### Differential Equations to Difference Equations

Chapter 2

Section 2.1

1. (a) $y$ is a function of $x$
Relationship: $y = 2x$
Domain $= [0, S]$, where $S$ is the maximum speed of the train
Range $= [0, 2S]$

(c) $y$ is a function of $x$
Domain $= [0, 8760]$, in hours
Range $= [221.5, 238.9]$, in thousands of miles

(d) $y$ is not a function of $x$

(f) $y$ is a function of $x$
Relationship: $y = 2\sqrt{\pi x}$
Domain $= (0, \infty)$
Range $= (0, \infty)$

(g) $y$ is a function of $x$
Relationship: $\displaystyle{y = \begin{cases}41,& \text{if } 0 < x \le 1,\\ 58,& \text{if } 1 < x \le 2,\\ 75,& \text{if } 2 < x \le 3,\\ 92,& \text{if } 3 < x \le 3.5, \end{cases}}$ with $x$ in ounces and $y$ in cents (as of 14 May, 2007)
Domain $= (0, 3.5]$
Range $= \{41, 58, 75, 92\}$

3. (a) $(-\infty, \infty)$

(c) $\{t : t \le -3 \text{ or } t \ge 2\}$

(e) $\{s : s \ne -3, s \ne 3\}$

(g) $\{s : s < -3-2\sqrt{2} \text{ or } s > -3 + 2\sqrt{2}\}$

4. (a) $f \circ g(x) = 20x + 4$, $g \circ f(x) = 20x + 58$, $f \circ g(3) = 64$, $g \circ f(3) = 118$

(c) $\displaystyle{f \circ g(x) = \frac{6x - 55}{(x - 9)^2}}$, $\displaystyle{g \circ f(x) = -\frac{1}{x^2 - 6x + 9}}$, $\displaystyle{f \circ g(3) = -\frac{37}{36}}$, $g \circ f$ is not defined at $3$

9. (a) Domain $= (-\infty, \infty)$
Range $= \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$

(b) Domain $= (-\infty, \infty)$
Range $= \{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$

Section 2.2

1. (a) $\displaystyle{\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}}$, $\displaystyle{\cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}}$, $\displaystyle{\tan\left(\frac{4\pi}{3}\right) = \sqrt{3}}$, $\displaystyle{\sec\left(\frac{4\pi}{3}\right) = -2}$

(c) $\displaystyle{\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}}$, $\displaystyle{\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}}$, $\displaystyle{\tan\left(\frac{4\pi}{3}\right) = -\sqrt{3}}$, $\displaystyle{\sec\left(\frac{4\pi}{3}\right) = -2}$

(e) $\displaystyle{\sin\left(-\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2}}$, $\displaystyle{\cos\left(-\frac{2\pi}{3}\right) = -\frac{1}{2}}$, $\displaystyle{\tan\left(-\frac{2\pi}{3}\right) = \sqrt{3}}$, $\displaystyle{\sec\left(-\frac{2\pi}{3}\right) = -2}$

(g) $\displaystyle{\sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2}}$, $\displaystyle{\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2}}$, $\displaystyle{\tan\left(\frac{11\pi}{6}\right) = -\frac{1}{\sqrt{3}}}$, $\displaystyle{\sec\left(\frac{11\pi}{6}\right) = \frac{2}{\sqrt{3}}}$

2. (a) Amplitude $= 1$, period $= \displaystyle{\frac{2\pi}{3}}$, phase angle $= 0$

(c) Amplitude $= 1$, period $= 2$, phase angle $= \pi$

(e) Amplitude $= 4$, period $= 2$, phase angle $= 0$

(e) Amplitude $= 5$, period = $= \pi$, phase angle $= -\displaystyle{\frac{\pi}{2}}$

4. (a) $\sin(2x) = \sin(x + x) = \sin(x)\cos(x) + \cos(x)\sin(x)$ $= 2\sin(x)\cos(x)$

5. (a) This follows from: $\cos(2x) = \cos^2(x) - \sin^2(x) = \cos^2(x) - (1 - \cos^2(x)) = 2\cos^2(x) - 1.$

