Calculus/Precalculus/Solutions

Convert to interval notation

1. ${\displaystyle \{x:-4
${\displaystyle \mathbf {(-4,2)} }$
${\displaystyle \mathbf {(-4,2)} }$
2. ${\displaystyle \{x:-{\frac {7}{3}}\leq x\leq -{\frac {1}{3}}\}}$
${\displaystyle \mathbf {[-{\frac {7}{3}},-{\frac {1}{3}}]} }$
${\displaystyle \mathbf {[-{\frac {7}{3}},-{\frac {1}{3}}]} }$
3. ${\displaystyle \{x:-\pi \leq x<\pi \}}$
${\displaystyle \mathbf {[-\pi ,\pi )} }$
${\displaystyle \mathbf {[-\pi ,\pi )} }$
4. ${\displaystyle \{x:x\leq 17/9\}}$
${\displaystyle \mathbf {(-\infty ,{\frac {17}{9}}]} }$
${\displaystyle \mathbf {(-\infty ,{\frac {17}{9}}]} }$
5. ${\displaystyle \{x:5\leq x+1\leq 6\}}$
${\displaystyle 4\leq x\leq 5}$
${\displaystyle \mathbf {[4,5]} }$
${\displaystyle 4\leq x\leq 5}$
${\displaystyle \mathbf {[4,5]} }$
6. ${\displaystyle \{x:x-1/4<1\}\,}$
${\displaystyle x<1{\frac {1}{4}}={\frac {5}{4}}}$
${\displaystyle \mathbf {(-\infty ,{\frac {5}{4}})} }$
${\displaystyle x<1{\frac {1}{4}}={\frac {5}{4}}}$
${\displaystyle \mathbf {(-\infty ,{\frac {5}{4}})} }$
7. ${\displaystyle \{x:3>3x\}\,}$
${\displaystyle 1>x}$

${\displaystyle x<1}$

${\displaystyle \mathbf {(-\infty ,1)} }$
${\displaystyle 1>x}$

${\displaystyle x<1}$

${\displaystyle \mathbf {(-\infty ,1)} }$
8. ${\displaystyle \{x:0\leq 2x+1<3\}}$
${\displaystyle -1\leq 2x\leq 2}$

${\displaystyle -{\frac {1}{2}}\leq x<1}$

${\displaystyle \mathbf {[-{\frac {1}{2}},1)} }$
${\displaystyle -1\leq 2x\leq 2}$

${\displaystyle -{\frac {1}{2}}\leq x<1}$

${\displaystyle \mathbf {[-{\frac {1}{2}},1)} }$
9. ${\displaystyle \{x:5
This is equivalent to ${\displaystyle 5
${\displaystyle \mathbf {(5,6)} }$
This is equivalent to ${\displaystyle 5
${\displaystyle \mathbf {(5,6)} }$
10. ${\displaystyle \{x:5
It helps to draw a picture to determine the set of numbers described:

A number in the set can be on either the red or blue line, so the entire number line is included.

${\displaystyle \mathbf {(-\infty ,\infty )} }$
It helps to draw a picture to determine the set of numbers described:

A number in the set can be on either the red or blue line, so the entire number line is included.

${\displaystyle \mathbf {(-\infty ,\infty )} }$

State the following intervals using set notation

11. ${\displaystyle [3,4]\,}$
${\displaystyle \mathbf {\{x:3\leq x\leq 4\}} }$
${\displaystyle \mathbf {\{x:3\leq x\leq 4\}} }$
12. ${\displaystyle [3,4)\,}$
${\displaystyle \mathbf {\{x:3\leq x<4\}} }$
${\displaystyle \mathbf {\{x:3\leq x<4\}} }$
13. ${\displaystyle (3,\infty )}$
${\displaystyle \mathbf {\{x:x>3\}} }$
${\displaystyle \mathbf {\{x:x>3\}} }$
14. ${\displaystyle (-{\frac {1}{3}},{\frac {1}{3}})\,}$
${\displaystyle \mathbf {\{x:-{\frac {1}{3}}
${\displaystyle \mathbf {\{x:-{\frac {1}{3}}
15. ${\displaystyle (-\pi ,{\frac {15}{16}})\,}$
${\displaystyle \mathbf {\{x:-\pi
${\displaystyle \mathbf {\{x:-\pi
16. ${\displaystyle (-\infty ,\infty )}$
${\displaystyle \mathbf {\{x:x\in \Re \}} }$
${\displaystyle \mathbf {\{x:x\in \Re \}} }$

Which one of the following is a true statement?

