Learning math: relations and functions 2

Table of contents

These are my solutions to the problems in the book Precalculus, 3rd edition (Stitz, Zeager).

Term: set

A collection of objects that is well-defined (i.e. there's no ambiguity about what the elements are).

Ways to describe sets

  • Verbal/semantic: Let AA be the set of positive integers up to and including 5
    • "Let A be..." is common language among mathematicians. They could say "A is...", but the common convention is "Let A be...".
  • Roster notation: A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}
  • Set builder notation: A={x∣1≤x≤5}A = \{x \mid 1 \le x \le 5\}
    • The vertical line should be read as: "such that"
  • Interval notation: [1,5][1, 5]
    • Use parentheses for non-inclusion: (2,9](2, 9]
Sets of numbers

See https://en.wikipedia.org/wiki/Set_(mathematics)#Special_sets_of_numbers_in_mathematics.

Intersection and union notation
  • Intersection: A∩BA \cap B
  • Union: A∪BA \cup B
Describe all the ways in which a point can be reflected symmetrically along the axis in the Cartesian coordinate plane
  • Along the xx axis: replace yy with −y-y
  • Along the yy axis: replace xx with −x-x
  • Along the origin: reflect the point both along the xx and the yy axis
How to simplify sqrt(45)?

(Am)n=AmnAmn=Am/n45=451/2AB=(AB)1/2=A1/2B1/2=AB45=3∗3∗5=3∗35=325=35 (A^m)^n = A^{mn} \\ \sqrt[n]{A^m} = A^{m/n} \\ \sqrt{45} = 45^{1/2} \\ \sqrt{AB} = (AB)^{1/2} = A^{1/2} B^{1/2} = \sqrt{A} \sqrt{B} \\ \sqrt{45} = \sqrt{3 * 3 * 5} = \sqrt{3 * 3} \sqrt{5} = \sqrt{3^2} \sqrt{5} = 3 \sqrt{5}

Prime factorization: factor the number inside the square root into prime numbers, and then simplify (if you get two or more of the same factor, like we got for 33 above).

What is a "principal square root"?

It's \sqrt{}.

\sqrt{} is always only the non-negative square root.

In other words, 4=2\sqrt{4} = 2.

Principal square root is a function. A function has only one output for a specific input. E.g. f(x)=xf(x) = \sqrt{x} is a map from xx to a single output value, the positive square root of xx.

On the other hand, I'm not sure if 41/2={2,−2}=±2=±44^{1/2} = \{2, -2\} = \pm 2 = \pm \sqrt{4}. Apparently not. Apparently 41/2=44^{1/2} = \sqrt{4}. I need to investigate deeper. Here's some discussion:

How to use ± properly with square roots?

Remember to use ±\pm properly. An example:

y2=4±y2=±4y=±2 y^2 = 4 \\ \pm\sqrt{y^2} = \pm\sqrt{4} \\ y = \pm 2

The most verbose (and equivalently correct) way to write this would be ±y=±2\pm y = \pm 2, but that's redundant (see this and this):

y=2y=−2−y=2Same as y=−2, because you can multiply both sides with −1−y=−2Same as y=2, because you can multiply both sides with −1 y = 2 \\ y = -2 \\ -y = 2 \quad \text{Same as }y = -2\text{, because you can multiply both sides with }-1 \\ -y = -2 \quad \text{Same as }y = 2\text{, because you can multiply both sides with }-1

Writing y=±2y = \pm 2 is equivalent to the 4 equations above, and shorter.

But to make sure you never forget to add ±\pm, I think it's best that you always put the ±\pm in front of the \sqrt{} (e.g. when you're converting y2=xy^2 = x into ±y=±x\pm y = \pm\sqrt{x}), at least mentally. And then mentally remove it from one side, since putting it on both sides is redundant, as shown above.

