Domain and Range

Part 1

Absolute value function

Definition:
For any real number $x$ the absolute value is denoted by $|x|$ and is defined as
$|x| = \begin{cases} x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x>0 \\ -x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x<0 \end{cases}$
And the graph of it is

Properties:

1) $|x| \geq 0$
2) $|x|=0 \longleftrightarrow x=0$
3) $|xy|=|x||y|$
4) $|x+y| \leq |x|+|y|$
5) $||x||=|x|$
6) $|-x|=|x|$
7) $|x-y|=0 \longleftrightarrow x=y$
8) $|x-y| \leq |x-z|+|z-y|$
9) $|\dfrac{x}{y}|=\dfrac{|x|}{|y|} \,\,\,\,\, y \neq 0$
10) $||x|-|y|| \leq |x-y|$

For finding domain and range of a function which consists of absolute value, one should use above properties.
Example:
Determine domain and range of $f(x)=\dfrac{x+2}{|x|-2}$
Solution:
$|x|-2=0 \rightarrow |x|=2 \rightarrow x=\pm 2$
Therefore
$D_f=\mathbb{R} - \lbrace \pm 2 \rbrace$
On the other hand
$f(x)=\dfrac{x+2}{|x|-2}= \begin{cases} \dfrac{x+2}{x-2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x \geq 0 \\ \\ \dfrac{x+2}{-x-2}=-1 \,\,\,\,\,\,\,\,\, x<0 \end{cases}$
So
$x \geq 0 \rightarrow y=\dfrac{x+2}{x-2} \rightarrow x=\dfrac{2(y+1)}{y-1} \geq 0$
Therefore
$\begin{cases} x \geq 0 \,\,\,\,\,\,\,\, y\in (-\infty,-1] \cup (1,+\infty) \\ \\ x<0 \,\,\,\,\,\,\,\, y=-1 \end{cases}$
$\rightarrow R_f=(-\infty,-1] \cup (1,+\infty)$
Here is the graph of $f$

Example:
Determine domain and range of $f(x)=\dfrac{1}{\sqrt{|x+1|-4}}$.
Solution:
$|x+1|-4 >0 \,\, \rightarrow|x+1|>4 \rightarrow$ $\begin{cases} x+1>4 \rightarrow x>3 \\ x+1<-4 \rightarrow x<-5 \end{cases}$
$D_f=(-\infty,-5) \cup (3,+\infty)$
Notice that
$y=\dfrac{1}{\sqrt{|x+1|-4}}>0$
On the other hand
$y^2=\dfrac{1}{|x+1|-4} \rightarrow |x+1|=\dfrac{1}{y^2}+4>4 \rightarrow \dfrac{1}{y^2}>0 \rightarrow y \in (-\infty,0)\cup(0,+\infty)$
Therefore
$y \in (\mathbb{R}-\lbrace 0 \rbrace ) \cap ( \mathbb{R} ^+ )$
So
$R_f=(0,+\infty)=\mathbb{R}^+$
Here is graph of $f$

Exercises

Determine domain and range.

1) $y=\dfrac{x}{|x-1|}$
2) $y=\dfrac{x-4}{|x|-4}$
3) $y=\dfrac{\sqrt{\sqrt{(x+1)^2}-1}}{\sqrt{|x+1|-1}}$
4) $y=\dfrac{\sqrt{(x-1)^2}}{x-1}$
5) $y=\sqrt{-|x+1|}$
6) $y=\dfrac{\sqrt{(x^2-3x+2)^2}}{\sqrt{(x-2)^2}}$
7) $y=|x-1|+|x|+|x+1|$

Exponential function

General form of an exponential function is $y=a^{u(x)}$ which $a>0$ and $u(x)$ is a function. Defining domain and range of an exponential function depends on $u(x)$.
In Special form if $a=e \simeq 2.71828\cdots$ then $y=e^{u(x)}$. For having better concept of $y=a^{u(x)}$ it can be rewritten as $y=e^{u(x)\log_e a}$. Notice that $\log_e a$ is denoted by $\ln a$. Therefore
$y=e^{u(x)\ln a}$
According to this definition $a>0$ is an enough condition for defining the exponential function if $u(x)$ is a real valued function.