17. ${\displaystyle |x+y|=|x|+|y|\,}$
Let ${\displaystyle x=-5,y=5}$ . Then

${\displaystyle |x+y|=|-5+5|=|0|=0}$ , and
${\displaystyle |x|+|y|=|-5|+|5|=5+5=10}$
Thus, ${\displaystyle |x+y|\neq |x|+|y|}$

false
Let ${\displaystyle x=-5,y=5}$ . Then

${\displaystyle |x+y|=|-5+5|=|0|=0}$ , and
${\displaystyle |x|+|y|=|-5|+|5|=5+5=10}$
Thus, ${\displaystyle |x+y|\neq |x|+|y|}$

false
18. ${\displaystyle |x+y|\geq |x|+|y|}$
Using the same example as above, we have ${\displaystyle |x+y|\ngeq |x|+|y|}$ .
false
Using the same example as above, we have ${\displaystyle |x+y|\ngeq |x|+|y|}$ .
false
19. ${\displaystyle |x+y|\leq |x|+|y|}$
true
true

Evaluate the following expressions

20. ${\displaystyle 8^{1/3}\,}$
${\displaystyle (2^{3})^{1/3}=2^{1}=\mathbf {2} }$
${\displaystyle (2^{3})^{1/3}=2^{1}=\mathbf {2} }$
21. ${\displaystyle (-8)^{1/3}\,}$
${\displaystyle (-2^{3})^{1/3}=-2^{1}=\mathbf {-2} }$
${\displaystyle (-2^{3})^{1/3}=-2^{1}=\mathbf {-2} }$
22. ${\displaystyle {\bigg (}{\frac {1}{8}}{\bigg )}^{1/3}\,}$
${\displaystyle ({\frac {1}{2^{3}}})^{1/3}=(2^{-3})^{1/3}=2^{-1}=\mathbf {\frac {1}{2}} }$
${\displaystyle ({\frac {1}{2^{3}}})^{1/3}=(2^{-3})^{1/3}=2^{-1}=\mathbf {\frac {1}{2}} }$
23. ${\displaystyle (8^{2/3})(8^{3/2})(8^{0})\,}$
${\displaystyle 8^{{\frac {2}{3}}+{\frac {3}{2}}+0}=8^{{\frac {4}{6}}+{\frac {9}{6}}}=8^{\frac {13}{6}}=(2^{3})^{\frac {13}{6}}=\mathbf {2^{13/2}} }$
${\displaystyle 8^{{\frac {2}{3}}+{\frac {3}{2}}+0}=8^{{\frac {4}{6}}+{\frac {9}{6}}}=8^{\frac {13}{6}}=(2^{3})^{\frac {13}{6}}=\mathbf {2^{13/2}} }$
24. ${\displaystyle {\bigg (}{\bigg (}{\frac {1}{8}}{\bigg )}^{1/3}{\bigg )}^{7}}$
${\displaystyle (({\frac {1}{2^{3}}})^{1/3})^{7}=((2^{-3})^{1/3})^{7}=(2^{-1})^{7}=2^{-7}={\frac {1}{2^{7}}}=\mathbf {\frac {1}{128}} }$
${\displaystyle (({\frac {1}{2^{3}}})^{1/3})^{7}=((2^{-3})^{1/3})^{7}=(2^{-1})^{7}=2^{-7}={\frac {1}{2^{7}}}=\mathbf {\frac {1}{128}} }$
25. ${\displaystyle {\sqrt[{3}]{\frac {27}{8}}}}$
${\displaystyle ({\frac {27}{8}})^{1/3}=({\frac {3^{3}}{2^{3}}})^{1/3}={\frac {3^{1}}{2^{1}}}=\mathbf {\frac {3}{2}} }$
${\displaystyle ({\frac {27}{8}})^{1/3}=({\frac {3^{3}}{2^{3}}})^{1/3}={\frac {3^{1}}{2^{1}}}=\mathbf {\frac {3}{2}} }$
26. ${\displaystyle {\frac {4^{5}\cdot 4^{-2}}{4^{3}}}}$
${\displaystyle 4^{5-2-3}=4^{0}=\mathbf {1} }$
${\displaystyle 4^{5-2-3}=4^{0}=\mathbf {1} }$
27. ${\displaystyle {\bigg (}{\sqrt {27}}{\bigg )}^{2/3}}$
${\displaystyle ((3^{3})^{1/2})^{2/3}=(3^{\frac {3}{2}})^{\frac {2}{3}}=3^{1}=\mathbf {3} }$
${\displaystyle ((3^{3})^{1/2})^{2/3}=(3^{\frac {3}{2}})^{\frac {2}{3}}=3^{1}=\mathbf {3} }$
28. ${\displaystyle {\frac {\sqrt {27}}{\sqrt[{3}]{9}}}}$
${\displaystyle {\frac {(3^{3})^{1/2}}{(3^{2})^{1/3}}}={\frac {3^{\frac {3}{2}}}{3^{\frac {2}{3}}}}=3^{{\frac {3}{2}}-{\frac {2}{3}}}=3^{{\frac {9}{6}}-{\frac {4}{6}}}=\mathbf {3^{5/6}} }$
${\displaystyle {\frac {(3^{3})^{1/2}}{(3^{2})^{1/3}}}={\frac {3^{\frac {3}{2}}}{3^{\frac {2}{3}}}}=3^{{\frac {3}{2}}-{\frac {2}{3}}}=3^{{\frac {9}{6}}-{\frac {4}{6}}}=\mathbf {3^{5/6}} }$

