Chapter 2
Exercise 2.1
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Exercise 2.2
Q1
(Trang-Paige Tran a,c,e,g)
(a)\[ dom(f) = (-\infty,+\infty) \]
(b)\[ 5x+6 \neq 0 \\ x \neq -\frac{6}{5} \\ dom(f) = (-\infty,-\frac{6}{5}) \cup (-\frac{6}{5},+\infty)\]
(c)\[ x^2-5 \neq 0 \\ x \neq \pm \sqrt{5} \\ dom(f) = (-\infty,-\sqrt{5}) \cup (-\sqrt{5},\sqrt{5}) \cup (\sqrt{5},+\infty)\]
(d)\[ x^2-2x-3 \neq 0 \\ (x-3)(x+1) \neq 0 \\ x \neq -1 \ or \ x \neq 3 \\ dom(f) = (-\infty,-1) \cup (-1,3) \cup (3,+\infty)\]
(e)\[ \left\{\begin{aligned} 2x-3 \geq 0 \\ 2x-3 \neq 0 \\ \end{aligned}\right. \\ 2x-3>0 \\ x > \frac{3}{2} \\ dom(f)=(\frac{3}{2},+\infty)\]
(f)\[ \left\{\begin{aligned} 1-2x \neq 0 \\ x+3 \geq 0 \\ \end{aligned}\right. \\ \left\{\begin{aligned} x \neq \frac{1}{2} \\ x \geq -3 \\ \end{aligned}\right.\\ dom(f)=[-3,\frac{1}{2}) \cup (\frac{1}{2},+\infty)\]
(g)\[ \left\{\begin{aligned} 1-x^2 \neq 0 \\ 2x+5 \geq 0 \\ \end{aligned}\right. \\ \left\{\begin{aligned} x \neq \pm 1 \\ x \geq -\frac{5}{2} \\ \end{aligned}\right.\\ dom(f)=[-\frac{5}{2},-1) \cup (-1,1) \cup (1,+\infty)\]
(h)\[ \left\{\begin{aligned} x^2+3x-10 \neq 0 \\ x^2+3x-10 \geq 0 \\ \end{aligned}\right. \\ x^2+3x-10 > 0 \\ (x+5)(x-2) > 0 \\ x < -5 \ or \ x > 2 \\ dom(f)=(-\infty,-5) \cup (2,+\infty)\]
Q2
(Trang-Paige Tran a,c,e,g)
(a)\[ f(x)=y=x^2-5 \\ x^2 = y+5 \\ only \ if \ y+5 \geq 0, x = \pm \sqrt{y+5} \\ we \ get \ y \geq -5 \\ ran(f) = [-5,+\infty)\] (b)This is a parabola with an upward opening, and the minimum value is : \[ f(x) = x^2-2x-3 \\ = (x-1)^2-4 \] Thus the minimum value is -4, when x=1. \[ ran(f) = [-4,+\infty)\] (c)\[ f(x)=y=\frac{2}{5x+6} \\ 5x+6 = \frac{2}{y} \\ 5x = \frac{2}{y} - 6 \\ x = \frac{2}{5y}-\frac{6}{5} \\ x \ can \ be \ solved \ if \ and \ only \ if \ y \neq 0 \\ ran(f) = (-\infty,0) \cup (0, +\infty)\] (d)\[ f(x)=y=3-\frac{1}{2x-1} \\ \frac{1}{2x-1} = 3-y \\ 2x-1 = \frac{1}{3-y} \\ 2x = \frac{1}{3-y}+1 \\ x = \frac{1}{2(3-y)}+\frac{1}{2} \\ x \ can \ be \ solved \ if \ and \ only \ if \ 2(3-y) \neq 0 \\ y \neq 3 \\ ran(f) = (-\infty,3) \cup (3, +\infty)\]
(e)\[ f(x)=y=\frac{1}{\sqrt{2x-3}} \\
\sqrt{2x-3} = \frac{1}{y} \\
only \ if \ y \geq 0, 2x-3 = \frac{1}{y^2} \\
2x = \frac{1}{y^2}+3 \\
x = \frac{1}{2y^2}+\frac{3}{2}\\
x \ can \ be \ solved \ if \ and \ only \ if \ y \neq 0 \\
Thus \ y>0 \\
ran(f) = (0, +\infty)\]
(f)\[ f(x)=y=\frac{1}{x^2-5} \\
x^2-5 = \frac{1}{y} \\
x^2 = \frac{1}{y}+5 \\