Tip:
$y=e^x=\lim_{n \rightarrow \infty} (1+\dfrac{1}{n})^nx=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\,\,\,\,\, (n \in \mathbb{N})$

Example:
Determine domain and range of $f(x)=2^{-x^{-2}}$.
Solution:
Notice that if $x=0$ then the denominator of the fraction is not defined. Therefore
$D_f=\mathbb{R}-\lbrace 0 \rbrace$
Defining range:
First notice that $y>0$. On the other hand
$\log y=-\dfrac{1}{x^2}\log 2 \rightarrow x^2=-\dfrac{\log 2}{\log y} \rightarrow x=\pm \sqrt{-\dfrac{\log 2}{\log y}} \rightarrow \dfrac{\log 2}{-\log y}>0 \rightarrow$

$\log y < 0 \rightarrow y < 1 \rightarrow y < 1$ and $y > 0 \rightarrow 0 < y < 1$
So
$R_f=(0,1)$
The graph of this function is

Example:
Determine domain and range of $f(x)=3^{-x}$.
Solution:
It is clear that $D_f=\mathbb{R}$. On the other hand
$y=\dfrac{1}{3^x} \rightarrow 3^x=\dfrac{1}{y} \rightarrow x=\log_3 \dfrac{1}{y}$
$R_f= \lbrace y| y \in \mathbb{R},\dfrac{1}{y}>0 \rbrace = \lbrace y \in \mathbb{R} | y>0 \rbrace =(0,+\infty)$
$R_f=\mathbb{R^+}$
Graph of $f$ is

Exercises

Determine domain and range.

1) $y=e^{-\dfrac{1}{\sqrt{x-\lfloor x \rfloor}}}$
2) $y=3^{\dfrac{\sqrt{8}}{2}}$
3) $y=5^{-x^2}$
4) $y=5^{\lfloor x \rfloor + \lfloor -x \rfloor}$
5) $y=3^{3^{\log_3 x}}$
6) $y=3^{\dfrac{x^3-x^2}{x^2-x^3}}$
7) $y=2^{\dfrac{\sqrt{9-x^2}}{\sqrt{x^2-9}}}$

Logarithmic function

General form of a logarithmic function is
$y=\log_a A(x)$
which
$A(x)>0\,\,\,\,\,\,\ a \neq 1 \,\,\,\,\,\,\ a>0$
Therefore for defining domain of a logarithmic function one should just check the above constraints.
Example:
Determine domain and range of $f(x)=\log \dfrac{x-2}{x+2}$.

Solution:
First let's determine domain
$D_f=\lbrace x \in \mathbb{R}| \dfrac{x-2}{x+2}>0 \rbrace$
$D_f=(-\infty, -2) \cup (2,+\infty)$
Now for defining the range, one should follow next steps
$\dfrac{x-2}{x+2}=10^y \rightarrow x-2=10^y \times x +2 \times 10^y$
$\rightarrow x=\dfrac{2(1+10^y)}{1-10^y} , x \in D_f \rightarrow \dfrac{2(1+10^y)}{1-10^y}>2 \,\,\,$ or $\,\,\, \dfrac{2(1+10^y)}{1-10^y}<-2$
So
$\dfrac{10^y}{1-10^y}<0\,\,\,$ or $\,\,\, \dfrac{1}{1-10^y} < 0$
$\rightarrow 1-10^y=0 \rightarrow 10^y=1 \rightarrow y=0$
Therefore
$R_f=(-\infty,0) \cup (0,+\infty)= \mathbb{R}- \lbrace 0 \rbrace$
Graph of $f$ is

Example:
Determine domain and range of $f(x)=\log \dfrac{1}{x}$.

Solution:
First let's simplify the function.
$f(x)=\log \dfrac{1}{x}=-\log x$
Now
$D_f=\lbrace x \in \mathbb{R}|x>0 \rbrace =(0,+\infty)= \mathbb{R^+}$
On the other hand
$y=-\log x \rightarrow \log x=-y \rightarrow x=10^{-y} \,\,\, x>0 \rightarrow y \in \mathbb{R} \rightarrow R_f=\mathbb{R}$
Graph of $f$ is

Example:
Determine domain and range of $f(x)=\log \sqrt{1-x^2}$.

Solution:
Notice
$D_f=\lbrace x|x\in \mathbb{R}, 1-x^2>0 \rbrace = \lbrace x \in \mathbb{R} | x^2<1 \rbrace$
$=\lbrace x \in \mathbb{R} | -1 < x < 1 \rbrace = (-1,1)$
Also
$y=\log \sqrt{1-x^2} \rightarrow \sqrt{1-x^2}=10^y \rightarrow 1-x^2=10^{2y}$

$\rightarrow x^2=1-10^{2y} \rightarrow x=\pm \sqrt{1-10^{2y}} \rightarrow R_f=\lbrace y | y \in \mathbb{R}, 1-10^{2y} \geq 0 \rbrace$

$= \lbrace y \in \mathbb{R}|10^{2y} \leq 1 \rbrace = \lbrace y \in \mathbb{R}| y \leq 0 \rbrace=(-\infty,0]$

Exercises

Determine domain and range.