6. (a) $\displaystyle{\sin\left(x-\frac{\pi}{2}\right) = \sin(x)\cos\left(-\frac{\pi}{2}\right) + \cos(x)\sin\left(-\frac{\pi}{2}\right) = -\cos(x)}$

(c) $\displaystyle{\sin\left(x+\frac{\pi}{2}\right) = \sin(x)\cos\left(\frac{\pi}{2}\right) + \cos(x)\sin\left(\frac{\pi}{2}\right) = \cos(x)}$

8. $\displaystyle{\tan(t + \pi) = \frac{\sin(t + \pi)}{\cos(t + \pi)} = \frac{\sin(t)\cos(\pi) + \cos(t)\sin(\pi)}{\cos(t)\cos(\pi) - \sin(t)\sin(\pi)}}$ $\displaystyle{= \frac{-\sin(t)}{-\cos(t)} = \tan(t)}$

Section 2.3

1. (a) $\displaystyle{\lim_{x \to 2}(4x^2 - 3x) = 10}$

(c) $\displaystyle{\lim_{t \to 1} \frac{t^2 - 3}{t + 5} = -\frac{1}{3}}$

2. (a) $\displaystyle{\lim_{x \to 2} \frac{x^3 - 8}{x + 2} = 0}$

(c) $\displaystyle{\lim_{s \to -1} \frac{s^3 + 1}{s^4 - 1} = -\frac{3}{4}}$

(e) $\displaystyle{\lim_{t \to 2} \frac{t^3 - 8}{t -2} = 12}$

(g) $\displaystyle{\lim_{u \to 4} \frac{u}{(u - 4)^2} = \infty}$

3. (a) $\displaystyle{\lim_{x \to 1^+}(3x^2 + 4) = 7}$

(c) $\displaystyle{\lim_{x \to 3^+} \frac{1}{x - 3} = \infty}$

(e) $\displaystyle{\lim_{t \to -2^-} \frac{t}{t + 2} = \infty}$

4. (a) $\displaystyle{\lim_{x \to 2^+}\lfloor x \rfloor = 2}$

(b) $\displaystyle{\lim_{x \to 2^-}\lfloor x \rfloor = 1}$

(c) $\displaystyle{\lim_{x \to 3^-} \lceil x \rceil = 3}$

(e) $\displaystyle{\lim_{x \to 0^+}\lfloor \cos(x) \rfloor = 0}$

5. (b) $\displaystyle{\lim_{z \to 2^-}g(z) = 5}$

(c) $\displaystyle{\lim_{z \to 2^+}g(z) = 5}$

(d) Yes: $\displaystyle{\lim_{z \to 2}g(z) = 5}$

7. (a) $\displaystyle{\lim_{x \to \infty} (3x + 4) = \infty}$

(b) $\displaystyle{\lim_{x \to \infty} \frac{x^3 + 3x - 1}{2x^3 - x^2 + 21} = \frac{1}{2}}$

(c) $\displaystyle{\lim_{u \to \infty} \frac{u^4 + 3u - 6}{3u^2 + 1} = \infty}$

(e) $\displaystyle{\lim_{x \to -\infty} \frac{x^5 - 6x + 13}{x^2 + 18x - 25} = -\infty}$

(g) $\displaystyle{\lim_{v \to \infty} \sqrt{\frac{2v + 1}{v - 2}} = \sqrt{2}}$

(i) $\displaystyle{\lim_{x \to \infty} \frac{3x + 1}{\sqrt{4x^2 + 5}} = \frac{3}{2}}$

(j) $\displaystyle{\lim_{x \to -\infty} \frac{3x + 1}{\sqrt{4x^2 + 5}} = -\frac{3}{2}}$

9. $\displaystyle{\lim_{x \to \frac{\pi}{2}^-} \tan(x) = \infty}$, $\displaystyle{\lim_{x \to \frac{\pi}{2}^+} \tan(x) = -\infty}$, $\displaystyle{\lim_{x \to \frac{\pi}{2}} \tan(x)}$ does not exist

11. (a) Use the fact that $-1 \le \sin(x) \le 1$ for all $x > 0$.