x = \pm \sqrt{\frac{1}{y}+5} \\
x \ can \ be \ solved \ if \ and \ only \ if \ y \neq 0 \ and \ \frac{1}{y}+5 \geq 0\\
y > 0 \ or \ y \leq -\frac{1}{5} \\
ran(f) = (-\infty,-\frac{1}{5}] \cup (0, +\infty)\]
(g)As in (b), \[f(x)=x^2-2x-3\]
we get \[ ran(f) = [-4,+\infty) \]
Suppose \[ h = x^2-2x-3, h \in [-4,+\infty) \]
Then we get
\[ f(x)=y=\frac{1}{x^2-2x-3} = \frac{1}{h} \\
h = \frac{1}{y} \\
x \ can \ be \ solved \ if \ and \ only \ if \ y \neq 0 \\
As \ h \in [-4,+\infty), \ Thus \ y > 0 \ or \ y \leq -\frac{1}{4} \\
ran(f) = (-\infty,-\frac{1}{4}] \cup (0, +\infty)\]
Exercise 2.3
Q1
(Ju-yeon Sally Choi)
\[ F(x,y)= 2x^2+3y^2-4 \] As for x-intercepts, \[ y=0 \ and \ F(x,y)=0\\ F(x,y) = 2x^2+3 \times 0^2 - 4 = 2x^2 - 4 = 0\\ x^2 = 2\\ x = \pm \sqrt{2}\] Thus the x-intercepts are \[(-\sqrt{2},0) \ or \ (\sqrt{2},0) \] As for y-intercepts, \[ x=0 \ and \ F(x,y)=0\\ F(x,y) = 2 \times 0^2+3y^2 - 4 = 3y^2 - 4 = 0\\ y^2 = \frac{4}{3}\\ y = \pm \frac{2\sqrt{3}}{3}\] Thus the x-intercepts are \[(0,-\frac{2\sqrt{3}}{3}) \ or \ (0,\frac{2\sqrt{3}}{3}) \]
Q2
(Ju-yeon Sally Choi)
\[ \left\{\begin{aligned} 0 = a \times 2^2 + b \times 2 + c \\ 0 = a \times (-3)^2 + b \times (-3) + c \\ -6 = a \times 0^2 + b \times 0 + c \\ \end{aligned}\right. \\ \left\{\begin{aligned} 0 = 4a + 2b + c \\ 0 = 9a - 3b + c \\ -6 = c \\ \end{aligned}\right. \\ \left\{\begin{aligned} a = 1 \\ b = 1 \\ c = -6\\ \end{aligned}\right. \\\]
Q3
(Ju-yeon Sally Choi)
(a)As for x-intercepts, y = 0 \[ y = 0 = x^2+4x+5 \] \[ b^2-4ac = 4^2-4 \times 1 \times5 = -4 < 0 \] Thus the x-intercepts don’t exist.
As for y-intercepts, x = 0 \[ y =0^2+4 \times 0 +5 = 5 \] Thus the y-intercept is \[ (0,5) \]
Q4
(Oyu Enkhbold)
(a)As \[ 2x+y-3=0 \] We get \[ y = 3-2x \] Putting into \[ x^2+y^2 =5\] We get \[ x^2+(3-2x)^2 = 5 \\ x^2 + 9 - 12x + 4x^2 -5 = 0 \\ 5x^2 - 12x +4 = 0\\ (x-2)(5x-2)=0\\ x =2 \ or \ x=\frac{2}{5}\\ When \ x=2, \ y = -1 \\ When \ x= \frac{2}{5}, \ y = \frac{11}{5}\] Thus we get \[ L \cap C = \lbrace{ (2,-1),(\frac{2}{5},\frac{11}{5})}\rbrace \]
(b)As \[ 2x+y-3=0 \] We get \[ y = 3-2x \] Putting into \[ x^2+2y^2 =6\] We get \[ x^2+2(3-2x)^2 = 6 \\ x^2 + 18 - 24x + 8x^2 -6 = 0 \\ 9x^2 - 24x +12 = 0\\ 3x^2 - 8x +4 = 0\\ (x-2)(3x-2)=0\\ x =2 \ or \ x=\frac{2}{3}\\ When \ x=2, \ y = -1 \\ When \ x= \frac{2}{3}, \ y = \frac{5}{3}\] Thus we get \[ L \cap E = \lbrace{ (2,-1),(\frac{2}{3},\frac{5}{3})}\rbrace \]
