1.5 Answers to Exercises in Chapter 1

1.1.8

The answers in problems 1–16 require either \(x\neq a\) or \(h\neq 0\) (as appropriate).

1. (a) \(4 a + 4 x-5 \); (b) \(8a + 4 h-5\).
2. (a) \(4(x+a)(x^2+a^2) -3\); (b) \((2x+h)\left ((x+h)^2+x^2\right ) -3\).
3. (a) \(-\frac{1}{(2 a-1) (2 x-1)} \); (b) \(-\frac{1}{(2 a-1) (2 a+2 h-1)}\).
4. (a) \( \frac{2 a x+a+x}{(2 a+1) (2 x+1)}\); (b) \(\frac{2 a^2+2 a h+2 a+h}{(2 a+1) (2 a+2 h+1)}\).
5. (a) \( -\frac{4}{\sqrt{4x-3}\sqrt{4a-3}\left (\sqrt{4x-3}+\sqrt{4a-3}\right )}\); (b) \( -\frac{4}{\sqrt{4x+4h-3}\sqrt{4x-3}\left (\sqrt{4x+4h-3}+\sqrt{4x-3}\right )}\).
6. (a) \(\frac{3}{\sqrt{3x+2}+\sqrt{3a+2}} \); (b) \(\frac{3}{\sqrt{3x+3h+2}+\sqrt{3x+2}} \).
7. \( \frac{\sqrt{a}\sqrt{x}+ a + x +1}{\sqrt{x}+\sqrt{a}}\).
8. \(\left (x^2+xa + a^2\right ) \left ((x^3+a^3) -\frac{2}{x^3a^3}\right ) \).
9. \( -\frac{x+a}{\sqrt{4-x^2}+\sqrt{4-a^2}}\).
10. \( \frac{(x+a)\left (4-(x+a)\left (x^2+a^2\right )\right )}{x\sqrt{4-x^2}+a\sqrt{4-a^2}}\).
11. \(\frac{1}{\left (\sqrt[4]{x}+\sqrt[4]{a}\right )\left (\sqrt{x}+ \sqrt{a}\right )}\).
12. \(\frac{x+a}{x\sqrt[3]{x}+\sqrt[3]{x^2a^2}+a\sqrt[3]{a}}\) or \(\frac{\sqrt[3]{x}+ \sqrt[3]{a}}{\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}}\).
13. \(\frac{(x+a)\left (x^2+a^2\right )}{x^2\sqrt[3]{x^2}+ xa\sqrt[3]{xa}+ a^2\sqrt[3]{a^2}}\).
14. \(\frac{x^2+ xa ++a^2}{\left (\sqrt[4]{x}+\sqrt[4]{a}\right )\left (x\sqrt{x}+ a\sqrt{a}\right )}\) or \(\frac{\sqrt{x}+\sqrt[4]{xa}+\sqrt{a}}{\left (\sqrt[4]{x}+\sqrt[4]{a}\right )\left (\sqrt{x}+ \sqrt{a}\right )}\).
15. \(\frac{\sqrt[3]{x^2a^2}(x+a)}{x^2\sqrt[3]{a^2}+xa\sqrt[3]{xa}+a^2\sqrt[3]{x^2}}\).
16. \(\frac{\sqrt[4]{x^3a^3}\left (x^2+xa+a^2\right )}{\left (x\sqrt[4]{a}+a\sqrt[4]{x}\right )\left (x^2\sqrt{a}+ a^2\sqrt{x}\right )}\).

1.2. Prior Knowledge Exercises

1. (a) \(\frac{\sqrt{193}}{12}\); (b) \(\frac{\sqrt{269}}{10}\).
2. (a) \(\left (-\frac{5}{2},\frac{1}{24}\right )\); (b) \(\left (-\frac{3}{2}, -\frac{3}{20}\right )\).
3. \(\left \{\left (\frac{7}{5},\frac{6}{5}\right ), \left (-1,6\right )\right \}\).
4. \(\left \{\left (\frac{1}{2},1\right ), \left (-\frac{1}{2},2\right )\right \}\).
5. (a) \(y= – \frac{7}{12}x+\frac{17}{12}\); (b) \(y= \frac{13}{10}x+\frac{9}{5}\).
6. (a) \(y= \frac{12}{7}x+\frac{727}{168}\); (b) \(y= -\frac{10}{13}x+\frac{587}{312}\).