(b) $\displaystyle{\lim_{x \to \infty}\frac{\sin(x)}{x} = 0}$

Section 2.4

1. (a) Since $f$ is a polynomial, $f$ is continuous at $t = 2$

(b) Since $f$ is a rational function, and is defined at $x = 17$, $f$ is continuous at $x = 17$

(c) Since $f$ is not defined at $x = 16$, $f$ is not continuous at $x = 16$

(e) Since $h$ is not defined at $s = -1$, $h$ is not continuous at $s = -1$

2. Since $\displaystyle{\lim_{t \to 2} g(t) = 7 = g(2)}$, $g$ is continuous at $t = 2$.

3. (a) $g$ is continuous on $(-\infty, \infty)$.

(c) $g$ is continuous on $(-\infty, 0)$ and on $(0, \infty)$.

(e) $f$ is continuous on $(-\infty, -2]$ and on $[2, \infty)$.

4. $f$ is continuous on $(-\infty, 1)$ and on $[1, \infty)$, but is not continuous at $x = 1$.

5. $h$ is continuous on $(-\infty, -1]$ and on $(-1, \infty)$, but $h$ is not continuous at $z = -1$.

6. Yes, $f$ has a removable discontinuity at $t = 4$. Let $\displaystyle{g(t) = \begin{cases}\displaystyle{\frac{t^2 - 7t + 12}{t - 4}},& \text{if } t \ne 4,\\ 1,& \text{if } t = 4.\end{cases}}$

7. The discontinuity at $t = 5$ is not removable.

9. (a) $f$ is continuous on $(-\infty, \infty)$

(b) $g$ is continuous on $(-\infty, -1]$, $(-1, 1)$, and $[1, \infty)$; $g$ is not continuous at $x = -1$ and $x = 1$.

(c) For $n = 0, \pm 1, \pm 2, \ldots$, $h$ is continuous on intervals of the form $[2n, 2n + 1]$ and $(2n + 1, 2n)$; $h$ is not continuous at $t = 0, \pm 1, \pm 2, \ldots$.

11. For $n = 0, \pm 1, \pm 2, \ldots$, $f$ is continuous on intervals of the form $((2n+1)\pi, 2n\pi)$, $\Big[2n\pi, \big(2n + \frac{1}{2}\big)\pi\Big)$, and $\Big(\big(2n + \frac{1}{2}\big)\pi, (2n + 1)\pi\Big]$;$f$ is not continuous at the points $x = \Big(2n + \frac{1}{2}\Big)\pi$.

Section 2.5

1. (a) $1.418$

(c) $0.739$

2. (b) $2$

(c) $-0.739$ and $0.739$

3. (b) $\$51.28 $(c)$ 979 $units (d) If the producers raised the price above the equilibrium price, then demand would decrease while supply increased. With supply exceeding demand, the producers would eventually decrease the price. If the producers lowered the price below the equilibrium price, then demand would increase while supply decreased. With the demand exceeding the supply, the producers would eventually increase the price. (e) If the producers increased production, the supply curve would be higher and the equilibrium price would decrease. If the producers decreased production, the supply curve would be lower and the equilibrium price would increase. 4.$ 125 \text{ yards } \times 250 \text{ yards} $5.$\$10.00$

6. (a) No

(b) Yes, $-1$ is the minimum value of $h$ on $[-1, 2)$.

(c) Yes, because the Extreme Value Theorem does not apply in this case since the interval $[-1, 2)$ is not closed.

7. (a) No

(b) Yes, because the Intermediate Value Theorem does not apply in this case since the Heaviside function is not continuous on $[-1, 1]$.

(c) $H$ has a maximum value of $1$ and a minimum value of $-1$ on $[-1, 1]$.

(d) Yes; although the Extreme Value Theorem does not apply in this case (since $H$ is not continuous on $[-1, 1]$), that does not preclude $H$ from having maximum and minimum values on the interval.

9. Apply the Intermediate Value Theorem to the function $h(x) = f(x) - g(x)$.

10. Apply the Intermediate Value Theorem to the function $g(x) = f(x) - x$.