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Given the pair of functions \(f\) and \(F\), sketch the graph of \(y=F(x)\) by starting with the graph of \(y = f(x)\) and using Theorem linearlaurentlgraphs. Track at least two
points and the asymptotes. State the domain and range using interval notation.
Use the Desmos graph below with the settings \(f(x)=\frac {1}{x},\,h=2,\,k=1,\,a=1\)
Given the pair of functions \(f\) and \(F\), sketch the graph of \(y=F(x)\) by starting with the graph of \(y = f(x)\) and using Theorem linearlaurentlgraphs. Track at least two
points and the asymptotes. State the domain and range using interval notation.
Use the Desmos graph below with the settings \(f(x)=\frac {1}{x},\,h=-1,\,k=2,\,a=-2\)
Given the pair of functions \(f\) and \(F\), sketch the graph of \(y=F(x)\) by starting with the graph of \(y = f(x)\) and using Theorem linearlaurentlgraphs. Track at least two
points and the asymptotes. State the domain and range using interval notation.
Use the Desmos graph below with the settings \(f(x)=\frac {1}{x},\,h=-\frac {1}{2},\,k=2,\,a=-1\)
Given the pair of functions \(f\) and \(F\), sketch the graph of \(y=F(x)\) by starting with the graph of \(y = f(x)\) and using Theorem linearlaurentlgraphs. Track at least two
points and the asymptotes. State the domain and range using interval notation.
\(\ds {\lim _{x \rightarrow - \infty } f(x) = 0}\) More specifically, as \(x \rightarrow -\infty , f(x) \rightarrow 0^{-}\)
\(\ds {\lim _{x \rightarrow \infty } f(x) = 0}\) More specifically, as \(x \rightarrow \infty , f(x) \rightarrow 0^{+}\)
Consider the function \(g(t) = \dfrac {t}{t^{2} + 1}\)
State the domain.
\((-\infty , \infty )\)
Identify any vertical asymptotes of the graph.
No vertical asymptotes
Identify any holes in the graph.
No holes in the graph
Find the horizontal asymptote, if it exists. \(y = \answer {0}\)
Find the slant asymptote, if it exists.
No slant asymptote
Graph the function using a graphing utility and describe the behavior near the asymptotes.
Horizontal asymptote:
\(\ds {\lim _{t \rightarrow - \infty } g(t) = 0}\) More specifically, as \(t \rightarrow -\infty , g(t) \rightarrow 0^{-}\)
\(\ds {\lim _{t \rightarrow \infty } g(t) = 0}\) More specifically, as \(t \rightarrow \infty , g(t) \rightarrow 0^{+}\)
Consider the function \(g(t) = \dfrac {t + 7}{(t + 3)^{2}}\)
State the domain.
\((-\infty , -3) \cup (-3, \infty )\)
Identify any vertical asymptotes of the graph. \(x=\answer {-3}\)
Identify any holes in the graph.
No holes in the graph
Find the horizontal asymptote, if it exists. \(y = \answer {0}\)
Find the slant asymptote, if it exists.
No slant asymptote
Graph the function using a graphing utility and describe the behavior near the asymptotes.
Vertical asymptote:
\(\ds {\lim _{t \rightarrow -3} g(t) = \infty }\)
Horizontal asymptote:
\(\ds {\lim _{t \rightarrow - \infty } g(t) = 0}\)(This is hard to see on the calculator, but trust me, the graph is below the \(t\)-axis to the left of \(t = -7\).) More specifically,
as \(t \rightarrow -\infty , g(t) \rightarrow 0^{-}\)
\(\ds {\lim _{t \rightarrow \infty } g(t) = 0}\) More specifically, as \(t \rightarrow \infty , g(t) \rightarrow 0^{+}\)
Consider the function \(g(t) = \dfrac {t^{3} + 1}{t^{2} - 1}\)
State the domain.
\((-\infty , -1) \cup (-1, 1) \cup (1, \infty )\)
Identify any vertical asymptotes of the graph. \(\answer {t = 1}\)
Identify any holes in the graph. Hole at \(\answer {(-1, -\frac {3}{2})}\)
Find the horizontal asymptote, if it exists.
No horizontal asymptote
Find the slant asymptote, if it exists. \(y= \answer {t}\)
Graph the function using a graphing utility and describe the behavior near the asymptotes.