Gnuplot Produced by GNUPLOT 5.4 patchlevel 2 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 -5 -4 -3 -2 -1 0 1 2 3 4 5 y x f(x) f(x) g(x) g(x) 
set xtics 1
set ytics 1
set xrange [-5:5]
set yrange [-7:7]

f(x) = 2
g(x) = -2
plot f(x), g(x)

1.1.5 example

c2=(x2−x1)2+(y2−y1)242=(3−1)2+(2−y1)2±42−22=2−y1y1=2±12=2±223=2±43=2±23 c^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 \\ 4^2 = (3 - 1)^2 + (2 - y_1)^2 \\ \pm \sqrt{4^2 - 2^2} = 2 - y_1 \\ y_1 = 2 \pm \sqrt{12} = 2 \pm \sqrt{2}\sqrt{2}\sqrt{3} = 2 \pm \sqrt{4}\sqrt{3} = 2 \pm 2 \sqrt{3}

Midpoint of a line

M=(x1+x22,y1+y22)M = (\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2})

1.1.4

1.

{x∣−1≤x<5}[−1,5){x∣0≤x<3}[0,3){x∣2<x≤7}(2,7]{x∣−5<x≤0}(−5,0]{x∣−3<x<3}(−3,3){x∣5≤x≤7}[5,7]{x∣x≤3}(−∞,3]{x∣x<9}(−∞,9){x∣4<x}(4,∞){x∣x≥−3}[−3,∞) \{x \mid -1 \le x < 5\} \qquad [-1, 5) \\ \{x \mid 0 \le x < 3\} \qquad [0, 3) \\ \{x \mid 2 < x \le 7\} \qquad (2, 7] \\ \{x \mid -5 < x \le 0\} \qquad (-5, 0] \\ \{x \mid -3 < x < 3\} \qquad (-3, 3) \\ \{x \mid 5 \le x \le 7\} \qquad [5, 7] \\ \{x \mid x \le 3\} \qquad (-\infty, 3] \\ \{x \mid x < 9\} \qquad (-\infty, 9) \\ \{x \mid 4 < x\} \qquad (4, \infty) \\ \{x \mid x \ge -3\} \qquad [-3, \infty)

2. [0,5][0, 5]

3. (−1,6](-1, 6]

4. (0,4](0, 4]

5. {}\{\} Empty set

6. (−∞,0)∪[1,5](-\infty, 0) \cup [1, 5]

7. 5\cancel{5} {5}\{5\}

See https://math.stackexchange.com/questions/1728991/is-a-set-containing-only-one-element-equal-to-the-element-itself.

8. (−∞,5)∪(5,∞)(-\infty, 5) \cup (5, \infty)

10. (−∞,−3)∪(−3,4)∪(4,∞)(-\infty, -3) \cup (-3, 4) \cup (4, \infty)

14. (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty)

18. (2,∞)∪{1,−1}(2, \infty) \cup \{1, -1\}

22.

a2+b2=d2Pythagorean theoremd=(y2−y1)2+(x2−x1)2Distance equation a^2 + b^2 = d^2 \qquad \text{Pythagorean theorem} \\ d = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} \qquad \text{Distance equation}

d=(5−2)2+(−3−1)2=9+16=5d = \sqrt{(5 - 2)^2 + (-3 - 1)^2} = \sqrt{9 + 16} = 5

M=(x1+x22,y1+y22)Midpoint of a line M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) \qquad \text{Midpoint of a line}

24. d=(−1−4)2+(32−12)2=25+1=213d = \sqrt{(-1 - 4)^2 + (\frac{3}{2} - \frac{1}{2})^2} = \sqrt{25 + 1} = \sqrt{2}\sqrt{13}

26. d=(−195−65)2+(−115−245)2=(−255)2+(−355)2=−52+(−72)=74d = \sqrt{(-\frac{19}{5} - \frac{6}{5})^2 + (-\frac{11}{5} - \frac{24}{5})^2} = \sqrt{(-\frac{25}{5})^2 + (-\frac{35}{5})^2} = \sqrt{-5^2 + (-7^2)} = \sqrt{74}