1.2.3

1. (a) Only \(B\) is on \(\mathcal{C}\); (b) \(P\left(-1, 3\pm \sqrt{2}\right)\); (c) \(P\left(-2\pm \frac{\sqrt{11}}{2}\right)\).
2. (a) \(C\left(1,-2\right)\), \(r=\sqrt{13}\); (b) \(C\left(\frac{1}{2},-\frac{3}{2}\right)\), \(r=\sqrt{\frac{17}{2}}\); (c) \(C\left(-\frac{1}{2},-\frac{3}{4}\right)\), \(r=2\); (d) \(C\left(\frac{1}{3},-\frac{1}{2}\right)\), \(r=\sqrt{3}\).
3. (a) \(y=-\frac{2}{3}x+7\); (b) \(y=\frac{1}{3}x+1\); (c) \(y=-\frac{1}{2}x\); (d) \(y=-\frac{1}{3}x-\frac{1}{3}\).
4. Note that \(x^2+y^2 +4x+2by=-12 \Longleftrightarrow (x+2)^2 +(y+b)^2 = b^2-8\). Thus,

   (i) if \( b^2-8 < 0\), the graph is the empty set,

   (ii) if \( b^2-8 = 0\), that is if \(b=\pm 2\sqrt{2}\) the graph is the point \(\left(-2,\pm 2\sqrt{2}\right)\),

   (iii) if \( b^2-8 > 0\), that is if \(b\in \left(\infty, – 2\sqrt{2}\right)\cup \left(2\sqrt{2},\infty\right)\) the graph is a circle \(C\left(-2,-b\right)\) and \(r=\sqrt{b^2-8}\).

5. (a) \(x = -1 – \sqrt{ -6+6y -y^2} \Longleftrightarrow \left\{
   \begin{array}{l}
   (x+1)^2+(y-3)^2=3,\\
   x \le -1.
   \end{array}
   \right.\) Semicircle to the left of the center.

   (b) \(y = 2 + \sqrt{ -5-6x -x^2} \Longleftrightarrow \left\{
   \begin{array}{l}
   (x+3)^2+(y-2)^2=4,\\
   y \ge 2.
   \end{array}
   \right.\) Semicircle above the center.

6. Note that \((x-3)(x+1)+(y-5)(y-4)=0
   \Longleftrightarrow x^2-2x-3+y^2-9y+20=0 \Longleftrightarrow \) \(\Longleftrightarrow (x-1)^2+\left(y-\frac{9}{2}\right)^2-9y=3-20+1+\frac{81}{4}\). Thus, the equation represents a circle with center at

   \(C\left(1,\frac{9}{2}\right)\). But, the midpoint of \(AB\) is given by \(\left(\frac{3+(-1)}{2}, \frac{5+4}{2}\right)=\left(1,\frac{9}{2}\right)\), which is the center of the circle.

7. (a) \(d(A,P)=2d(B,P) \Longleftrightarrow (x+2)^2+\left (y-\frac{1}{3}\right )^2=\frac{52}{9}\). Circle with \(C\left (-2,\frac{1}{3}\right )\) and \(r=\frac{2}{3}\sqrt{13}\).

   (b) \(d(A,P)=\frac{1}{2}d(B,P) \Longleftrightarrow (x-3)^2+\left (y-\frac{2}{3}\right )^2=\frac{40}{9}\). Circle with \(C\left (3,\frac{2}{3}\right )\) and \(r=\frac{2}{3}\sqrt{10}\).

8. (a) \(d(O,P)^2=2x \Longleftrightarrow x^2+y^2=2x\). Circle with \(C(1,0)\) and \(r=1\).

   (b) \begin{align*}
   d(O,P)^2=2\left|x\right| &\Longleftrightarrow \left\{
   \begin{array}{ll}
   x^2+y^2=2x,& x>0,\\
   x^2+y^2=-2x,& x<0,\\
   y^2=0,& x=0,
   \end{array}
   \right.
   \Longleftrightarrow \left\{
   \begin{array}{ll}
   (x-1)^2+y^2=1,& x>0,\\
   (x+1)^2+y^2=1,& x<0,\\
   y=0,& x=0,
   \end{array}
   \right.\\
   &\Longleftrightarrow \left\{
   \begin{array}{ll}
   \text{Circle with } C(1,0) \text{ and } r=1,\\
   \text{Circle with } C(-1,0) \text{ and } r=1,\\
   \text{The origin } (0,0).
   \end{array}
   \right.
   \end{align*}

1.2.5

1. \(y=-\frac{1}{2}x-\frac{5}{2}\) and \(y=-\frac{1}{2}x+\frac{15}{2}\).
2. (a) \((0,2)\), \(\left(\frac{8}{5},-\frac{6}{5}\right)\).
   (b) \((1,-\sqrt{3})\), \((1,\sqrt{3})\).
3. (a) \(y=x-1\); (b) \((-1,-2)\); (c) \(A=8\).
4. \(y= -\frac{9}{4}x-3\) tangent at \(\left (-\frac{2}{3},-\frac{3}{2}\right )\); \(y= -\frac{9}{4}x+3\) tangent at \(\left (\frac{2}{3},\frac{3}{2}\right )\).
5. \(\left (\pm\sqrt{2},2\right )\).
6. \(\left \{(-3,3),(3, 27)\right \}\).