Simplify the following

29. ${\displaystyle x^{3}+3x^{3}\,}$
${\displaystyle \mathbf {4x^{3}} }$
${\displaystyle \mathbf {4x^{3}} }$
30. ${\displaystyle {\frac {x^{3}+3x^{3}}{x^{2}}}}$
${\displaystyle \mathbf {4x} }$
${\displaystyle \mathbf {4x} }$
31. ${\displaystyle (x^{3}+3x^{3})^{3}\,}$
${\displaystyle \mathbf {64x^{9}} }$
${\displaystyle \mathbf {64x^{9}} }$
32. ${\displaystyle {\frac {x^{15}+x^{3}}{x}}}$
${\displaystyle \mathbf {x^{14}+x^{2}} }$
${\displaystyle \mathbf {x^{14}+x^{2}} }$
33. ${\displaystyle (2x^{2})(3x^{-2})\,}$
${\displaystyle \mathbf {6} }$
${\displaystyle \mathbf {6} }$
34. ${\displaystyle {\frac {x^{2}y^{-3}}{x^{3}y^{2}}}}$
${\displaystyle \mathbf {\frac {1}{xy^{5}}} }$
${\displaystyle \mathbf {\frac {1}{xy^{5}}} }$
35. ${\displaystyle {\sqrt {x^{2}y^{4}}}}$
${\displaystyle \mathbf {xy^{2}} }$
${\displaystyle \mathbf {xy^{2}} }$
36. ${\displaystyle {\bigg (}{\frac {8x^{6}}{y^{4}}}{\bigg )}^{1/3}}$
${\displaystyle \mathbf {\frac {2x^{2}}{y^{\frac {4}{3}}}} }$
${\displaystyle \mathbf {\frac {2x^{2}}{y^{\frac {4}{3}}}} }$

Find the roots of the following polynomials

37. ${\displaystyle x^{2}-1\,}$
${\displaystyle x=\pm 1}$
${\displaystyle x=\pm 1}$
38. ${\displaystyle x^{2}+2x+1\,}$
${\displaystyle x=-1}$
${\displaystyle x=-1}$
39. ${\displaystyle x^{2}+7x+12\,}$
${\displaystyle x=-3,x=-4}$
${\displaystyle x=-3,x=-4}$
40. ${\displaystyle 3x^{2}-5x-2\,}$
${\displaystyle x=2,x=-{\frac {1}{3}}}$
${\displaystyle x=2,x=-{\frac {1}{3}}}$
41. ${\displaystyle x^{2}+5/6x+1/6\,}$
${\displaystyle x=-{\frac {1}{3}},x=-{\frac {1}{2}}}$
${\displaystyle x=-{\frac {1}{3}},x=-{\frac {1}{2}}}$
42. ${\displaystyle 4x^{3}+4x^{2}+x\,}$
${\displaystyle x=0,x=-{\frac {1}{2}}}$
${\displaystyle x=0,x=-{\frac {1}{2}}}$
43. ${\displaystyle x^{4}-1\,}$
${\displaystyle x=\pm i,x=\pm 1}$
${\displaystyle x=\pm i,x=\pm 1}$
44. ${\displaystyle x^{3}+2x^{2}-4x-8\,}$
${\displaystyle x=\pm 2}$
${\displaystyle x=\pm 2}$