28.

d=(27−12)2+(20−245)2(27−12)2=27−22712+12=39−29326=39−662=3(20−245)2=20−42045+4(45)=200−4900=80d=3+80=83 d = \sqrt{(\sqrt{27} - \sqrt{12})^2 + (\sqrt{20} - 2\sqrt{45})^2} \\ (\sqrt{27} - \sqrt{12})^2 = 27 - 2\sqrt{27}\sqrt{12} + 12 = 39 - 2\sqrt{9}\sqrt{3}\sqrt{2}\sqrt{6} = 39 - 6\sqrt{6}^2 = 3 \\ (\sqrt{20} - 2\sqrt{45})^2 = 20 - 4\sqrt{20}\sqrt{45} + 4(45) = 200 - 4\sqrt{900} = 80 \\ d = \sqrt{3 + 80} = \sqrt{83}

29. d=x2+y2d = \sqrt{x^2 + y^2}

M=(x2,y2)M = (\frac{x}{2}, \frac{y}{2})

30.

x2=3y2=2y1=−1(3−x)2+32=423−x=±16−9x={3−7,3+7}(3−7,−1),(3+7,−1) x_2 = 3 \\ y_2 = 2 \\ y_1 = -1 \\ (3 - x)^2 + 3^2 = 4^2 \\ 3 - x = \pm\sqrt{16 - 9} \\ x = \{3 - \sqrt{7}, 3 + \sqrt{7}\} \\ (3 - \sqrt{7}, -1), (3 + \sqrt{7}, -1)

31.

x1=0x2=−5y1=352=(−5−0)2+(y−3)2(y−3)2=0y−3=0y=3(0,3) x_1 = 0 \\ x_2 = -5 \\ y_1 = 3 \\ 5^2 = (-5 - 0)^2 + (y - 3)^2 \\ (y - 3)^2 = 0 \\ y - 3 = 0 \\ y = 3 \\ (0, 3)

33.

y=−x(x−x1)2+(y−y1)2=12x1=0y1=0x2+(−x)2=12x2=(−x)2x2+x2=122x2=12±2x2=1±2x2=1±2x=1x=±12≈±0.707(12,−12),(−12,12) y = -x \\ (x - x_1)^2 + (y - y_1)^2 = 1^2 \\ x_1 = 0 \\ y_1 = 0 \\ x^2 + (-x)^2 = 1^2 \\ x^2 = (-x)^2 \\ x^2 + x^2 = 1^2 \\ 2x^2 = 1^2 \\ \pm\sqrt{2x^2} = 1 \\ \pm\sqrt{2}\sqrt{x^2} = 1 \\ \pm\sqrt{2}x = 1 \\ x = \pm\frac{1}{\sqrt{2}} \approx \pm 0.707 \\ (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

35.

x0=ax1=ad=(x1−x0)2+(y1−y0)2d=(a−a)2+(y1−y0)2d=(y1−y0)2d=y1−y0 x_0 = a \\ x_1 = a \\ d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} \\ d = \sqrt{(a - a)^2 + (y_1 - y_0)^2} \\ d = \sqrt{(y_1 - y_0)^2} \\ d = y_1 - y_0

36.

P=(x1,y1)Q=(x2,y2)M=(x1+x22,y1+y22) P = (x_1, y_1) \\ Q = (x_2, y_2) \\ M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})

dpq=(x2−x1)2+(y2−y1)2dpq2=(x2−x1)2+(y2−y1)22 d_{pq} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\ \frac{d_{pq}}{2} = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2}

