Calculus/Precalculus/Solutions

1. ${\displaystyle \{x:-4<x<2\}\,}$

${\displaystyle \mathbf {(-4,2)} }$

2. ${\displaystyle \{x:-{\frac {7}{3}}\leq x\leq -{\frac {1}{3}}\}}$

${\displaystyle \mathbf {[-{\frac {7}{3}},-{\frac {1}{3}}]} }$

3. ${\displaystyle \{x:-\pi \leq x<\pi \}}$

${\displaystyle \mathbf {[-\pi ,\pi )} }$

4. ${\displaystyle \{x:x\leq 17/9\}}$

${\displaystyle \mathbf {(-\infty ,{\frac {17}{9}}]} }$

5. ${\displaystyle \{x:5\leq x+1\leq 6\}}$

${\displaystyle 4\leq x\leq 5}$
${\displaystyle \mathbf {[4,5]} }$

6. ${\displaystyle \{x:x-1/4<1\}\,}$

${\displaystyle x<1{\frac {1}{4}}={\frac {5}{4}}}$
${\displaystyle \mathbf {(-\infty ,{\frac {5}{4}})} }$

7. ${\displaystyle \{x:3>3x\}\,}$

${\displaystyle 1>x}$

${\displaystyle x<1}$

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

8. ${\displaystyle \{x:0\leq 2x+1<3\}}$

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

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

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

9. ${\displaystyle \{x:5<x{\mbox{ and }}x<6\}\,}$

This is equivalent to ${\displaystyle 5<x<6}$
${\displaystyle \mathbf {(5,6)} }$

10. ${\displaystyle \{x:5<x{\mbox{ or }}x<6\}\,}$

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.

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

11. ${\displaystyle [3,4]\,}$

${\displaystyle \mathbf {\{x:3\leq x\leq 4\}} }$

12. ${\displaystyle [3,4)\,}$

${\displaystyle \mathbf {\{x:3\leq x<4\}} }$

13. ${\displaystyle (3,\infty )}$

${\displaystyle \mathbf {\{x:x>3\}} }$

14. ${\displaystyle (-{\frac {1}{3}},{\frac {1}{3}})\,}$

${\displaystyle \mathbf {\{x:-{\frac {1}{3}}<x<{\frac {1}{3}}\}} }$

15. ${\displaystyle (-\pi ,{\frac {15}{16}})\,}$

${\displaystyle \mathbf {\{x:-\pi <x<{\frac {15}{16}}\}} }$

16. ${\displaystyle (-\infty ,\infty )}$

${\displaystyle \mathbf {\{x:x\in \Re \}} }$

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

18. ${\displaystyle |x+y|\geq |x|+|y|}$

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

19. ${\displaystyle |x+y|\leq |x|+|y|}$

true

20. ${\displaystyle 8^{1/3}\,}$

${\displaystyle (2^{3})^{1/3}=2^{1}=\mathbf {2} }$

21. ${\displaystyle (-8)^{1/3}\,}$

${\displaystyle (-2^{3})^{1/3}=-2^{1}=\mathbf {-2} }$

22. ${\displaystyle {\bigg (}{\frac {1}{8}}{\bigg )}^{1/3}\,}$

${\displaystyle ({\frac {1}{2^{3}}})^{1/3}=(2^{-3})^{1/3}=2^{-1}=\mathbf {\frac {1}{2}} }$

23. ${\displaystyle (8^{2/3})(8^{3/2})(8^{0})\,}$

${\displaystyle 8^{{\frac {2}{3}}+{\frac {3}{2}}+0}=8^{{\frac {4}{6}}+{\frac {9}{6}}}=8^{\frac {13}{6}}=(2^{3})^{\frac {13}{6}}=\mathbf {2^{13/2}} }$

24. ${\displaystyle {\bigg (}{\bigg (}{\frac {1}{8}}{\bigg )}^{1/3}{\bigg )}^{7}}$

${\displaystyle (({\frac {1}{2^{3}}})^{1/3})^{7}=((2^{-3})^{1/3})^{7}=(2^{-1})^{7}=2^{-7}={\frac {1}{2^{7}}}=\mathbf {\frac {1}{128}} }$

25. ${\displaystyle {\sqrt[{3}]{\frac {27}{8}}}}$

${\displaystyle ({\frac {27}{8}})^{1/3}=({\frac {3^{3}}{2^{3}}})^{1/3}={\frac {3^{1}}{2^{1}}}=\mathbf {\frac {3}{2}} }$

26. ${\displaystyle {\frac {4^{5}\cdot 4^{-2}}{4^{3}}}}$

${\displaystyle 4^{5-2-3}=4^{0}=\mathbf {1} }$

27. ${\displaystyle {\bigg (}{\sqrt {27}}{\bigg )}^{2/3}}$

${\displaystyle ((3^{3})^{1/2})^{2/3}=(3^{\frac {3}{2}})^{\frac {2}{3}}=3^{1}=\mathbf {3} }$

28. ${\displaystyle {\frac {\sqrt {27}}{\sqrt[{3}]{9}}}}$

${\displaystyle {\frac {(3^{3})^{1/2}}{(3^{2})^{1/3}}}={\frac {3^{\frac {3}{2}}}{3^{\frac {2}{3}}}}=3^{{\frac {3}{2}}-{\frac {2}{3}}}=3^{{\frac {9}{6}}-{\frac {4}{6}}}=\mathbf {3^{5/6}} }$

