1.A: Answers

Important Note about Precision of Answers: In many of the problems in this book you are required to read information from a graph and to calculate with that information. You should take reasonable care to read the graphs as accurately as you can (a small straightedge is helpful), but even skilled and careful people make slightly different readings of the same graph. That is simply one of the drawbacks of graphical information. When answers are given to graphical problems, the answers should be viewed as the best approximations we could make, and they usually include the word “approximately” or the symbol “≈” meaning “approximately equal to.” Your answers should be close to the given answers, but you should not be concerned if they differ a little. (Yes those are vague terms, but it is all we can say when dealing with graphical information.)

Section 1.0

1. 1. \(m = \frac{y - 9}{x - 3}\). If \(x = 2.97\), \(m = \frac{-0.1791}{-0.03} = 5.97\). If \(x = 3.001\), \(m = \frac{0.006001}{0.001} = 6.001\). If \(x = 3 + h\), \(\displaystyle m = \frac{(3+h)^2 - 9}{(3+h) - 3} = \frac{9 + 6h + h^2 - 9}{h} = 6 + h\)
   2. When \(h\) is close to \(0\), \(6 + h\) is close to \(6\).
3. 1. \(m = \frac{y - 4}{x - 2}\). If \(x = 1.99\), \(m = \frac{-0.0499}{-0.01} = 4.99\). If \(x = 2.004\), \(m = \frac{0.020016}{0.004} = 5.004\). If \(x = 2 + h\), \(\displaystyle m = \frac{\left[(2+h)^2 + (2+h) - 2 \right] - 4}{(2+h) - 2} = 5 + h\)
   2. When \(h\) is very small, \(5 + h\) is very close to \(5\).
5. All of these answers are approximate. Your answers should be close to these numbers.
   1. average rate of temperature change \(\displaystyle \approx \frac{80^{\circ} - 64^{\circ}}{1\text{ p.m.} - 9\text{ a.m.}} = \frac{16^{\circ}}{4\text{ hours}} = 4  \frac{\circ}{\text{hour}} \)
   2. At 10 a.m., temperature was rising about \(5^{\circ}\) per hour; at 7 p.m., its was rising about \(-10^{\circ}\)/hr (falling about \(10^{\circ}\)/hr).
7. All of these answers are approximate. Your answers should be close to these numbers.
   1. average velocity \(\displaystyle \approx \frac{300\mbox{ ft } - 0\mbox{ ft }}{20\mbox{ sec } - 0\mbox{ sec }} = 15 \, \frac{\mbox{ft}}{\mbox{sec}}\)
   2. average velocity \(\displaystyle \approx \frac{100\mbox{ ft } - 200\mbox{ ft }}{30\mbox{ sec } - 10\mbox{ sec }} = -5 \, \frac{\mbox{ft}}{\mbox{sec}}\)
   3. At \(t = 10\) seconds, velocity \(\approx 30\) feet per second (between \(20\) and \(35\) ft/sec); at \(t = 20\) seconds, velocity \(\approx -1\) feet per second; at \(t = 30\) seconds, velocity \(\approx -40\) feet per second.
9. 1. \(A(0) = 0\), \(A(1) = 3\), \(A(2) = 6\), \(A(2.5) = 7.5\), \(A(3) = 9\)
   2. The area of the rectangle bounded below by the \(t\)-axis, above by the line \(y\) = 3, on the left by the vertical line \(t = 1\) and on the right by the vertical line \(t = 4\).
   3. Graph of \(y = 3x\).