dpm=(x2+x12−x1)2+(y2+y12−y1)2dpm=(x2+x12−2x12)2+(y2+y12−2y12)2dpm=(x2−x12)2+(y2−y12)2dpm=(x2−x1)24+(y2−y1)24dpm=(x2−x1)2+(y2−y1)24=(x2−x1)2+(y2−y1)24dpm=(x2−x1)2+(y2−y1)22dpm=dpq2 d_{pm} = \sqrt{(\frac{x_2 + x_1}{2} - x_1)^2 + (\frac{y_2 + y_1}{2} - y_1)^2} \\ d_{pm} = \sqrt{(\frac{x_2 + x_1}{2} - \frac{2x_1}{2})^2 + (\frac{y_2 + y_1}{2} - \frac{2y_1}{2})^2} \\ d_{pm} = \sqrt{(\frac{x_2 - x_1}{2})^2 + (\frac{y_2 - y_1}{2})^2} \\ d_{pm} = \sqrt{\frac{(x_2 - x_1)^2}{4} + \frac{(y_2 - y_1)^2}{4}} \\ d_{pm} = \sqrt{\frac{(x_2 - x_1)^2 + (y_2 - y_1)^2}{4}} = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{\sqrt{4}} \\ d_{pm} = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2} \\ d_{pm} = \frac{d_{pq}}{2}

dmq=(x2−x2+x12)2+(y2−y2+y12)2dmq=(2x22−x2+x12)2+(2y22−y2+y12)2dmq=(2x2−(x2+x1)2)2+(2y2−(y2+y1)2)2dmq=(x2−x1)2+(y2−y1)24dmq=(x2−x1)2+(y2−y1)22 d_{mq} = \sqrt{(x_2 - \frac{x_2 + x_1}{2})^2 + (y_2 - \frac{y_2 + y_1}{2})^2} \\ d_{mq} = \sqrt{(\frac{2x_2}{2} - \frac{x_2 + x_1}{2})^2 + (\frac{2y_2}{2} - \frac{y_2 + y_1}{2})^2} \\ d_{mq} = \sqrt{(\frac{2x_2 - (x_2 + x_1)}{2})^2 + (\frac{2y_2 - (y_2 + y_1)}{2})^2} \\ d_{mq} = \sqrt{\frac{(x_2 - x_1)^2 + (y_2 - y_1)^2}{4}} \\ d_{mq} = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2}

dmq=dpm=dpq2 d_{mq} = d_{pm} = \frac{d_{pq}}{2}

38.

D=(x,y)x=xb−(xc−xa)=4−(0−(−3))=1y=yb−(yc−ya)=0−(−3−1)=4D=(1,4) D = (x, y) \\ x = x_b - (x_c - x_a) = 4 - (0 - (-3)) = 1 \\ y = y_b - (y_c - y_a) = 0 - (-3 - 1) = 4 \\ D = (1, 4)

All corners of a square are right angles and the same length. Therefore the x,yx,y difference between AA and CC is the same as the x,yx,y difference between DD and BB.

The diagonal line from AA to BB should also be the same length as the diagonal line from CC to DD. Checking:

dab=(4−(−3))2+(0−1)2=50=52dcd=(1−0)2+(−3−4)2=50=52 d_{ab} = \sqrt{(4 - (-3))^2 + (0 - 1)^2} = \sqrt{50} = 5\sqrt{2} \\ d_{cd} = \sqrt{(1 - 0)^2 + (-3 - 4)^2} = \sqrt{50} = 5\sqrt{2}

Converse of the Pythagorean theorem

If a2+b2=c2a^2 + b^2 = c^2, then the angle between sides aa and bb is a right angle.

https://en.wikipedia.org/wiki/Pythagorean_theorem#Converse

Relations

Term: relation

A relation is a set of points in the plane.

Therefore a relation can be described in all the ways that sets can be described.

A point is an ordered pair of numbers in the Euclidean plane.