29. ${\displaystyle x^{3}+3x^{3}\,}$

${\displaystyle \mathbf {4x^{3}} }$

30. ${\displaystyle {\frac {x^{3}+3x^{3}}{x^{2}}}}$

${\displaystyle \mathbf {4x} }$

31. ${\displaystyle (x^{3}+3x^{3})^{3}\,}$

${\displaystyle \mathbf {64x^{9}} }$

32. ${\displaystyle {\frac {x^{15}+x^{3}}{x}}}$

${\displaystyle \mathbf {x^{14}+x^{2}} }$

33. ${\displaystyle (2x^{2})(3x^{-2})\,}$

${\displaystyle \mathbf {6} }$

34. ${\displaystyle {\frac {x^{2}y^{-3}}{x^{3}y^{2}}}}$

${\displaystyle \mathbf {\frac {1}{xy^{5}}} }$

35. ${\displaystyle {\sqrt {x^{2}y^{4}}}}$

${\displaystyle \mathbf {xy^{2}} }$

36. ${\displaystyle {\bigg (}{\frac {8x^{6}}{y^{4}}}{\bigg )}^{1/3}}$

${\displaystyle \mathbf {\frac {2x^{2}}{y^{\frac {4}{3}}}} }$

37. ${\displaystyle x^{2}-1\,}$

${\displaystyle x=\pm 1}$

38. ${\displaystyle x^{2}+2x+1\,}$

${\displaystyle x=-1}$

39. ${\displaystyle x^{2}+7x+12\,}$

${\displaystyle x=-3,x=-4}$

40. ${\displaystyle 3x^{2}-5x-2\,}$

${\displaystyle x=2,x=-{\frac {1}{3}}}$

41. ${\displaystyle x^{2}+5/6x+1/6\,}$

${\displaystyle x=-{\frac {1}{3}},x=-{\frac {1}{2}}}$

42. ${\displaystyle 4x^{3}+4x^{2}+x\,}$

${\displaystyle x=0,x=-{\frac {1}{2}}}$

43. ${\displaystyle x^{4}-1\,}$

${\displaystyle x=\pm i,x=\pm 1}$

44. ${\displaystyle x^{3}+2x^{2}-4x-8\,}$

${\displaystyle x=\pm 2}$

45. ${\displaystyle 4a^{2}-ab-3b^{2}\,}$

${\displaystyle (4a+3b)(a-b)}$

46. ${\displaystyle (c+d)^{2}-4\,}$

${\displaystyle (c+d+2)(c+d-2)}$

47. ${\displaystyle 4x^{2}-9y^{2}\,}$

${\displaystyle (2x+3y)(2x-3y)}$

48. ${\displaystyle {\frac {x^{2}-1}{x+1}}\,}$

${\displaystyle x-1,x\neq -1}$

49. ${\displaystyle {\frac {3x^{2}+4x+1}{x+1}}\,}$

${\displaystyle 3x+1,x\neq -1}$

50. ${\displaystyle {\frac {4x^{2}-9}{4x^{2}+12x+9}}\,}$

${\displaystyle {\frac {2x-3}{2x+3}}}$

51. ${\displaystyle {\frac {x^{2}+y^{2}+2xy}{x(x+y)}}\,}$

${\displaystyle {\frac {x+y}{x}},x\neq -y}$

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

a. Compute ${\displaystyle f(0)}$ and ${\displaystyle f(2)}$.

${\displaystyle f(0)=0}$, ${\displaystyle f(2)=4}$

b. What are the domain and range of ${\displaystyle f}$?

The domain is ${\displaystyle (-\infty ,\infty )}$; the range is ${\displaystyle [0,\infty )}$,

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, ${\displaystyle f(-1)=f(1)=1}$.

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

a. Give formulae for

i. ${\displaystyle f+g}$

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

ii. ${\displaystyle f-g}$

${\displaystyle (f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x}$.

iii. ${\displaystyle g-f}$

${\displaystyle (g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x}$.

iv. ${\displaystyle f\times g}$

${\displaystyle (f\times g)(x)=(x+2)/x}$.

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!

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.

vii. ${\displaystyle f\circ g}$

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

viii. ${\displaystyle g\circ f}$

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

b. Compute ${\displaystyle f(g(2))}$ and ${\displaystyle g(f(2))}$.

${\displaystyle f(g(2))=5/2}$; ${\displaystyle g(f(2))=1/4}$.

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.

54. Does this graph represent 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?

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

b. What is the range?

${\displaystyle {(-1/9,\infty )}}$

c. Where is ${\displaystyle f}$ continuous?

${\displaystyle {x>0}}$

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?

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

b. What is the range?

${\displaystyle {(-1,\infty )}}$

c. Where is ${\displaystyle f}$ continuous?

${\displaystyle {x>0}}$

57. Consider the following function

${\displaystyle f(x)={\frac {\sqrt {2x-3}}{x-10}}}$

a. What is the domain?

${\displaystyle {(3/2,10)\cup (10,\infty )}}$

b. What is the range?

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

c. Where is ${\displaystyle f}$ continuous?

${\displaystyle {(3/2,10)and(x>10)}}$

58. Consider the following function

${\displaystyle f(x)={\frac {x-7}{x^{2}-49}}}$

a. What is the domain?

${\displaystyle {(-\infty ,-7)\cup (-7,\infty )}}$

b. What is the range?

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

c. Where is ${\displaystyle f}$ continuous?

${\displaystyle {(-\infty ,-7)and(-7,\infty )}}$

59. Find the equation of the line that passes through the point (1,-1) and has slope 3.

${\displaystyle 3x-y=4}$

60. Find the equation of the line that passes through the origin and the point (2,3).

${\displaystyle 3x-2y=0}$