Section 1.1

1. 1. \(2\)
   2. \(1\)
   3. DNE (does not exist)
   4. \(1\)
3. 1. \(1\)
   2. \(-1\)
   3. \(-1\)
   4. \(2\)
5. 1. \(-7\)
   2. \(\frac{13}{0}\) (DNE)
7. 1. \(0.54\) (radian mode!)
   2. \(-0.318\)
   3. \(-0.54\)
9. 1. \(0\)
   2. \(0\)
   3. \(0\)
11. 1. \(0\)
    2. \(-1\)
    3. DNE
13. The one- and two-sided limits agree at \(x = 1\), \(x = 4\) and \(x = 5\), but not at \(x = 2\): \[ \displaystyle \lim_{x\to 1^{-}} \, g(x) = 1 \qquad \lim_{x\to 1^{+}} \, g(x) = 1 \qquad \lim_{x\to 1} \, g(x) = 1\nonumber\] \[ \displaystyle \lim_{x\to 2^{-}} \, g(x) = 1 \qquad \lim_{x\to 2^{+}} \, g(x) = 4 \qquad \lim_{x\to 2} \, g(x) = \text{DNE}\nonumber\] \[ \displaystyle \lim_{x\to 4^{-}} \, g(x) = 2 \qquad \lim_{x\to 4^{+}} \, g(x) = 2 \qquad \lim_{x\to 4} \, g(x) = 2\nonumber\] \[ \displaystyle \lim_{x\to 5^{-}} \, g(x) = 1 \qquad \lim_{x\to 5^{+}} \, g(x) = 1 \qquad \lim_{x\to 5} \, g(x) = 1\nonumber\]
15. 1. \(1.0986\)
    2. \(1\)
17. 1. \(0.125\)
    2. \(3.5\)
19. 1. \(A(0) = 0\), \(A(1) = 2.25\), \(A(2) = 5\), \(A(3) = 8.25\)
    2. \(A(x) = 2x + \frac14 x^2\)
    3. The area of the trapezoid bounded below by the \(t\)-axis, above by the line \(y = \frac12 t + 2\), on the left by the vertical line \(t = 1\) and on the right by the vertical line \(t = 3\).

Section 1.2

1. 1. \(2\)
   2. \(0\)
   3. DNE
   4. \(1.5\)
3. 1. \(1\)
   2. \(3\)
   3. \(1\)
   4. \(\approx 0.8\)
5. 
7. 1. \(2\)
   2. \(-1\)
   3. DNE
   4. \(2\)
   5. \(2\)
   6. \(2\)
   7. \(1\)
   8. \(2\)
   9. DNE
9. 1. When \(v = 0\), \(L = A\).
   2. \(0\)
11. 1. \(4\)
    2. \(1\)
    3. \(2\)
    4. \(0\)
    5. \(1\)
    6. \(1\)
13. 1. Slope of the line tangent to the graph of \(y = \cos(x)\) at the point \((0,1)\).
    2. slope \(= 0\)
15. 1. \(\approx 1\)
    2. \(\approx 3.43\)
    3. \(\approx 4\)
17. At \(x = -1\): “connected and smooth”; at \(x = 0\): “connected with a corner”; at \(x = 1\): “simple hole”; at \(x = 2\): “vertical jump”; at \(x = 3\): “simple hole”; at \(x = 4\): “corner”; at \(x = 5\): “smooth”
19. Many lists will work. Here is one example: Put \(a_n = 2 + \frac{1}{n}\) so \(a_n\) approaches \(2\) and \(\frac{\left|a_n-2\right|}{a_n-2} = 1\) for all \(n\). Put \(b_n = 2 - \frac{1}{n}\) so \(b_n \rightarrow 2\) and \(\frac{\left|b_n-2\right|}{b_n-2} = -1\) for all \(n\).
21. \(\displaystyle -x^2 \leq x^2\cos\left(\frac{1}{x^2}\right) \leq x^2\) so limit is \(0\)
23. \(\displaystyle -x^2 \leq x^2\sin\left(\frac{1}{x}\right) \leq x^2\) so limit is \(3\)
25. \(\displaystyle \frac{1}{x^2} -1 < \left\lfloor \frac{1}{x^2} \right\rfloor \leq \frac{1}{x^2}\) so limit is \(1\)