\(C(25) = 590\) means it costs $590 to remove 25% of the fish and and \(C(95)= 33630\) means it would cost $33630 to remove 95% of the fish from
the pond.
What does the vertical asymptote at \(x = 100\) mean within the context of the problem?
The vertical asymptote at \(x = 100\) means that as we try to remove 100% of the fish from the pond, the cost increases without
bound; i.e., it’s impossible to remove all of the fish.
What percentage of the Ippizuti fish can you remove for $40000?
For $40000 you could remove about 95.76% of the fish.
In the scenario of Example averagevelocityrocketex, \(s(t) = -5t^2+100t\), \(0 \leq t \leq 20\) gives the height of a model rocket above the Moon’s surface, in feet, \(t\) seconds after
liftoff. For each of the times \(t_{0}\) listed below, find and simplify a the formula for the average velocity \(\overline {v}(t)\) between \(t\)
and \(t_{0}\) (see Definition averagevelocitydefn) and use \(\overline {v}(t)\) to find and interpret the instantaneous velocity of the rocket at \(t = t_{0}\) (See Example
averagevelocityrocketex).
\(t_{0} = 5\)
\(\overline {v}(t) = \frac {s(t) - s(5)}{t - 5} = \frac {-5t^2+100t-375}{t-5} = -5t+75\), \(t \neq 5\). The instantaneous velocity of the rocket when \(t_{0} = 5\) is \(-5(5)+75 = 50\) meaning it is traveling \(50\) feet per second upwards.
\(t_{0} = 9\)
\(\overline {v}(t) = \frac {s(t) - s(9)}{t - 9} = \frac {-5t^2+100t-495}{t-9} = -5t+55\), \(t \neq 9\). The instantaneous velocity of the rocket when \(t_{0} = 9\) is \(-5(9)+55 = 10\), so the rocket has slowed to \(10\) feet per second (but still heading up.)
\(t_{0} = 10\)
\(\overline {v}(t) = \frac {s(t) - s(10)}{t - 10} = \frac {-5t^2+100t-495}{t-10} = -5t+50\), \(t \neq 10\). The instantaneous velocity of the rocket when \(t_{0} = 10\) is \(-5(10)+50 = 0\), so the rocket has momentarily stopped! In Example ARCRocketExample, we
learned the rocket reaches its maximum height when \(t = 10\) seconds, which means the rocket must change
direction from heading up to coming back down, so it makes sense that for this instant, its velocity is \(0\).
\(t_{0} = 11\)
\(\overline {v}(t) = \frac {s(t) - s(11)}{t - 11} = \frac {-5t^2+100t-495}{t-11} = -5t+45\), \(t \neq 11\). The instantaneous velocity of the rocket when \(t_{0} = 11\) is \(-5(11)+45 = -10\) meaning the rocket has, indeed, changed direction and is heading
downwards at a rate of \(10\) feet per second. (Note the symmetry here between this answer and our answer when \(t=9\).)
The population of Sasquatch in Portage County \(t\) years after the year 1803 is modeled by the function
\[P(t) = \frac {150t}{t + 15}.\]
Find and interpret the
horizontal asymptote of the graph of \(y = P(t)\) and explain what it means.
The horizontal asymptote of the graph of \(P(t) = \frac {150t}{t + 15}\) is \(y = 150\) and it means that the model predicts the population of Sasquatch in Portage
County will never exceed 150.
The cost in dollars, \(C(x)\) to make \(x\) dOpi media players is \(C(x) = 100x+2000\), \(x \geq 0\). You may wish to review the concepts of fixed and variable costs
introduced in Example PortaBoyCost in Section LinearFunctions.
Find a formula for the average cost \(\overline {C}(x)\).
Find and interpret \(\overline {C}(1)\) and \(\overline {C}(100)\).
\(\overline {C}(1) = 2100\) and \(\overline {C}(100) = 120\). When just \(1\) dOpi is produced, the cost per dOpi is \(\$2100\), but when \(100\) dOpis are produced, the cost per dOpi is \(\$120\).
How many dOpis need to be produced so that the average cost per dOpi is \(\$ 200\)?
\(\overline {C}(x) = 200\) when \(x = 20\). So to get the cost per dOpi to \(\$200\), \(20\) dOpis need to be produced.
We find \(\ds {\lim _{x \rightarrow 0^{+}} \overline {C}(x) = \infty }\). This means that as fewer and fewer dOpis are produced, the cost per dOpi becomes unbounded.
