# 3.2 Domain and range  (Page 8/11)

 Page 8 / 11

## Verbal

Why does the domain differ for different functions?

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

How do we determine the domain of a function defined by an equation?

Explain why the domain of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[3]{x}\text{\hspace{0.17em}}$ is different from the domain of $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[]{x}.$

There is no restriction on $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[3]{x}\text{\hspace{0.17em}}$ because you can take the cube root of any real number. So the domain is all real numbers, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).\text{\hspace{0.17em}}$ When dealing with the set of real numbers, you cannot take the square root of negative numbers. So $\text{\hspace{0.17em}}x$ -values are restricted for $\text{\hspace{0.17em}}f\left(x\right)=\sqrt[]{x}\text{\hspace{0.17em}}$ to nonnegative numbers and the domain is $\text{\hspace{0.17em}}\left[0,\infty \right).$

When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket?

How do you graph a piecewise function?

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the $\text{\hspace{0.17em}}x$ -axis and $\text{\hspace{0.17em}}y$ -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate $\text{\hspace{0.17em}}-\infty \text{\hspace{0.17em}}$ or Combine the graphs to find the graph of the piecewise function.

## Algebraic

For the following exercises, find the domain of each function using interval notation.

$f\left(x\right)=-2x\left(x-1\right)\left(x-2\right)$

$f\left(x\right)=5-2{x}^{2}$

$\left(-\infty ,\infty \right)$

$f\left(x\right)=3\sqrt{x-2}$

$f\left(x\right)=3-\sqrt{6-2x}$

$\left(-\infty ,3\right]$

$f\left(x\right)=\sqrt{4-3x}$

$\begin{array}{l}\\ f\left(x\right)=\sqrt[]{{x}^{2}+4}\end{array}$

$\left(-\infty ,\infty \right)$

$f\left(x\right)=\sqrt[3]{1-2x}$

$f\left(x\right)=\sqrt[3]{x-1}$

$\left(-\infty ,\infty \right)$

$f\left(x\right)=\frac{9}{x-6}$

$f\left(x\right)=\frac{3x+1}{4x+2}$

$\left(-\infty ,-\frac{1}{2}\right)\cup \left(-\frac{1}{2},\infty \right)$

$f\left(x\right)=\frac{\sqrt{x+4}}{x-4}$

$f\left(x\right)=\frac{x-3}{{x}^{2}+9x-22}$

$\left(-\infty ,-11\right)\cup \left(-11,2\right)\cup \left(2,\infty \right)$

$f\left(x\right)=\frac{1}{{x}^{2}-x-6}$

$f\left(x\right)=\frac{2{x}^{3}-250}{{x}^{2}-2x-15}$

$\left(-\infty ,-3\right)\cup \left(-3,5\right)\cup \left(5,\infty \right)$

$\frac{5}{\sqrt{x-3}}$

$\frac{2x+1}{\sqrt{5-x}}$

$\left(-\infty ,5\right)$

$f\left(x\right)=\frac{\sqrt{x-4}}{\sqrt{x-6}}$

$f\left(x\right)=\frac{\sqrt{x-6}}{\sqrt{x-4}}$

$\left[6,\infty \right)$

$f\left(x\right)=\frac{x}{x}$

$f\left(x\right)=\frac{{x}^{2}-9x}{{x}^{2}-81}$

$\left(-\infty ,-9\right)\cup \left(-9,9\right)\cup \left(9,\infty \right)$

Find the domain of the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{2{x}^{3}-50x}\text{\hspace{0.17em}}$ by:

1. using algebra.
2. graphing the function in the radicand and determining intervals on the x -axis for which the radicand is nonnegative.

## Graphical

For the following exercises, write the domain and range of each function using interval notation.

domain: $\text{\hspace{0.17em}}\left(2,8\right],\text{\hspace{0.17em}}$ range $\text{\hspace{0.17em}}\left[6,8\right)\text{\hspace{0.17em}}$

domain: range:

domain: range: $\text{\hspace{0.17em}}\left[0,2\right]$

domain: $\text{\hspace{0.17em}}\left(-\infty ,1\right],\text{\hspace{0.17em}}$ range: $\text{\hspace{0.17em}}\left[0,\infty \right)\text{\hspace{0.17em}}$

domain: $\text{\hspace{0.17em}}\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right];\text{\hspace{0.17em}}$ range: $\text{\hspace{0.17em}}\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right]\text{\hspace{0.17em}}$

domain: range: $\text{\hspace{0.17em}}\left[0,\infty \right)\text{\hspace{0.17em}}$

For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation.