Section 1.3

1. Discontinuous at \(1\), \(3\) and \(4\).
3. 1. Discontinuous at \(x = 3\): fails condition (i) there.
   2. At \(x = 2\): fails (i).
   3. Where \(\cos(x) < 0\) (for example, at \(x = \pi\)): fails (i).
   4. Where \(x^2\) is an integer (for example, at \(x = 1\) or \(x = 2\)): fails (ii).
   5. Where \(\sin(x) = 0\) (for example, at \(x = 0\), \(x = \pm \pi\), \(\pm 2\pi\ldots\)): fails (i).
   6. At \(x = 0\): fails (i).
   7. At \(x = 0\): fails (i).
   8. At \(x = 3\): fails (i).
   9. At \(x = \frac{\pi}{2}\): fails (i).
5. 1. \(f(x) = 0\) for at least \(3\) values of \(x\) in the interval \(0 \leq x \leq 5\).
   2. \(1\)
   3. \(3\)
   4. \(2\)
   5. Yes. (It does not have to happen, but it is possible.)
7. 1. \(f(0) = 0\), \(f(3) = 9\) and \(0 \leq 2 \leq 9\); \(c = \sqrt{2} \approx 1.414\)
   2. \(f(-1) = 1\), \(f(2) = 4\) and \(1 \leq 3 \leq 4\); \(c = \sqrt{3} \approx 1.732\)
   3. \(f(0) = 0\), \(f(\frac{\pi}{2} = 1\) and \(0 \leq \frac12 \leq 1\); \(c = \arcsin(\frac12) \approx 0.524\)
   4. \(f(0) = 0\), \(f(1) = 1\) and \(0 \leq \frac13 \leq 1\); \(c = \frac13\)
   5. \(f(2) = 2\), \(f(5) = 20\) and \(2 \leq 4 \leq 20\); \(c = \frac{1 + \sqrt{17}}{2} \approx 2.561\)
   6. \(f(1) = 0\), \(f(10) \approx 2.30\) and \(0 \leq 2 \leq 2.30\); \(c = e^2 \approx 7.389\)
9. Neither student is correct. The bisection algorithm converges to the root labeled \(C\).
11. 1. \(D\)
    2. \(D\)
    3. hits \(B\)
13. \(\left[ -0.9375, -0.875\right]\), root \(\approx -0.879\); \(\left[ 1.3125, 1.375\right]\), root \(\approx 1.347\); \)\left[ 2.5, 2.5625\right]\), root \(\approx 2.532\)
15. \(\left[ 2.3125, 2.375\right]\), root \(\approx 2.32\)
17. \(\left[ -0.375, -0.3125\right]\), root \(\approx -0.32\)
19. 
21. 1. \(A(2.1) - A(2)\) is the area of the region bounded below by the \(t\)-axis, above by the graph of \(y = f(t)\), on the left by the vertical line \(t = 2\), and on the right by the vertical line \(t = 2.1\). \(\frac{A(2.1) - A(2)}{0.1} \approx f(2)\mbox{ or }f(2.1) \quad \Rightarrow \quad \frac{A(2.1) - A(2)}{0.1} \approx 1\)
    2. \(A(4.1) - A(4)\) is the area of the region bounded below by the \(t\)-axis, above by the graph of \(y = f(t)\), on the left by the vertical line \(t = 4\), and on the right by the vertical line \(t = 4.1\). \(\frac{A(4.1) - A(4)}{0.1} \approx f(4) \approx 2\)
23. 1. Yes (you justify).
    2. Yes.
    3. Try it.