From one point of view, all of Precalculus can be thought of as studying sets of points in the plane.

https://www.cuemath.com/algebra/relations-in-math/

https://en.wikipedia.org/wiki/Relation_(mathematics)

Examples:

R={(x,y)∣y=5x+3}S={(3,1),(3,2),(5,7)} R = \{(x, y) \mid y = 5x + 3\} \\ S = \{(3,1), (3, 2), (5,7)\}

The graph of an equation is the set of all x,yx,y points that make the equation true (i.e. points that satisfy the equation).

Y-intercept, x-intercept, and testing for symmetry

The y-intercept or x-intercept is the point where xx or yy is 00 (respectively).

https://en.wikipedia.org/wiki/Y-intercept

You can figure out what the graph of an equation would look like by testing these things:

  • Y-intercept: set xx to 00 and solve
  • X-intercept: set yy to 00 and solve
  • Whether the graph is symmetric about the xx axis: set yy to −y-y and solve
  • Whether the graph is symmetric about the yy axis: set xx to −x-x and solve
  • Whether the graph is symmetric about the origin: set xx to −x-x and yy to −y-y and solve

1.2.2

21. A={(−4,−1),(−2,1),(0,3),(1,4)}A = \{(-4,-1), (-2,1), (0,3), (1,4)\}

22. B={(x,3)∣x≥−3}B = \{(x, 3) \mid x \ge -3\}

23. C={(2,y)∣y>−3}C = \{(2, y) \mid y > -3\}

24. D={(−2,y)∣−4≤y<3}D = \{(-2, y) \mid -4 \le y < 3\}

26. F={(x,y)∣x∈R,y≥0}F = \{(x, y) \mid x \in \Reals, y \ge 0\}

You can also omit the x∈Rx \in \Reals. Apparently in that case it's implied that x∈Rx \in \Reals.

28. H={(x,y)∣−3<x≤2,y∈R}H = \{(x, y) \mid -3 < x \le 2, y \in \Reals\}

You can also omit the y∈Ry \in \Reals. Apparently in that case it's implied that y∈Ry \in \Reals.

30. J={(x,y)∣−4<x<5,−3<y<2}J = \{(x, y) \mid -4 < x < 5, -3 < y < 2\}

41.

y=x2+1 y = x^2 + 1

y-intercept=1(0,1) \text{y-intercept} = \cancel{1} \quad (0,1) \\

Remember, y- and x-intercepts are points, so you need to specify them as (x,y)(x,y) pairs.

0=x2+1x=±−1No x-intercept 0 = x^2 + 1 \\ x = \pm\sqrt{-1} \qquad \text{No x-intercept}

y=(−x)2+1=x2+1Graph is symmetric about the y axis y = (-x)^2 + 1 = x^2 + 1 \quad \text{Graph is symmetric about the y axis}

y=x2+1−y≠yGraph is not symmetric about the x axis...2=x2+1...but to make sure, we test with point (x,2)...x=±1=±1Point (1,2)−y=x2+1...and then substitute y with -y, i.e. test the point (1,-2)...−(x2)−y=?1−1−2≠1Nope, graph is not symmetric about the x axis y = x^2 + 1 \\ -y \neq y \quad \text{Graph is not symmetric about the x axis...} \\ 2 = x^2 + 1 \quad \text{...but to make sure, we test with point (x,2)...} \\ x = \pm\sqrt{1} = \pm1 \qquad \text{Point }(1,2) \\ -y = x^2 + 1 \quad \text{...and then substitute y with -y, i.e. test the point (1,-2)...} \\ -(x^2) - y \overset{?}{=} 1 \\ -1 - 2 \neq 1 \quad \text{Nope, graph is not symmetric about the x axis}

−y=?(−x)2+1Test for symmetry about the origin−y=?x2+1−y≠yGraph is not symmetric about the origin -y \overset{?}{=} (-x)^2 + 1 \quad \text{Test for symmetry about the origin} \\ -y \overset{?}{=} x^2 + 1 \\ -y \neq y \quad \text{Graph is not symmetric about the origin}