In this situation, there is a fixed cost of \(\$2000\) (\(C(0) = 2000\)), we are trying to spread that \(\$2000\) over fewer and fewer dOpis.
Interpret the behavior of \(\overline {C}(x)\) as \(x \rightarrow \infty \).
As \(x \rightarrow \infty \), \(\overline {C}(x) \rightarrow 100^{+}\). This means that as more and more dOpis are produced, the cost per dOpi approaches \(\$100\), but is always a little
more than \(\$100\). Since \(\$100\) is the variable cost per dOpi (\(C(x) = \underline {100}x+2000\)), it means that no matter how many dOpis are produced, the average
cost per dOpi will always be a bit higher than the variable cost to produce a dOpi. As before, we can attribute this to
the \(\$2000\) fixed cost, which factors into the average cost per dOpi no matter how many dOpis are produced.
This exercise explores the relationships between fixed cost, variable cost, and average cost. The reader is encouraged to
revisit Example PortaBoyCost in Section LinearFunctions as needed. Suppose the cost in dollars \(C(x)\) to make \(x\) items is given by \(C(x) = mx + b\) where \(m\) and \(b\) are positive real
numbers.
Show the fixed cost (the money spent even if no items are made) is \(b\).
The cost to make \(0\) items is \(C(0) = m(0)+b = b\). Hence, so the fixed costs are \(b\).
Show the variable cost (the increase in cost per item made) is \(m\).
\(C(x) = mx+b\) is a linear function with slope \(m>0\). Hence, the cost increases at a rate of \(m\) dollars per item made. Hence, the variable cost
is \(m\).
Find a formula for the average cost when making \(x\) items, \(\overline {C}(x)\).
\(\overline {C}(x) = \frac {C(x)}{x} = \frac {mx+b}{x} = m + \frac {b}{x}\) for \(x > 0\).
Show \(\overline {C}(x) > m\) for all \(x>0\) and, moreover, \(\overline {C}(x) \rightarrow m^{+}\) as \(x \rightarrow \infty \).
Since \(b>0\), \(\overline {C}(x) = m + \frac {b}{x} > m\) for \(x > 0\). As \(x \rightarrow \infty \), \(\frac {b}{x} \rightarrow 0\) so \(\overline {C}(x) = m + \frac {b}{x} \rightarrow m\).
Interpret \(\overline {C}(x) \rightarrow m^{+}\) both geometrically and in terms of fixed, variable, and average costs.
Geometrically, the graph of \(y = \overline {C}(x)\) has a horizontal asymptote \(y = m\), the variable cost. In terms of costs, as more items are
produced, the affect of the fixed cost on the average cost, \(\frac {b}{x}\) falls away so that the average cost per item approaches the
variable cost to make each item.
Suppose the price-demand function for a particular product is given by \(p(x) = mx + b\) where \(x\) is the number of items made and sold for \(p(x)\)
dollars. Here, \(m<0\) and \(b>0\). If the cost (in dollars) to make \(x\) of these products is also a linear function \(C(x)\), show that the graph of the
average profit function \(\overline {P}(x)\) has a slant asymptote with slope \(m\) and interpret.
If \(p(x) = mx + b\) and \(C(x)\) is linear, say \(C(x) = rx+s\), then we can compute the the profit function (in general) as: \(P(x) = xp(x) - C(x) = x(mx+b) - (rx+s)\) which simplifies to \(P(x) = mx^2 + (b-r)x -s\). Hence, the average
profit \(\overline {P}(x) = \frac {P(x)}{x} = \frac {mx^2 + (b-r)x -s}{x} = mx + (b-r) - \frac {s}{x}\). We see that as \(x \rightarrow \infty \), \(\frac {s}{x} \rightarrow 0\) so \(\overline {P}(x) \approx mx + (b-r)\). Hence, \(y = mx + (b-r)\) is the slant asymptote to \(y = \overline {P}(x)\). This means that as more items are sold, the average profit is
decreasing at approximately the same rate as the price function is decreasing, \(m\) dollars per item. That is, to sell one
additional item, we drop the price \(p(x)\) by \(m\) dollars which results in a drop in the average profit by approximately \(m\) dollars.
In Exercise circuitexercisepoly in Section GraphsofPolynomials, we fit a few polynomial models to the following electric circuit data. The circuit was built with a
variable resistor. For each of the following resistance values (measured in kilo-ohms, \(k \Omega \)), the corresponding
power to the load (measured in milliwatts, \(mW\)) is given below. (The authors wish to thank Don Anthan and
Ken White of Lakeland Community College for devising this problem and generating the accompanying data
set.)
