5.1E Exercises
- Page ID
- 154497
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Tell whether the function is linear and write it in standard form if so, identifying the leading coefficient and the constant term.
- \( f(x) = 5+3x\)
- \( g(x) = 2x^2 + 1 \)
- \( h(x) = 9x + 1 - \sqrt{x} \)
- \( p(x) = c_1 x + c_0 \), where \(c_0, c_1 \) are real numbers.
- \( q(x) = 2(x+3) \)
- \( r(x) = \dfrac{2x+3}{x+1} \)
- \( C(x) = 30x + 300 \)
- \( R(x) = 150x \)
- \( P(x) = 200x - (1200 + 4x^2) \)
- \( T(t) = t + \dfrac{1}{t} \)
- Answer
-
- Yes, \( a_1 = 3\), \( a_0 = 5\).
- No.
- No.
- Yes, \(c_1\) is leading coefficient (gives slope) and \(c_0\) is the constant term.
- Yes, \( q(x) = 2x + 6 \), so \(a_1 = 2, a_0 = 6 \).
- No.
- Yes, \( a_1 = 30, a_0 = 300\).
- Yes, \(a_1 = 150, a_0 = 0 \).
- No.
- No.
If a function passes through the graphed points, tell whether you believe it is a linear function. If so, write the function in standard form.
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- Answer
-
- Yes, this is \( f(x) = -x + 10 \).
- Yes, this is \( f(x) = \frac{1}{3}x \).
- No.
- No.
Sketch the graph of the linear function (without using technology).
- \( f(x) = 5+3x\)
- \( g(x) = \frac{1}{2}x - 5 \)
- \( h(x) = -2x + 10 \)
- \( p(x) = 2(x-3) \)
- \( C(x) = 30x + 300 \)
- \( R(x) = 150x \)
- Answer
-
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Using the table of values, find the average rate of change of the function for the indicated intervals.
| \( t\) | \( f(t) \) |
| 0 | 0 |
| 1 | 2 |
| 2 | 2 |
| 3 | 5 |
| 4 | 6 |
- \( t = 0 \) to \(t = 1 \)
- \( t = 1\) to \( t = 2\)
- \( t = 0\) to \( t = 3 \)
- \( t = 1\) to \( t = 3\)
- \( t = 3 \) to \(t = 4 \)
- \( t = 2\) to \(t = 4 \)
- Answer
-
Identify the two points of interest on the table, \( (t_1, f(t_1)) \) and \( (t_2, f(t_2)) \), and then compute the average rate of change (slope calculation) between them.
- \(2\)
- \( 0\)
- \( \frac{5}{3} \)
- \(\frac{3}{2}\)
- \( 1\)
- \( 2\)
Recall that the average rate of change of a function between two points is the slope of the secant line drawn through those same points. Using the graph below, sketch the secant lines and find the average rate of change for the indicated intervals.
- \( t = 0 \) to \(t = 1 \)
- \( t = 1\) to \( t = 2\)
- \( t = 0\) to \( t = 3 \)
- \( t = 1\) to \( t = 3\)
- \( t = 3 \) to \(t = 4 \)
- \( t = 2\) to \(t = 4 \)
- Answer
-
You should get the same answers as the last problem in the end. The secant lines should look like this:
Note: the numbers in the problems below have been thoroughly cooked, so even though they're "applications," you should not be using a calculator. This kind of thing is totally reasonable on a test without calculators.
1. I used terrarium-grade silicone caulk rather than aquarium-grade in constructing my axolotl habitat, like a n00b. Now water is seeping out the seams at a constant rate of 0.25 gallon per hour. If the tank originally had 100 gallons in it, write the volume of water left in the tank as a function of time. If the tank leaks all night (for 8 hours), how much water will be in the tank in the morning? How long would it take to completely drain if I don't fix it asap?
2. Swimming at the surface of the ocean, the pressure is the same as the air pressure at sea level, 1 atmosphere (14.6 psi). Diving down, the pressure increases by 1 atmosphere every 10 meters deeper you go. Find a linear model that gives the pressure as a function of depth below the surface (in meters). At a depth of 50 meters, what will the pressure be? At what depth will the pressure be 101 atmospheres?
3. Back in the day before unlimited phone plans, my monthly phone bill charged a $15 flat fee for up to 100 text messages. For every text message over 100, it charged an extra $0.10. Denoting the number of texts sent in a month by \(x\), write a linear function to calculate the cost of my bill, assuming I always send more than 100 messages. If I send 120 texts in a month, what will the bill be?
- Answer
-
1. \( V(t) = 100 - 0.25t \). \( V(8) = 98 \)gal. It will take \(t = 400\)hrs for all the water to leak out, so I have time to fix the tank!
2. \( P(d) = 1+\frac{1}{10}d \). \(P(50) = 6\)atm. \(P = 101\)atm if \(d = 1000\)m.
3. \( C(x) = 15+0.1(x-100) \). \(C(120) = $17\).
From 2000 to 2020, the population growth rates of Podunk Town, Kansas and One-Horse Town, Kansas were both roughly linear. If Podunk had a population of 4100 in the year 2000, and the population grew by about 80 people each year, write a linear model for population as a function of time in years after 2000. Do the same for One-Horse, which started with 6500 people but only had net population growth of about 40 people each year. What was each town's population in 2010? Was there any year between 2000 and 2020 when Podunk's population overtook that of One-Horse?
- Answer
-
Podunk's population is \(P(t) = 4100+80t \) where \(t\) is number of years after 2000. Likewise, One-Horse's population is \(O(t) = 6500 + 40t\). 2010 is 10 years past 2000, so we find \( P(10) = 4900\) and \( O(10) = 6900 \).
Rather than checking the populations for each year to see if Podunk ever caught up with One-Horse, we could note that such a crossing-over would happen at an intersection of the two linear functions. We try setting the functions equal to each other to see if there is a reasonable value for \(t\) where they intersect.
\[ 4100+80t = 6500 + 40t \quad \longrightarrow \quad t = 60 \notag \]
Hmm, no, it would take 60 years after 2000 for Podunk to win out, but this model is only valid for 2000-2020. One-Horse always had the larger population in this time period.
The table below gives the monthly cost and revenue values for the production and sale of \(x\) washing machines. Identify whether cost and/or revenue are linear functions of \(x\). If so, find the function's algebraic definition.
| \( x\) | \( C(x)\) | \( R(x) \) |
| 0 | $2,000 | $0 |
| 20 | $2,582 | $22,400 |
| 50 | $3,429 | $56,000 |
| 100 | $4,800 | $112,000 |
| 200 | $7,434 | $224,000 |
- Answer
-
Cost does not demonstrate a linear relationship, but revenue does. The rate of change is a constant 1120, so the function is \(R(x) = 1120x \).