delivery charge. Write a piecewise-defined linear function which calculates the cost in dollars \(C(p)\) of \(p\) medium two-topping pizzas
delivered during a weekend.
A restaurant offers a buffet which costs \(\$15\) per person. For parties of \(10\) or more people, a group discount applies, and the cost is
\(\$12.50\) per person. Write a piecewise-defined linear function which calculates the total bill \(T(n)\) of a party of \(n\) people who all choose the
buffet.
\(T(n) = \left \{ \begin{array}{rcl} 15n & \mbox { if } & 1 \leq n \leq 9 \\ 12.5n & \mbox { if } & n \geq 10 \\ \end{array} \right .\)
A mobile plan charges a base monthly rate of \(\$10\) for the first \(500\) minutes of air time plus a charge of \(15\)¢ for each additional minute.
Write a piecewise-defined linear function which calculates the monthly cost in dollars \(C(m)\) for using \(m\) minutes of air
time.
You may wish to refer to number piecewisemobile for inspiration.
\(C(m) = \left \{ \begin{array}{rcl} 10 & \mbox { if } & 0 \leq m \leq 500 \\ 10+0.15(m-500) & \mbox { if } & m > 500 \end{array} \right .\)
The local pet shop charges \(12\)¢ per cricket up to 100 crickets, and \(10\)¢ per cricket thereafter. Write a piecewise-defined linear
function which calculates the price in dollars \(P(c)\) of purchasing \(c\) crickets.
\(P(c) = \left \{ \begin{array}{rcl} 0.12c & \mbox { if } & 1 \leq c \leq 100 \\ 12 + 0.1(c-100) & \mbox { if } & c > 100 \end{array} \right .\)
Here is a cross-section of a swimming pool.
Write a piecewise-defined linear function which describes the depth of the pool, \(D\) (in feet) as a function of:
the distance (in feet) from the edge of the shallow end of the pool, \(d\).
\[ D(d) = \left \{ \begin{array}{rcl} 8 & \mbox { if } & 0 \leq d \leq 15 \\ -\frac {1}{2} \, d + \frac {31}{2} & \mbox { if } & 15 \leq d \leq 27 \\ 2 & \mbox { if } & 27 \leq d \leq 37 \\ \end{array} \right .\]
the distance (in feet) from the edge of the deep end of the pool, \(s\).
\[ D(s) = \left \{ \begin{array}{rcl} 2 & \mbox { if } & 0 \leq s \leq 10 \\ \frac {1}{2} \, s -3 & \mbox { if } & 10 \leq s \leq 22 \\ 8 & \mbox { if } & 22 \leq s \leq 37 \\ \end{array} \right .\]
Graph each of the functions in (a) and (b). Discuss with your classmates how to transform one into the other and how
they relate to the diagram of the pool.
\(y=D(d)\)
\(y=D(s)\)
The function defined by \(I(x) = x\) is called the Identity Function. Thinking from a procedural perspective, explain a possible origin of
this name.
Since \(I(x) = x\) for all real numbers \(x\), the function \(I\) doesn’t change the ‘identity’ of the input at all.
Why must the graph of a function \(y = f(x)\) have at most one \(y\)-intercept?
Consider what would happen graphically if there were more than one …
If a graph contains more than one \(y\)-intercept, it would violate the Vertical Line Test since \(x=0\) would be matched with (at least)
two different \(y\)-values.
Why is a discussion of vertical lines omitted when discussing functions?
Vertical Lines fail the Vertical Line Test.
Find a formula for the \(x\)-intercept of the graph of \(f(x) = mx + b\). Assume \(m \neq 0\).
\(\left (- \frac {b}{m}, 0 \right )\). (Note the importance here of \(m \neq 0\).)
Suppose \((c,0)\) is the \(x\)-intercept of a linear function \(f\). Use the point-slope form of a linear function, Equation linearfunctionpointslope to show \(f(x) = m(x-c)\). This is the
‘slope \(x\)-intercept’ form of the linear function.
Plugging in \((c,0)\) for \((x_{0}, f(x_{0}))\), we get \(f(x) = f(x_{0}) + m (x - x_{0}) = 0 + m(x-c)\) or \(f(x) = m(x-c)\).
Prove that for all linear functions \(L\) with with slope \(3\), \(L(120) = L(100) + 60\).
Since \(L\) is linear with slope \(3\), \(L(x) = L(x_{0}) + m \Delta x = L(100) + (3)(120-100) = L(100)+60\).
Find the slopes between the following points from the data set given in Example timetempregressionex and compare them with the slope of the
corresponding regression line:
\((0, 64)\), \((4, 75)\)
\(m = \frac {75-64}{4-0} = 2.75\)
\((4, 75)\), \((8, 83)\)
\(m=\answer {2}\)
\((8, 83)\), \((10, 83)\)
\(m=\answer {0}\)
\((10, 83)\), \((12, 82)\)
\(m=\answer {-\frac {1}{2}}\)
The first two points contributed to a regression line slope of \(m = 2.55\); the last two points contributed to a regression line slope of \(m=-0.25\).
