# Calculus/Precalculus/Solutions

## Convert to interval notation

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

$x<1$

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

$x<1$

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

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

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

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

$\mathbf {[-{\frac {1}{2}},1)}$
9. $\{x:5
This is equivalent to $5
$\mathbf {(5,6)}$
This is equivalent to $5
$\mathbf {(5,6)}$
10. $\{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.

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

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

## State the following intervals using set notation

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

## Which one of the following is a true statement?

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

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

false
Let $x=-5,y=5$ . Then

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

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

## Evaluate the following expressions

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

## Simplify the following

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

## Find the roots of the following polynomials

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

## Factor the following expressions

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

## Simplify the following

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

## Functions

52. Let $f(x)=x^{2}$ .

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

53. Let $f(x)=x+2$ , $g(x)=1/x$ .

a. Give formulae for
i. $f+g$
$(f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x$ .
$(f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x$ .
ii. $f-g$
$(f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x$ .
$(f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x$ .
iii. $g-f$
$(g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x$ .
$(g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x$ .
iv. $f\times g$
$(f\times g)(x)=(x+2)/x$ .
$(f\times g)(x)=(x+2)/x$ .
v. $f/g$
$(f/g)(x)=x(x+2)$  provided $x\neq 0$ . Note that 0 is not in the domain of $f/g$ , since it's not in the domain of $g$ , and you can't divide by something that doesn't exist!
$(f/g)(x)=x(x+2)$  provided $x\neq 0$ . Note that 0 is not in the domain of $f/g$ , since it's not in the domain of $g$ , and you can't divide by something that doesn't exist!
vi. $g/f$
$(g/f)(x)=1/[x(x+2)]$ . 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 $1/[x(x+2)]$  either.
$(g/f)(x)=1/[x(x+2)]$ . 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 $1/[x(x+2)]$  either.
vii. $f\circ g$
$(f\circ g)(x)=1/x+2=(2x+1)/x$ .
$(f\circ g)(x)=1/x+2=(2x+1)/x$ .
viii. $g\circ f$
$(g\circ f)(x)=1/(x+2)$ .
$(g\circ f)(x)=1/(x+2)$ .
b. Compute $f(g(2))$  and $g(f(2))$ .
$f(g(2))=5/2$ ; $g(f(2))=1/4$ .
$f(g(2))=5/2$ ; $g(f(2))=1/4$ .
c. Do $f$  and $g$  have inverses? If so, find formulae for them.
Yes; $f^{-1}(x)=x-2$  and $g^{-1}(x)=1/x$ . Note that $g$  and its inverse are the same.
Yes; $f^{-1}(x)=x-2$  and $g^{-1}(x)=1/x$ . Note that $g$  and its inverse are the same.
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

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

56. Consider the following function

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

57. Consider the following function

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

58. Consider the following function

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

## Graphing

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