(c)As \[ x^2+y^2 =5 \] We get \[ x^2 = 5-y^2 \] Putting into \[ x^2+2y^2 =6\] We get \[ (5-y^2)+2y^2 = 6 \\ y^2=1\\ y =1 \ or \ y=-1\\ When \ y=1, \ x= \pm 2 \\ When \ y= -1, \ x= \pm 2\] Thus we get \[ C \cap E = \lbrace{ (2,1),(2,-1),(-2,1),(-2,-1)}\rbrace \]
Q5
(Oyu Enkhbold)
As \[ ax+y=2 \] We get \[ y = 2-ax \] Putting into \[ x^2+y^2 =1\] We get \[ x^2+(2-ax)^2 = 1 \\ x^2 + 4 - 4ax + a^2x^2 -1 = 0 \\ (1+a^2)x^2 - 4ax +3 = 0\] As there is only one element in the set, thus the equation only has one solution.As for the linear equation in two unknowns, that is \[ b^2-4ac = 0 \] then we can get \[ (-4a)^2 - 4 \times (1+a^2) \times 3 = 0 \\ 4a^2-12 = 0 \\ a = \pm \sqrt{3}\] Thus we get \[a = \pm \sqrt{3}\]
Exercise 2.4
Q4
(Jisong Kwon)
(a)When the object hit the ground, h(t)=0.
\[ h(t)=1+4t-5t^2=0 \\ (1-t)(1+5t)=0 \\ t=1 \ or\ t=-\frac{1}{5} \\ As\ t\ is\ more\ than\ 0\ or\ equal\ to\ 0,\ t=1.\\ Thus\ when\ t=1 ,\ the\ object\ hit\ the\ ground.\]
- method 1:
As for a quadratic equation in one variable, the abscissa corresponding to its maximum value is -b/2a(here a=-5, b=4).
Thus when t=-b/2a=2/5, we get the maximum height, that is \[ h(\frac{2}{5})= 1+4\times \frac{2}{5}-5\times (\frac{2}{5})^2 \\=\frac{9}{5} \]
method 2:differentiate
\[h'(t)= 4-10t=0\\ we\ get\ t=\frac{2}{5}\\ h''(t)=-10<0\\ Thus\ when\ t=\frac{2}{5},\ we\ get\ the\ maximum\ height,\ that \ is\ h(\frac{2}{5})= \frac{9}{5}\]
Q5
(Jisong Kwon)
(a)\[ R=(20000+500n)\times(80-n)\\
=-500n^2+20000n+1600000, domain={0,1,2,3,···,80}\]
(b)
The image of R=-500n^2+20000n+1600000(Here we suppose the range of n is the set of real numbers ) is as follows:
method 1:
As for a quadratic equation in one variable, the abscissa corresponding to its maximum value is -b/2a(here a=-500, b=20000).
Thus when n=-b/2a=20, we get the maximum revenue, that is \[ R= -500\times 20^2+20000\times20+1600000 \\=1800000 \]
method 2:differentiate
\[R'= -1000n+20000=0\\ we\ get\ n=20\\ R''=-1000<0\\ Thus\ when\ n=20,\ we\ get\ the\ maximum\ revenue,\ that \ is\ R=1800000\]
Exercise 2.5
Q2
(SeoWoog Lee)
(a)\[ As\ g(x)=x^r(r \neq 1)\ and\ g(f(x))=\sqrt{x^2+1}\\ Thus\ g(x)=\sqrt{x}= x^{\frac{1}{2}}\\ f(x)=x^2+1\] (b)\[ As\ g(x)=x^r(r \neq 1)\ and\ g(f(x))=\frac{1}{x+1}\\ Thus\ g(x)=\frac{1}{x}= x^{-1}\\ f(x)=x+1 \]
Exercise 2.6
Q2
(SeoWoog Lee)
(a)\[ y=3x-2\\ y+2=3x\\ x=\frac{y+2}{3}\\ f^{-1}(y)=\frac{y+2}{3},dom(f^{-1}(y))=(-\infty,\infty)\]