Resistance: (\(k \Omega \))
1.012
2.199
3.275
4.676
6.805
9.975
Power: (\(mW\))
1.063
1.496
1.610
1.613
1.505
1.314
Using some fundamental laws of circuit analysis mixed with a healthy dose of algebra, we can derive the actual formula
relating power \(P(x)\) to resistance \(x\):
\[P(x) = \frac {25x}{(x + 3.9)^2}, \quad x \geq 0.\]
Graph the data along with the function \(y = P(x)\) using a graphing utility.
Use a graphing utility to approximate the maximum power that can be delivered to the load. What is the
corresponding resistance value?
The maximum power is approximately \(1.603 \; mW\) which corresponds to \(3.9 \; k\Omega \).
Find and interpret the end behavior of \(P(x)\) as \(x \rightarrow \infty \).
As \(x \rightarrow \infty , \; P(x) \rightarrow 0^{+}\) which means as the resistance increases without bound, the power diminishes to zero.
Let \(f(x) = \dfrac {ax^2-c}{x+3}\). Find values for \(a\) and \(c\) so the graph of \(f\) has a hole at \((-3, 12)\).
\(a = -2\) and \(c = -18\) so \(f(x) = \dfrac {-2x^2+18}{x+3}\).
Let \(f(x) = \dfrac {ax^{n} -4}{2x^2+1}\).
Find values for \(a\) and \(n\) so the graph of \(y = f(x)\) has the horizontal asymptote \(y = 3\).
\(a=6\) and \(n=2\) so \(f(x) = \dfrac {6x^{2} -4}{2x^2+1}\)
Find values for \(a\) and \(n\) so the graph of \(y=f(x)\) has the slant asymptote \(y = 5x\).
\(a=10\) and \(n = 3\) so \(f(x) = \dfrac {10x^{3} -4}{2x^2+1}\) .
Suppose \(p\) is a polynomial function and \(a\) is a real number. Define \(r(x)= \dfrac {p(x) - p(a)}{x-a}\). Use the Factor Theorem, Theorem factorthm, to prove the graph of \(y = r(x)\)
has a hole at \(x =a\).
If we define \(f(x) = p(x) - p(a)\) then \(f\) is a polynomial function with \(f(a) = p(a) - p(a) = 0\). The Factor Theorem guarantees \((x-a)\) is a factor of \(f(x)\), that is, \(f(x) = p(x) - p(a) = (x-a)q(x)\) for some
polynomial \(q(x)\). Hence, \(r(x) = \frac {p(x)-p(a)}{x-a} = \frac {(x-a)q(x)}{x-a} = q(x)\) so the graph of \(y = r(x)\) is the same as the graph of the polynomial \(y = q(x)\) except for a hole when \(x = a\).
For each function \(f(x)\) listed below, compute the average rate of change over the indicated interval. (See
Definition arc in Section AverageRateofChange for a review of this concept, as needed.) What trends do you observe? How do your
answers manifest themselves graphically? How do you results compare with those of Exercise monomialarcexercise in Section
GraphsofPolynomials?
The slope of the curves near \(x=1\) matches the exponent on \(x\). This exactly what we saw in Exercise monomialarcexercise in Section
GraphsofPolynomials.
In his now famous 1919 dissertation The Learning Curve Equation, Louis Leon Thurstone presents a rational function which
models the number of words a person can type in four minutes as a function of the number of pages of practice one has
completed. (This paper, which is now in the public domain and can be found here, is from a bygone era when students
at business schools took typing classes on manual typewriters.) Using his original notation and original language, we have \(Y = \frac {L(X + P)}{(X + P) + R}\)
where \(L\) is the predicted practice limit in terms of speed units, \(X\) is pages written, \(Y\) is writing speed in terms of words in four
minutes, \(P\) is equivalent previous practice in terms of pages and \(R\) is the rate of learning. In Figure 5 of the paper, he graphs a
scatter plot and the curve \(Y = \frac {216(X + 19)}{X + 148}\). Discuss this equation with your classmates. How would you update the notation? Explain what the
horizontal asymptote of the graph means. You should take some time to look at the original paper. Skip over the
computations you don’t understand yet and try to get a sense of the time and place in which the study was conducted.
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