1) $y=\sqrt{\log_{\dfrac{1}{2}} {(9-x^2)}}$
2) $y=\sqrt{\log(\lfloor \dfrac{10x+10}{x^2-1} \times \dfrac{x^2-1}{x+1} \rfloor)+\cos x}$
3) $y=\log (\dfrac{x^2-1}{x-1})$
4) $y=\log (\dfrac{1-x^2}{x^2-1})$
5) $y=\dfrac{2}{\log x}$
6) $y=\sqrt{\log_{\dfrac{1}{2}} x}$
7) $y=\sqrt[3]{\log_2 \sqrt{x^2}}$
8) $y=\log(\dfrac{2-x}{x+2})$
9) $y=\log_x x^2$

Trigonometric function

Functions like $f(x)=\sin x, g(x)=\cos x, h(x)=\tan x, k(x)=\cot x$ are called trigonometric function. Domain of $f(x)=\sin x$ and $g(x)=\cos x$ is all the real numbers $\mathbb{R}$. On the other hand domain of $h(x)=\tan x$ and $k(x)=\cot x$ are as follow:
$h(x)=\tan x=\dfrac{\sin x}{\cos x}, \cos x=0 \rightarrow x=k\pi+\dfrac{\pi}{2} \rightarrow$

$D_h=\mathbb{R}-\lbrace x|x=k\pi+\dfrac{\pi}{2}, k \in \mathbb{Z} \rbrace$
$h(x)=\cot x=\dfrac{\cos x}{\sin x}, \sin x=0 \rightarrow x=k\pi \rightarrow$

$D_k=\mathbb{R}-\lbrace x|x=k\pi, k \in \mathbb{Z} \rbrace$
Also notice that $-1 \leq \sin x \leq 1$ and $-1 \leq \cos x \leq 1$. Therefore
$R_f=[-1,1] \,\,\,\,\,\, R_g=[-1,1]$
The range of $h(x)=\tan x$ and $k(x)=\cot x$ is all the real numbers $\mathbb{R}$.
Example:
Determine domain and range of $f(x)=\sin x+\cos x$.

Solution:
Domain of $\sin x$ and $\cos x$ is all the real numbers, therefore the domain of
$f(x)=\sin x+\cos x$
is all the real numbers, too. Now for finding the range of $f$, one should know the formula which is given below
$\sin x +\cos x=\sqrt{2}\cos (x-\dfrac{\pi}{4})$
So
$f(x)=\sin x+\cos x=\sqrt{2}\cos (x-\dfrac{\pi}{4})$
On the other hand
$-1 \leq \cos (x-\dfrac{\pi}{4}) \leq 1$
Therefore
$-\sqrt{2} \leq \sqrt{2} \cos (x-\dfrac{\pi}{4}) \leq \sqrt{2} \rightarrow -\sqrt{2} \leq y \leq \sqrt{2} \rightarrow$
$R_f=[-\sqrt{2},\sqrt{2}]$
Graph of $f$ is

Example:
Determine domain and range of $f(x)=\sin \pi x+\cos \pi x +\sqrt{-\sin^4 \pi x}$.

Solution:
Notice that $-\sin^4 \pi x \geq 0$ and also $\sin^4 \pi x \geq 0$, therefore
$\sin^4 \pi x = 0 \rightarrow \sin \pi x=0 \rightarrow \pi x=k \pi \rightarrow x=k \in \mathbb{Z}$
So
$D_f=\mathbb{Z}$
According to $D_f=\mathbb{Z}$, one can rewritten the function as
$f(x)=\cos \pi x=\pm 1$
Now it is clear that
$R_f= \lbrace \pm 1 \rbrace$

Example:
Determine domain and range of $f(x)=\sin (\log (\log x))$.

