Calculus/Precalculus/Solutions

< Calculus | Precalculus

Convert to interval notation

1.

\{x:-4<x<2\}\,

\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<x{\mbox{ and }}x<6\}\,

This is equivalent to

5<x<6

\mathbf {(5,6)}

This is equivalent to

5<x<6

\mathbf {(5,6)}

10.

\{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.

\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}}<x<{\frac {1}{3}}\}}

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

15.

(-\pi ,{\frac {15}{16}})\,

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

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

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.

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

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

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