# Calculus/Precalculus/Solutions

## Convert to interval notation

1. $\{x:-4

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

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

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

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

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

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

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

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

$4\leq x\leq5$
$\mathbf{[4, 5]}$

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

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

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

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

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

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

9. $\{x:5

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

## State the following intervals using set notation

11. $[3,4] \,$

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

12. $[3,4) \,$

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

13. $(3,\infty)$

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

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

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

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

$\mathbf{\{x:-\pi

16. $(-\infty,\infty)$

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

## 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

18. $|x+y| \geq |x| + |y|$

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

19. $|x+y| \leq |x| + |y|$

true

## Evaluate the following expressions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Simplify the following

29. $x^3 + 3x^3 \,$

$\mathbf{ 4x^3 }$

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

$\mathbf{ 4x }$

31. $(x^3+3x^3)^3 \,$

$\mathbf{ 64x^9 }$

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

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

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

$\mathbf{ 6 }$

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

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

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

$\mathbf{ xy^2 }$

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

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

## Functions

52. Let $f(x)=x^2$.

a. Compute $f(0)$ and $f(2)$.

$f(0)=0$, $f(2)=4$

b. What are the domain and range of $f$?

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

c. Does $f$ have an inverse? If so, find a formula for it.

No, since $f$ isn't one-to-one; for example, $f(-1)=f(1)=1$.

53. Let $f(x)=x+2$, $g(x)=1/x$.

a. Give formulae for
i. $f+g$

$(f+g)(x)=x+2+1/x=(x^2+2x+1)/x$.

ii. $f-g$

$(f-g)(x)=x+2-1/x=(x^2+2x-1)/x$.

iii. $g-f$

$(g-f)(x)=1/x-x-2=(1-x^2-2x)/x$.

iv. $f\times g$

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

v. $f/g$

$(f/g)(x)=x(x+2)$ provided $x\ne0$. 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.

vii. $f\circ g$

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

viii. $g\circ f$

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

b. Compute $f(g(2))$ and $g(f(2))$.

$f(g(2))=5/2$; $g(f(2))=1/4$.

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.

54. Does this graph represent 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?
b. What is the range?
c. Where is $f$ continuous?

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?
b. What is the range?
c. Where is $f$ continuous?

57. Consider the following function

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

58. Consider the following function

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