Find the least squares regression line for these data and comment on the goodness of fit. (We’ll develop
more sophisticated models for the growth of populations in Section ApplicationsofExponentialandLogarithmicFunctions. For the moment, we use a theorem from
Calculus to approximate those functions with lines.) Interpret the slope of the line of best fit.
\(y = 936.31x - 1645322.6\) with \(r=0.9696\) which indicates a good fit. The slope \(936.31\) indicates Lake County’s population is increasing at a rate of
(approximately) 936 people per year.
Use the regression line to predict the population of Lake County in 2010. (The recorded figure from the 2010 census is
\(230,\!041\))
According to the model, the population in 2010 will be \(236, \!660\).
Use the regression line to predict when the population of Lake County will reach \(250,\!000\).
According to the model, the population of Lake County will reach \(250,\!000\) sometime between 2024 and 2025.
Find the least squares regression line for these data and comment on the goodness of fit. Interpret the slope
of the line of best fit.
\(y = 796.8x - 1309762.5\) with \(r=0.8916\) which indicates a reasonable fit. The slope \(796.8\) indicates Lorain County’s population is increasing at a rate of
(approximately) 797 people per year.
Use the regression line to predict the population of Lorain County in 2010. (The recorded figure from the 2010 census is
\(301,\!356\))
According to the model, the population in 2010 will be \(291, \! 805\).
Use the regression line to predict when the population of Lake County will reach \(325,\!000\).
According to the model, the population of Lake County will reach \(325,\!000\) sometime between 2051 and 2052.
The chart below contains a portion of the fuel consumption information for a 2002 Toyota Echo that Jeffrey used to own. The
first row is the cumulative number of gallons of gasoline that I had used and the second row is the odometer reading when I
refilled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25
gallons of gasoline when the odometer read 1051 miles.
Gasoline Used
(Gallons)
0
9.26
19.03
28.25
36.45
44.64
53.57
62.62
71.93
81.69
90.43
Odometer
(Miles)
41
356
731
1051
1347
1631
1966
2310
2670
3030
3371
Find the least squares line for this data. Is it a good fit? What does the slope of the line represent? Do you and your
classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years
ago?
The regression line is \(y = 36.8x + 16.39\) with \(r = .99987\), so this is an excellent fit. The slope \(36.8\) represents mileage in miles per gallon.
Using the energy production data given below
Year
1950
1960
1970
1980
1990
2000
Production
(in Quads)
35.6
42.8
63.5
67.2
70.7
71.2
Plot the data using a graphing utility and explain why it does not appear to be linear.
Discuss with your classmates why ignoring the first two data points may be justified from a historical
perspective.
Find the least squares regression line for the last four data points and comment on the goodness of fit. Interpret
the slope of the line of best fit.
\(y = 0.266x - 459.86\) with \(r = 0.9607\) which indicates a good fit. The slope \(0.266\) indicates the country’s energy production is increasing at a rate of \(0.266\) Quad
per year.
Use the regression line to predict the annual US energy production in the year \(2010\).
According to the model, the production in 2010 will be \(74.8\) Quad.
Use the regression line to predict when the annual US energy production will reach \(100\) Quads.
According to the model, the production will reach \(100\) Quad in the year 2105.
The height of an object dropped from the roof of a building is modeled by: \(h(t) = -16t^2 + 64\), for \(0 \leq t \leq 2\). Here, \(h(t)\) is the height of the object off the
ground in feet \(t\) seconds after the object is dropped. Find and interpret the average rate of change of \(h\) over the interval
\([0,2]\).
The average rate of change is \(\frac {h(2) - h(0)}{2-0}=-32\). During the first two seconds after it is dropped, the object has fallen at an average rate of \(32\)
feet per second.
Using data from Bureau of Transportation Statistics, the average fuel economy \(F(t)\) in miles per gallon for passenger cars in the
US can be modeled by \(F(t) = -0.0076t^2+0.45t + 16\), \(0 \leq t \leq 28\), where \(t\) is the number of years since \(1980\). Find and interpret the average rate of change of \(F\) over the
interval \([0,28]\).
The average rate of change is \(\frac {F(28) - F(0)}{28-0}=0.2372\). From 1980 to 2008, the average fuel economy of passenger cars in the US increased, on
average, at a rate of \(0.2372\) miles per gallon per year.
The temperature \(T(t)\) in degrees Fahrenheit \(t\) hours after 6 AM is given by:
Find and interpret \(T(4)\), \(T(8)\) and \(T(12)\).
\(T(4) = 56\), so at 10 AM (4 hours after 6 AM), it is \(56^{\circ }\)F. \(T(8) = 64\), so at 2 PM (8 hours after 6 AM), it is \(64^{\circ }\)F. \(T(12) = 56\), so at 6 PM (12 hours after 6 AM), it
is \(56^{\circ }\)F.