(b)\[ y=x^5+3 \\ y-3=x^5\\ x=(y-3)^{\frac{1}{5}}\\ f^{-1}(y)=(y-3)^{\frac{1}{5}},dom(f^{-1}(y))=(-\infty,\infty)\]
(c)\[ y=1+2x^{\frac{1}{7}}\\ y-1=2x^{\frac{1}{7}}\\ x=(\frac{y-1}{2})^7\\ f^{-1}(y)=(\frac{y-1}{2})^7,dom(f^{-1}(y))=(-\infty,\infty)\]
(d)\[ y=\sqrt[3]{2x^3-1}\\ y^3=2x^3-1\\ x=\sqrt[3]{\frac{y^3+1}{2}}\\ f^{-1}(y)=\sqrt[3]{\frac{y^3+1}{2}},dom(f^{-1}(y))=(-\infty,\infty)\]
Exercise 2.7
Q1
(Sara)
\[ (2x+1)(x-2)=x(x-2)\\ When \ x-2=0,\ that \ is \ x=2,\ we \ get \ (2x+1)(x-2)=0=x(x-2);\\ When \ x-2 \neq 0,\ that \ is \ x \neq 2,we \ get \\ (2x+1)(x-2) \times \frac{1}{x-2}=x(x-2)\times \frac{1}{x-2} \\ 2x+1=x \\ x=-1 \\ Thus \ x=2 \ or \ x=-1.\]
\[ \frac{x}{x+2}-\frac{x}{x-2}=\frac{-4x}{x^2-4}\\ (\frac{x}{x+2}-\frac{x}{x-2})\times(x+2)(x-2)=\frac{-4x}{x^2-4}\times(x+2)(x-2)\\ x(x-2)-x(x+2)=-4x \ and \ x \neq \pm2 \\ (x^2-2x)-(x^2+2x)=-4x \ and \ x \neq \pm2\\ -4x=-4x \ and \ x \neq \pm2 \\ x \in (-\infty,-2) \cup (-2,2) \cup (2,\infty)\]
(f)\[ \sqrt{x^2-9}+x=9\\ \sqrt{x^2-9}=9-x \\ (\sqrt{x^2-9})^2=(9-x)^2\\ x^2-9=x^2-18x+81 \ and \ x \in (-\infty,-3] \cup [3,\infty) \\ 18x=90\ and \ x \in (-\infty,-3] \cup [3,\infty) \\ x=5\]
(h)\[ \sqrt{x+5}+1=2\sqrt{x}\\ \sqrt{x+5}=2\sqrt{x}-1\\ (\sqrt{x+5})^2=(2\sqrt{x}-1)^2\\ x+5=4x-4\sqrt{x}+1 \ and \ x \geq 0\\ 3x-4\sqrt{x}-4=0 \ and \ x \geq 0\\ (3\sqrt{x}+2)(\sqrt{x}-2)=0 \ and \ x \geq 0\\ \sqrt{x}=\frac{-2}{3}(invalid) \ or \ \sqrt{x}=2 \ and \ x \geq 0\\ x=4\]
Q2
(Sara)
(a)\[ \pi = R(q)-C(q)\\ = (10q-q^2)-(2q+12)\\ = -q^2+8q-12\] (b)\[ \pi = -q^2+8q-12 = 0 \\ -(q-2)(q-6)=0\\ q=2 \ or \ q=6\\\] Thus the break-even quantity is 2 or 6.
Q3
(Sara)
(a)\[ When \ x=35, we \ get \\ y = 0.056057x^2+1.06657x\\ = 0.056057 \times 35^2+1.06657 \times 35\\ = 105.999775\]
(b)\[ As \ y \leq 200,\\ y = 0.056057x^2+1.06657x \leq 200 \ and \ x \geq 0\\ 0.056057x^2+1.06657x -200 \leq 0 \ and \ x \geq 0\\ x \approx -69.997(invalid) \ or \ x \approx 50.9706 \\ \] Thus one can drive at most 50.9706 mph is one needs to be certain of stopping within 200 ft.
Q4
(Kwang Tae Kim)
We suppose the sides adjacent to the right angle are x and y(y<x).As the perimeter is 12 units, the hypotenuse is z=12-x-y.As the sides adjacent to the right angle differ by 1 unit, we get y=x-1. \[ x^2+y^2=z^2=(12-x-y)^2\\ x^2+(x-1)^2=[12-x-(x-1)]^2\\ x^2+(x-1)^2=(13-2x)^2\\ x^2+x^2-2x+1=169-52x+4x^2\\ 2x^2-50x+168=0\\ x=4 \ or \ x=21\\ As \ the \ perimeter \ is \ 12 \ units, x=21>12 \ is \ invalid.\\ Thus \ x=4,y=x-1=3,z=12-x-y=5\]