1.3.3

1. \(15\).
2. \(\frac{53}{4}\).
3. \(\frac{45}{2}\).
4. \(\frac{27}{2}\).
5. \(\frac{2+\pi}{8}\).
6. \(\frac{1}{6}\left (6 +3\sqrt{3}+ 7\pi\right )\).

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7. \(\frac{35}{3}\).
8. \(32\).
9. \(6\).
10. \(\frac{64}{3}\).
11. \(\frac{3}{4}\left (4\pi-3\sqrt{3}\right )\).
12. (b) \(6\).

1.3.5

1. (c) \(\frac{\pi}{4}\left (18+\sqrt{2}\right ) \).
2. (c) \(\frac{\pi}{6}\left (8 + \sqrt{3}\right ) \).
3. (c) \(\frac{\pi}{18}\left (9 + 4\,\sqrt{3}\right ) \).
4. (c) \( \frac{3\pi}{4}\).
5. (c) \(\left(4\ln 2 +3\ln 3 +\ln 5 + \ln 7 + 5 \right)\cdot e \).

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6. (c) \(\frac{1}{4}\left(1+\sqrt[4]{e}+\sqrt[4]{e^2}+\sqrt[4]{e^3}\right) \sqrt[4]{e^3}\).
7. (c) \(\sqrt{e}\left (\frac{5}{2}+4\ln 2 + 3\ln 2 +\ln 5\right ) \).
8. (c) \( \frac{2\pi}{\sqrt{3}}\).
9. (c) \(\frac{3\pi}{2} \).
10. (c) \(\left (4 – \frac{\sqrt{3}}{3}\right )\pi \).

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11. (b) \(\frac{101}{8} \); (c) overestimate by \(\frac{101}{8}- 12 = \frac{5}{8}\).
12. (b) \(8\) (c) \(\frac{20}{3}\).

1.4.2

1. Assume \(a\neq x\): (a) \(-(x+a) \left (3 x^2 + 3 a^2-2\right )\); (b) \(\frac{2 ax (x+a)+3 \left ( a^2 +x^2\right )+3 a x}{(2 a+3) (2 x+3)}\); (c) \(\frac{x+a}{\sqrt{4 x^2 + 1}\,\sqrt{4 a^2 + 1}\left (x\, \sqrt{4 a^2 + 1} + a\,\sqrt{4 x^2 + 1}\right )} \); (d) \(\frac{x+a}{x\sqrt[3]{x}+\sqrt[3]{x^2a^2}+a\sqrt[3]{a}} + \frac{\sqrt[3]{x^2a^2}(x+a)}{x^2\sqrt[3]{a^2}+xa\sqrt[3]{xa}+a^2\sqrt[3]{x^2}}\) or \(\frac{\sqrt[3]{x}+ \sqrt[3]{a}}{\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2}} + \frac{\sqrt[3]{x^2a^2}(x+a)}{x^2\sqrt[3]{a^2}+xa\sqrt[3]{xa}+a^2\sqrt[3]{x^2}}\).
2. (a) \((x-3)^2+(y-1)^2 = 17\); (b) \(y=-\frac{1}{4}x-\frac{5}{2}\).
3. \((0,2)\) and \(\left (\frac{6}{5},-\frac{8}{5}\right )\).
4. (a) \(\frac{5\pi}{3}\); (b) \(\frac{5\pi}{4}\); (c) \(\sqrt{3}\,\left (3 + 6\ln 2 + 2\ln 3\right )\); (d) \(\frac{1}{3}\sqrt[3]{e^2}\left (1 + \sqrt[3]{e}+ \sqrt[3]{e^2}\right )\).
5. \((x-2)(x+2)+(y-3)(y-1)=0 \Longleftrightarrow x^2+y^2-4 y-1 =0 \Longleftrightarrow x^2+(y-2)^2=5\Longrightarrow C(0,1)\) and \(r=\sqrt{5}\). Since the midpoint of the segment \(AB\) is \((0,2)\) and since \(\frac{1}{2}|AB|=\sqrt{5}\), the result follows (why).
6. (a) \(\sqrt{3}+\pi\); (b) \(\frac{1}{6}\left (6 – 3\sqrt{3}+\pi\right )\).
7. \(A = 18\); see Figure #7 for the graph.


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8. (a) \(14\); (b) \(6\); (c) \(8\); see Figures #8 (a), (b), (c) for the graphs.
9. \(\left (-4,38\right )\) and \(\left (4,6\right )\); \(y=-12 x-10\) and \(y=4 x-10\).

11. (a) \(x\)-int.: \((-3,0)\), \((2,0)\); \(y\)-int.: \((0, 6)\); (b) \(y=-\left (x+\frac{1}{2}\right )^2+\frac{25}{4}\), \(V\left (-\frac{1}{2},\frac{25}{4}\right )\); (c) \((0,6)\); (d) \( \frac{32}{3}\); (e) \( \frac{125}{6} \).


#4(c)

12. \(13 x – 3 y – 11=0\).