Find and interpret the average rate of change of \(T\) over the interval \([4,8]\).
The average rate of change is \(\frac {T(8)-T(4)}{8-4}=2\). Between 10 AM and 2 PM, the temperature increases, on average, at a rate of \(2^{\circ }\)F per
hour.
Find and interpret the average rate of change of \(T\) from \(t=8\) to \(t=12\).
The average rate of change is \(\frac {T(12)-T(8)}{12-8}=-2\). Between 2 PM and 6 PM, the temperature decreases, on average, at a rate of \(2^{\circ }\)F per
hour.
Find and interpret the average rate of temperature change between 10 AM and 6 PM.
The average rate of change is \(\frac {T(12)-T(4)}{12-4}=0\). Between 10 AM and 6 PM, the temperature, on average, remains constant.
Suppose \(C(x) = x^2-10x+27\) represents the costs, in hundreds, to produce \(x\)thousand pens. Find and interpret the average rate of change as
production is increased from making 3000 to 5000 pens.
The average rate of change is \(\frac {C(5)-C(3)}{5-3}=-2\). As production is increased from 3000 to 5000 pens, the cost decreases at an average rate of
\(\$200\) per 1000 pens produced (20¢ per pen.)
Recall from Example ARCRocketExample The formula \(s(t) = -5t^2+100t\) for \(0 \leq t \leq 20\) gives the height, \(s(t)\), measured in feet, of a model rocket above the Moon’s surface as
a function of the time after lift-off, \(t\), in seconds.
Find and interpret the average rate of change of \(s\) over the following intervals:
\([14.9, 15]\)
\(-49.5\) so the average velocity of the rocket between \(14.9\) and \(15\) seconds after lift off is \(-49.5\) feet per second (\(49.5\) feet per second
directed downwards.)
\([15, 15.1]\)
\(-50.5\) so the average velocity of the rocket between \(14\) and \(15.1\) seconds after lift off is \(-50.5\) feet per second. (\(50.5\) feet per second
directed downwards.)
\([14.99, 15]\)
\(-49.95\) so the average velocity of the rocket between \(14.99\) and \(15\) seconds after lift off is \(-49.95\) feet per second. (\(49.95\) feet per second
directed downwards.)
\([15, 15.01]\)
\(-50.05\) so the average velocity of the rocket between \(15.01\) and \(15\) seconds after lift off is \(-50.05\) feet per second. (\(50.05\) feet per second
directed downwards.)
What value does the average rate of change appear to be approaching as the interval shrinks closer to the value
\(t=15\)?
The average rate of change seems to be approaching \(-50\).
Find the equation of the line containing \((15, 375)\) with slope \(m = -50\) and graph it along with \(s\) on the same set of axes using a graphing
utility. What happens as you zoom in near \((15, 375)\)?
Line: \(y = -50(t-15) + 375\) or \(y = -50t + 1125\). Graphing this line along with the \(s\) on a graphing utility we find the two graphs become indistinguishable as
we zoom in near \((15, 375)\).
Show the average rate of change of a function of the form \(f(x) = mx+b\) over any interval is \(m\).
Why doesn’t the graph of the vertical line \(x = b\) in the \(xy\)-plane represent \(y\) as a function of \(x\)?
With help from a graphing utility, graph the following pairs of functions on the same set of axes: (See Example greatestintegerdefn for the
definition of \(\lfloor x \rfloor \).)
\(f(x) = 2-x\) and \(g(x) = \lfloor 2-x \rfloor \)
\(f(x) = x^2-4\) and \(g(x) = \lfloor x^2 -4\rfloor \)
\(f(x) = x^3\) and \(g(x) = \lfloor x^3 \rfloor \)
Choose more functions \(f(x)\) and graph \(y = f(x)\) alongside \(y = \lfloor f(x) \rfloor \) until you can explain how, in general, one would obtain the graph of \(y = \lfloor f(x) \rfloor \) given the
graph of \(y = f(x)\).
The Lagrange Interpolate function \(L\) for two points \((x_{0}, y_{0})\) and \((x_{1}, y_{1})\) where \(x_{0} \neq x_{1}\) is given by:
For each of the following pairs of points, find \(L(x)\) using the formula above and verify each of the points lies on the
graph of \(y = L(x)\).
\((-1,3)\), \((2,3)\)
\(L(x) = 3\)
\((-3,-2)\), \((5,-2)\)
\(L(x) = -2\)
\((-3,-2)\), \((0,1)\)
\(L(x) = x+1\)
\((-1,5)\), \((2,-1)\)
\(L(x) = -2x+3\)
Verify that, in general, \(L(x_{0}) = y_{0}\) and \(L(x_{1}) = y_{1}\).
Show the point-slope form of a linear function, Equation linearfunctionpointslope is equivalent to the formula given for \(L(x)\) after making the
identifications: \(f(x_{0}) = y_{0}\) and \(m = \dfrac {y_{1} - y_{0}}{x_{1} - x_{0}}\).
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