$f\left(x\right)=\left\{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \\ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array}$

$f\left(x\right)=\left\{\begin{array}{lll}2x-1\hfill & \text{if}\hfill & x<1\hfill \\ 1+x\hfill & \text{if}\hfill & x\ge 1\hfill \end{array}$

domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$

$f\left(x\right)=\left\{\begin{array}{c}x+1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\\ x-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x>0\end{array}$

$f\left(x\right)=\left\{\begin{array}{ccc}3& \text{if}& x<0\\ \sqrt{x}& \text{if}& x\ge 0\end{array}$

domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$

$f\left(x\right)=\left\{\begin{array}{r}\hfill \begin{array}{r}\hfill {x}^{2}\\ \hfill x+2\end{array}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{l}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\hfill \\ \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 0\hfill \end{array}$

domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$

$f\left(x\right)=\left\{\begin{array}{ccc}x+1& \text{if}& x<1\\ {x}^{3}& \text{if}& x\ge 1\end{array}$

$f\left(x\right)=\left\{\begin{array}{c}|x|\\ 1\end{array}\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<2\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 2\hfill \end{array}$

domain: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$

## Numeric

For the following exercises, given each function $f,$ evaluate $f\left(-3\right),\text{\hspace{0.17em}}f\left(-2\right),\text{\hspace{0.17em}}f\left(-1\right),$ and $f\left(0\right).$

$f\left(x\right)=\left\{\begin{array}{lll}x+1\hfill & \text{if}\hfill & x<-2\hfill \\ -2x-3\hfill & \text{if}\hfill & x\ge -2\hfill \end{array}$

$\begin{array}{cccc}f\left(-3\right)=1;& f\left(-2\right)=0;& f\left(-1\right)=0;& f\left(0\right)=0\end{array}$

For the following exercises, given each function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ evaluate $f\left(-1\right),\text{\hspace{0.17em}}f\left(0\right),\text{\hspace{0.17em}}f\left(2\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(4\right).$

$f\left(x\right)=\left\{\begin{array}{lll}7x+3\hfill & \text{if}\hfill & x<0\hfill \\ 7x+6\hfill & \text{if}\hfill & x\ge 0\hfill \end{array}$

$\begin{array}{cccc}f\left(-1\right)=-4;& f\left(0\right)=6;& f\left(2\right)=20;& f\left(4\right)=34\end{array}$

$f\left(x\right)=\left\{\begin{array}{ccc}{x}^{2}-2& \text{if}& x<2\\ 4+|x-5|& \text{if}& x\ge 2\end{array}$

$f\left(x\right)=\left\{\begin{array}{ccc}5x& \text{if}& x<0\\ 3& \text{if}& 0\le x\le 3\\ {x}^{2}& \text{if}& x>3\end{array}$

$\begin{array}{cccc}f\left(-1\right)=-5;& f\left(0\right)=3;& f\left(2\right)=3;& f\left(4\right)=16\end{array}$

For the following exercises, write the domain for the piecewise function in interval notation.

domain: $\text{\hspace{0.17em}}\left(-\infty ,1\right)\cup \left(1,\infty \right)$

$f\left(x\right)=\left\{\begin{array}{c}2x-3\\ -3{x}^{2}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x<0\\ \text{if}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x\ge 2\end{array}$

## Technology

Graph $\text{\hspace{0.17em}}y=\frac{1}{{x}^{2}}\text{\hspace{0.17em}}$ on the viewing window $\text{\hspace{0.17em}}\left[-0.5,-0.1\right]\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left[0.1,0.5\right].\text{\hspace{0.17em}}$ Determine the corresponding range for the viewing window. Show the graphs.

window: $\text{\hspace{0.17em}}\left[-0.5,-0.1\right];\text{\hspace{0.17em}}$ range:

window: range:

Graph $\text{\hspace{0.17em}}y=\frac{1}{x}\text{\hspace{0.17em}}$ on the viewing window $\text{\hspace{0.17em}}\left[-0.5,-0.1\right]\text{\hspace{0.17em}}$ and Determine the corresponding range for the viewing window. Show the graphs.

## Extension

Suppose the range of a function $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is What is the range of $\text{\hspace{0.17em}}|f\left(x\right)|?$

Create a function in which the range is all nonnegative real numbers.

Create a function in which the domain is $\text{\hspace{0.17em}}x>2.$

Many answers. One function is $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{\sqrt{x-2}}.$

## Real-world applications

The height $\text{\hspace{0.17em}}h\text{\hspace{0.17em}}$ of a projectile is a function of the time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ it is in the air. The height in feet for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ seconds is given by the function $h\left(t\right)=-16{t}^{2}+96t.$ What is the domain of the function? What does the domain mean in the context of the problem?

The domain is it takes 6 seconds for the projectile to leave the ground and return to the ground

The cost in dollars of making $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ items is given by the function $\text{\hspace{0.17em}}C\left(x\right)=10x+500.$

1. The fixed cost is determined when zero items are produced. Find the fixed cost for this item.
2. What is the cost of making 25 items?
3. Suppose the maximum cost allowed is \$1500. What are the domain and range of the cost function, $\text{\hspace{0.17em}}C\left(x\right)?$

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1