Factor the following expressions

45. ${\displaystyle 4a^{2}-ab-3b^{2}\,}$
${\displaystyle (4a+3b)(a-b)}$
${\displaystyle (4a+3b)(a-b)}$
46. ${\displaystyle (c+d)^{2}-4\,}$
${\displaystyle (c+d+2)(c+d-2)}$
${\displaystyle (c+d+2)(c+d-2)}$
47. ${\displaystyle 4x^{2}-9y^{2}\,}$
${\displaystyle (2x+3y)(2x-3y)}$
${\displaystyle (2x+3y)(2x-3y)}$

Simplify the following

48. ${\displaystyle {\frac {x^{2}-1}{x+1}}\,}$
${\displaystyle x-1,x\neq -1}$
${\displaystyle x-1,x\neq -1}$
49. ${\displaystyle {\frac {3x^{2}+4x+1}{x+1}}\,}$
${\displaystyle 3x+1,x\neq -1}$
${\displaystyle 3x+1,x\neq -1}$
50. ${\displaystyle {\frac {4x^{2}-9}{4x^{2}+12x+9}}\,}$
${\displaystyle {\frac {2x-3}{2x+3}}}$
${\displaystyle {\frac {2x-3}{2x+3}}}$
51. ${\displaystyle {\frac {x^{2}+y^{2}+2xy}{x(x+y)}}\,}$
${\displaystyle {\frac {x+y}{x}},x\neq -y}$
${\displaystyle {\frac {x+y}{x}},x\neq -y}$

Functions

52. Let ${\displaystyle f(x)=x^{2}}$ .

a. Compute ${\displaystyle f(0)}$  and ${\displaystyle f(2)}$ .
${\displaystyle f(0)=0}$ , ${\displaystyle f(2)=4}$
${\displaystyle f(0)=0}$ , ${\displaystyle f(2)=4}$
b. What are the domain and range of ${\displaystyle f}$ ?
The domain is ${\displaystyle (-\infty ,\infty )}$ ; the range is ${\displaystyle [0,\infty )}$ ,
The domain is ${\displaystyle (-\infty ,\infty )}$ ; the range is ${\displaystyle [0,\infty )}$ ,
c. Does ${\displaystyle f}$  have an inverse? If so, find a formula for it.
No, since ${\displaystyle f}$  isn't one-to-one; for example, ${\displaystyle f(-1)=f(1)=1}$ .
No, since ${\displaystyle f}$  isn't one-to-one; for example, ${\displaystyle f(-1)=f(1)=1}$ .