Gnuplot Produced by GNUPLOT 5.4 patchlevel 2 -1 0 1 2 3 4 5 -4 -3 -2 -1 0 1 2 3 4 5 y x f(x) f(x) 
set xtics 1
set ytics 1
set xrange [-4:5]
set yrange [-1:5]

f(x) = x*x + 1
plot f(x)

50.

y2=x2−1 y^2 = x^2 - 1

y=0±−1No y-intercept y = 0 \pm \sqrt{-1} \quad \text{No y-intercept}

x-intercept=(±1,0)=(±1,0) \text{x-intercept} = (\pm\sqrt{1}, 0) = (\pm 1, 0)

y2=(−x)2−1=x2−1Is symmetric about the y axis y^2 = (-x)^2 - 1 = x^2 - 1 \quad \text{Is symmetric about the y axis}

(−y)2=?x2−1=y2Test for symmetry about the x axis(−y)2=y2Is symmetric about the x axis (-y)^2 \overset{?}{=} x^2 - 1 = y^2 \quad \text{Test for symmetry about the x axis} \\ (-y)^2 = y^2 \quad \text{Is symmetric about the x axis}

(−y)2=(−x)2−1=?y2Test for symmetry about the origin(−y)2=(−x)2−1=y2=x2−1Is symmetric about the origin (-y)^2 = (-x)^2 - 1 \overset{?}{=} y^2 \quad \text{Test for symmetry about the origin} \\ (-y)^2 = (-x)^2 - 1 = y^2 = x^2 - 1 \quad \text{Is symmetric about the origin}

Gnuplot Produced by GNUPLOT 5.4 patchlevel 2 -3 -2 -1 0 1 2 3 4 5 -5 -4 -3 -2 -1 0 1 2 3 4 5 y x f(x) f(x) g(x) g(x) 
set xtics 1
set ytics 1
set xrange [-5:5]
set yrange [-3:5]

f(x) = sqrt(x*x - 1)
g(x) = -sqrt(x*x - 1)
plot f(x), g(x)

Functions

Term: function

A special kind of relation that maps xx to only one yy.

In other words, a function is a many-to-one or one-to-one relation.

f(x)=2xIs a functionf(x)=5Is a functionf(x)=x2Is a functiony=±xNot a functionR1={(1,2),(1,5),(2,3)}Not a functionR2={(1,2),(2,−2),(3,3)}Is a function f(x) = 2x \qquad \text{Is a function} \\ f(x) = 5 \qquad \text{Is a function} \\ f(x) = x^2 \qquad \text{Is a function} \\ y = \pm x \qquad \text{Not a function} \\ R_1 = \{(1, 2), (1, 5), (2, 3)\} \qquad \text{Not a function} \\ R_2 = \{(1, 2), (2, -2), (3, 3)\} \qquad \text{Is a function}

Vertical line test

If two or more points lie on the same vertical line, then the relation is not a function.

Term: domain and range
  • Domain: the set of xx values in the function RR
  • Range: the set of yy values in the function RR

R={(1,3),(2,2),(3,8),(4,7)}R={(1,3),(2,2),(3,8),(4,8)}DomainR={(1,3),(2,2),(3,8),(4,8)}Range R = \{(1,3), (2, 2), (3,8), (4,7)\} \\ R = \{(\boxed{1},3), (\boxed{2},2), (\boxed{3},8), (\boxed{4},8)\} \quad \text{Domain} \\ R = \{(1,\boxed{3}), (2,\boxed{2}), (3,\boxed{8}), (4,\boxed{8})\} \quad \text{Range}

Domain={x∣1≤x≤4}, i.e. [1,4]Range={2,3,8} \text{Domain} = \{x \mid 1 \le x \le 4\}\text{, i.e. }[1, 4] \\ \text{Range} = \{2, 3, 8\}

Term: projecting a curve to an axis

It's basically like casting a shadow onto the xx or yy axis. You "flatten" the curve onto the xx or yy axis, thus getting the domain or range.