Section 1.4

1. 1. If \(x\) is within \(\frac12\) unit of \(3\) then \(2x+1\) is within \(1\) unit of \(7\).
   2. \(0.3\)
   3. \(0.02\)
   4. \(\frac{\epsilon}{2}\)
3. 1. If \(x\) is within \(\frac14\) unit of \(2\) then \(4x-3\) is within \(1\) unit of \(5\).
   2. \(0.1\)
   3. \(0.02\)
   4. \(\frac{\epsilon}{4}\)\end{enumerate*}
5. In 1, \(m = 2\), \(\delta = \frac{\epsilon}{2}\); in 2, \(m = 3\), \(\delta = \frac{\epsilon}{3}\); in 3, \(m = 4\), \(\delta = \frac{\epsilon}{4}\); in 4, \(m = 5\), \(\delta = \frac{\epsilon}{5}\). In general: \(\delta = \frac{\epsilon}{\left|m\right|}\)
7. \(0.02\) inches
9. 1. Any value of \(x\) between \(\sqrt[3]{7.5} \approx 1.957\) and \(\sqrt[3]{8.5} \approx 2.043\): If \(x\) is within \(0.043\) units of \(2\) then \(x^3\) will be within \(0.5\) units of \(8\).
   2. Any \(x\) between \(\sqrt[3]{7.95} \approx 1.9958\) and \(\sqrt[3]{8.05}\approx 2.0042\): If \(x\) is within \(0.0042\) units of \(2\) then \(x^3\) will be within \(0.05\) units of \(8\).
11. 1. Any value of \(x\) between \(0\) and \(8\): If \(x\) is within \(3\) units of \(3\) then \(\sqrt{1+x}\) will be within \(1\) unit of \(2\).
    2. Any \(x\) between \(2.99920004\) and \(3.00080004\): If \(x\) is within \(0.00079996\) units of \(3\) then \(\sqrt{1+x}\) will be within \(0.0002\) units of \(2\).
13. \(0.0059964\) inches
15. 
17. On your own.
19. Assume \(\displaystyle \lim_{x\to 2} \, f(x)\) exists. Let \(\epsilon = 0.1\), so there must be a \(\delta\) so that:\[2-\delta < x < 2+\delta \Rightarrow L-0.1 < f(x) < L+0.1\]Now, \(f(x) = 4\) for any \(x\) with \(2-\delta < x < 2\), so it must be true that: \begin{align*}2-\delta < x < 2 &\Rightarrow L-0.1 < 4 < L+0.1\\ &\Rightarrow L > 3.9\end{align*} Similarly, \(f(x) = 3\) for any \(x\) with \(2 < x< 2+\delta\), so it must be true that: \begin{align*}2< x < 2+\delta &\Rightarrow L-0.1 < 3 < L+0.1\\ &\Rightarrow L<3.1\end{align*} But no value of \(L\) can simultaneously satisfy \(L>3.9\) and \(L<3.1\), so we have reached a contradiction and our assumption must be false: \(\displaystyle \lim_{x\to 2} \, f(x)\) does not exist.
21. Assume \(\displaystyle \lim_{x\to 2} \, f(x)\) exists. Let \(\epsilon = 0.1\), so there must be a \(\delta\) so that:\[2-\delta < x < 2+\delta \Rightarrow L-0.1 < f(x) < L+0.1\]We can assume that \(\delta \leq 1\) (if not, we can replace our initial \(\delta\) with \(\delta = 1\) because any smaller value will also work). Now, \(f(x) = x\) for any \(x\) with \(2-\delta < x < 2\), so it must be true that: \begin{align*}2-\delta < x < 2 &\Rightarrow L-0.1 < x < L+0.1\\ &\Rightarrow L < 0.1 + x < 2.1\end{align*} On the other hand, \(f(x) = 6-x\) for any \(x\) with \(2 < x< 2+\delta\leq 3\), so it must be true that: \begin{align*}2< x < 2+\delta &\Rightarrow L-0.1 < 6-x < L+0.1\\ &\Rightarrow L>5.9-x\geq 2.9\end{align*} (because \(x\leq 3\)). But no value of \(L\) can simultaneously satisfy \(L<2.1\) and \(L>2.9\), so we have reached a contradiction and our assumption must be false: \(\displaystyle \lim_{x\to 2} \, f(x)\) does not exist.
23. Given any \(\epsilon > 0\), we know \(\frac{\epsilon}{2} > 0\), so there is a number \(\delta_{f} > 0\) such that \(\left| x - a \right| < \delta_{f} \Rightarrow \left| f(x) - L \right| < \frac{\epsilon}{2}\). Likewise, there is a number \(\delta_{g} > 0\) such that \(\left| x - a \right| < \delta_{g} \Rightarrow \left| g(x) - L \right| < \frac{\epsilon}{2}\). Let \(\delta\) be the smaller of \(\delta_{f}\) and \(\delta_{g}\). If \(\left| x - a \right| < \delta\) then \(\left| f(x) - L \right| < \frac{\epsilon}{2}\) and \(\left| g(x) - M \right| < \frac{\epsilon}{2}\) so: \begin{align*} \left| (f(x) - g(x))\right. &- \left. (L - M)) \right|\\ &= \left| (f(x) - L) - (g(x) - M) \right| \\ &\leq \left| f(x) - L \right| + \left| g(x) - M \right|\\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \\ \end{align*} so \(f(x)-g(x)\) is within \(\epsilon\) of \(L-M\) whenever \(x\) is within \(\delta\) of \(a\).