53. Let ${\displaystyle f(x)=x+2}$ , ${\displaystyle g(x)=1/x}$ .

a. Give formulae for
i. ${\displaystyle f+g}$
${\displaystyle (f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x}$ .
${\displaystyle (f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x}$ .
ii. ${\displaystyle f-g}$
${\displaystyle (f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x}$ .
${\displaystyle (f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x}$ .
iii. ${\displaystyle g-f}$
${\displaystyle (g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x}$ .
${\displaystyle (g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x}$ .
iv. ${\displaystyle f\times g}$
${\displaystyle (f\times g)(x)=(x+2)/x}$ .
${\displaystyle (f\times g)(x)=(x+2)/x}$ .
v. ${\displaystyle f/g}$
${\displaystyle (f/g)(x)=x(x+2)}$  provided ${\displaystyle x\neq 0}$ . Note that 0 is not in the domain of ${\displaystyle f/g}$ , since it's not in the domain of ${\displaystyle g}$ , and you can't divide by something that doesn't exist!
${\displaystyle (f/g)(x)=x(x+2)}$  provided ${\displaystyle x\neq 0}$ . Note that 0 is not in the domain of ${\displaystyle f/g}$ , since it's not in the domain of ${\displaystyle g}$ , and you can't divide by something that doesn't exist!
vi. ${\displaystyle g/f}$
${\displaystyle (g/f)(x)=1/[x(x+2)]}$ . Although 0 is still not in the domain, we don't need to state it now, since 0 isn't in the domain of the expression ${\displaystyle 1/[x(x+2)]}$  either.
${\displaystyle (g/f)(x)=1/[x(x+2)]}$ . Although 0 is still not in the domain, we don't need to state it now, since 0 isn't in the domain of the expression ${\displaystyle 1/[x(x+2)]}$  either.
vii. ${\displaystyle f\circ g}$
${\displaystyle (f\circ g)(x)=1/x+2=(2x+1)/x}$ .
${\displaystyle (f\circ g)(x)=1/x+2=(2x+1)/x}$ .
viii. ${\displaystyle g\circ f}$
${\displaystyle (g\circ f)(x)=1/(x+2)}$ .
${\displaystyle (g\circ f)(x)=1/(x+2)}$ .
b. Compute ${\displaystyle f(g(2))}$  and ${\displaystyle g(f(2))}$ .
${\displaystyle f(g(2))=5/2}$ ; ${\displaystyle g(f(2))=1/4}$ .
${\displaystyle f(g(2))=5/2}$ ; ${\displaystyle g(f(2))=1/4}$ .
c. Do ${\displaystyle f}$  and ${\displaystyle g}$  have inverses? If so, find formulae for them.
Yes; ${\displaystyle f^{-1}(x)=x-2}$  and ${\displaystyle g^{-1}(x)=1/x}$ . Note that ${\displaystyle g}$  and its inverse are the same.
Yes; ${\displaystyle f^{-1}(x)=x-2}$  and ${\displaystyle g^{-1}(x)=1/x}$ . Note that ${\displaystyle g}$  and its inverse are the same.
54. Does this graph represent a function?
As pictured, by the Vertical Line test, this graph represents a function.
As pictured, by the Vertical Line test, this graph represents a function.

55. Consider the following function

${\displaystyle f(x)={\begin{cases}-{\frac {1}{9}}&{\mbox{if }}x<-1\\2&{\mbox{if }}-1\leq x\leq 0\\x+3&{\mbox{if }}x>0.\end{cases}}}$
a. What is the domain?
${\displaystyle {(-\infty ,\infty )}}$
${\displaystyle {(-\infty ,\infty )}}$
b. What is the range?
${\displaystyle {(-1/9,\infty )}}$
${\displaystyle {(-1/9,\infty )}}$
c. Where is ${\displaystyle f}$  continuous?
${\displaystyle {x>0}}$
${\displaystyle {x>0}}$

56. Consider the following function

${\displaystyle f(x)={\begin{cases}x^{2}&{\mbox{if }}x>0\\-1&{\mbox{if }}x\leq 0.\end{cases}}}$
a. What is the domain?
${\displaystyle {(-\infty ,\infty )}}$
${\displaystyle {(-\infty ,\infty )}}$
b. What is the range?
${\displaystyle {(-1,\infty )}}$
${\displaystyle {(-1,\infty )}}$
c. Where is ${\displaystyle f}$  continuous?
${\displaystyle {x>0}}$
${\displaystyle {x>0}}$

57. Consider the following function

${\displaystyle f(x)={\frac {\sqrt {2x-3}}{x-10}}}$
a. What is the domain?
${\displaystyle {(3/2,10)\cup (10,\infty )}}$
${\displaystyle {(3/2,10)\cup (10,\infty )}}$
b. What is the range?
${\displaystyle {(-\infty ,\infty )}}$
${\displaystyle {(-\infty ,\infty )}}$
c. Where is ${\displaystyle f}$  continuous?
${\displaystyle {(3/2,10)and(x>10)}}$
${\displaystyle {(3/2,10)and(x>10)}}$

58. Consider the following function

${\displaystyle f(x)={\frac {x-7}{x^{2}-49}}}$
a. What is the domain?
${\displaystyle {(-\infty ,-7)\cup (-7,\infty )}}$
${\displaystyle {(-\infty ,-7)\cup (-7,\infty )}}$
b. What is the range?
${\displaystyle {(-\infty ,\infty )}}$
${\displaystyle {(-\infty ,\infty )}}$
c. Where is ${\displaystyle f}$  continuous?
${\displaystyle {(-\infty ,-7)and(-7,\infty )}}$
${\displaystyle {(-\infty ,-7)and(-7,\infty )}}$

Graphing

59. Find the equation of the line that passes through the point (1,-1) and has slope 3.
${\displaystyle 3x-y=4}$
${\displaystyle 3x-y=4}$
60. Find the equation of the line that passes through the origin and the point (2,3).
${\displaystyle 3x-2y=0}$
${\displaystyle 3x-2y=0}$