1.3.1

    1. No
    1. Yes. Domain: {x∣x is irrational}\{x \mid x\text{ is irrational}\}. Range: {1}\{1\}
    1. Yes. Domain: {x∣x=N12}\{x \mid x = \natnums_1^2\}. Range: N0\natnums_0
    • Mistake: domain should be {x∣x=2n,n∈N1}\{x \mid x = 2^n, n \in \natnums_1\}
    1. Yes. Domain: Z\Z. Range: {y∣y=x2,x∈Z}\{y \mid y = x^2, x \in \Z\}
    • Here it would be a bit clearer to use nn in the domain, i.e. {y∣y=n2,n∈Z}\{y \mid y = n^2, n \in \Z\}, just so that the range can "stand alone" and not refer even accidentally to the domain (where we usually use the variable name xx)
    1. No
    1. Yes. Domain: [−2,4)[-2, 4). Range: {3}\{3\}
    1. Yes. Domain: R\Reals. Range: {y∣y∈R,y≥0}\{y \mid y \in \Reals, y \ge 0\}
    • The book says that the domain is (−∞,∞)(-\infty, \infty) and that the range is [0,∞)[0, \infty). Are we both correct? I think so.
    1. No. y=±xy = \pm \sqrt{x} gives two outputs for a single input.
    1. Yes. Domain: {x∣−5≤x<3,x≠3}\{x \mid -5 \le x < 3, x \neq 3\}. Range: {y∣−2<y<−1}∪{y∣0≤y<4}\{y \mid -2 < y < -1\} \cup \{y \mid 0 \le y < 4\}
    • Here it would be less verbose to use the range syntax instead of the set builder syntax
    1. Yes. Domain: R\Reals. Range: (−∞,4](-\infty, 4]
    1. Yes. Domain: R\Reals. Range: (−∞,4](-\infty, 4]
    1. Yes. Domain: R\Reals. Range: R\Reals
    1. No
    1. Yes. Domain: R\Reals. Range: {2}\{2\}
    1. Yes
    1. Yes. I don't see a ±\pm in front of the \sqrt{}, so I must assume this is a function. I assume the domain to be {x∣x≥2}\{x \mid x \ge 2\} and the range to be {y∣y≥0}\{y \mid y \ge 0\}. Although maybe I shouldn't make those assumptions. Can we allow irrational numbers (i.e. −n\sqrt{-n}) in the range and still have this be a function? I think so.
    1. Yes. y=−4x3y = \frac{-4}{x^3}
    1. No. y=±x2−1y = \pm \sqrt{x^2 - 1}
    1. Yes
    • Btw, I tried to simplify this into y=1x−9xy = \frac{1}{x - \frac{9}{x}}, but that's a mistake: it changes the domain and range. See before versus after. A better simplification is x(x−3)(x+3)\frac{x}{(x - 3)(x + 3)} (once you realize the difference of two squares in the denominator).
    1. No
    1. No. y=±x−4y = \pm \sqrt{x - 4}
    1. Yes
    1. Yes. I don't see a ±\pm.
    1. No. y=±xy = \pm x
    1. P(t)P(t). Range: {y∣y≥0}\{y \mid y \ge 0\} (population can't go below zero).
  • 50 - 53.

What if I set xx to 00? We get y3=0y^3 = 0, which isn't of much help. What if I set xx to 22? I get 8+y3−6y=08 + y^3 - 6y = 0, so y3−6y=−8y^3 - 6y = -8. Doesn't help.

Can I assume there exists a vertical line x=qx = q that intersects with the curve at two points? How can I represent this intersection mathematically? Perhaps with a system of equations?

(The book doesn't offer solutions for 48.-53.)

Can sqrt[n]{-x} ever result in a non-imaginary number?

Yes: taking an odd root of a negative will result in a negative number.

Example: −273=−3\sqrt[3]{-27} = -3

It's basically only even roots of negative numbers that you need to be careful of.