Solution:
$D_f= \lbrace x| x \in \mathbb{R}; \log x>0,x>0 \rbrace$
$= \lbrace x| x\in \mathbb{R}, x>1,x>0 \rbrace =(1,+\infty)$
Also notice that
$|\sin (\log (\log x))| \leq 1 \rightarrow |y| \leq 1 \rightarrow -1 \leq y \leq 1$
So
$R_f=[-1,1]$
Graph of $f$ is

Inverse function

Definition:
Let $f$ be a function whose domain is $D_f$. The function $f$ is injective if and only if for all $x_1$ and $x_2$ in $D_f$, if $f(x_1)=f(x_2)$, then $x_1=x_2$.
$\forall x_1,x_2 \in D_f \,\,\, f(x_1)=f(x_2) \rightarrow x_1=x_2$

Definition:
Let $f$ be a function whose domain is $D_f$ and whose range is $R_f$. Then $f$ is invertible if there exists a function $g$ with domain $R_f$ and range $D_f$ such that
$f(x)=y \leftrightarrow g(y)=x$
If $f$ is invertible, the function $g$ is unique. That function $g$ is called the inverse of $f$, and usually denoted as $f^{-1}$.

Notice that $f$ is invertible if and only if $f$ is injective. For example the inverse of function $f=\lbrace (-1,4),(1,3),(2,1),(3,4) \rbrace$ which is $R=\lbrace (4,-1),(3,1),(1,2),(4,3) \rbrace$ is not a function, because $f$ is not invertible.
Example:
If $f=\lbrace (1,3),(-1,5),(3,1) \rbrace$, then determine domain and range of $f^{-1}$.
Solution:
It is clear that $f$ is injective and therefore it is invertible. Also
$D_f= \lbrace 1,-1,3 \rbrace \,\,\,\,\,\,\, R_f=\lbrace 1,3,5 \rbrace$
Therefore domain and range of $f^{-1}$ are
$D_{f^{-1}}=R_f=\lbrace 1,3,5 \rbrace \,\,\,\,\,\,\, R_{f^{-1}}=D_f=\lbrace -1,1,3 \rbrace$
So
$f^{-1}=\lbrace (3,1),(5,-1),(1,3) \rbrace$

Example:
Determine domain and range of $g(x)=\arccos x$.
Solution:
It is clear that $g$ is the inverse of $f(x)=\cos x$. $f$ is an injective function in
$\cdots \,\,\,$or$\,\,\,[-2\pi,\pi]\,\,\,$or$\,\,\,[-\pi,0]\,\,\,$or$\,\,\,[0,\pi]\,\,\,$or$\,\,\,[\pi,2\pi]\,\,\,$or$\,\,\,\cdots$
Therefore each one of above intervals can be selected as the range of $g$.
On the other hand
$D_g=R_f=[-1,1]$
Graph of $f$ is

Exercises

Determine domain and range.

1) $y=2\sin (4x)$
2) $y=\sin (2x)+\cos(2x)$
3) $y=5\cos (\dfrac{x^3-1}{x^2+x+1})$
4) $y=\cos (\dfrac{-x^5}{x^4})$
5) $y=2\tan(\sqrt{\dfrac{x-1}{1-x}})$
6) $y=3\tan x-4\cot x$
7) $y=\tan \sqrt[4]{x^3} \times \cot \sqrt[4]{x^3}$
8) $y=\sin (\sqrt{\sin \sqrt{\dfrac{x-2}{2-x}}})$
9) $y=\cos (\log_2 (\log_3 x))$
10) $y=\sqrt[6]{\dfrac{\cos^4 x}{\sin^4 x-\sin^2 x}}$

Composition of Functions

Composition of two functions like $f$ and $g$ is denoted as $f \circ g$. In general domain of $f \circ g$ is
$D_{f \circ g} = \lbrace x \in D_g |g(x) \in D_f \rbrace$
It is clear that one can finds domain and range of $f \circ g$, after compositing $f$ and $g$.
Example:
If $f=\lbrace (2,1),(3,5),(1,6) \rbrace$ and $g=\lbrace (1,3),(2,4),(4,1) \rbrace$, then determine domain and range of $f \circ g$.
Solution:
Therefore
$f \circ g=\lbrace (1,5),(4,6) \rbrace$
Now it is obvious that
$D_{f \circ g} = \lbrace 1,4 \rbrace \,\,\, R_{f \circ g}= \lbrace 5,6 \rbrace$

Example:
Let $g(x)=\dfrac{1}{x}$ and $f(x)=10^{-10 ^{\log(\log x)}}$. Determine domain and range of $f \circ g$ and $g \circ f$.
Solution:
First let's determine domain of $f$ and $g$
$D_f=\lbrace x | x\in \mathbb{R}, \log x>0 \rbrace=\lbrace x \in \mathbb{R}|x>1 \rbrace = (1,+\infty)$
$D_g=\mathbb{R}-\lbrace 0 \rbrace$
If $f$ is simplified, then
$f(x)=\dfrac{1}{x}$
Therefore
$D_{f \circ g}=\lbrace x \in D_g | g(x) \in D_f \rbrace = \lbrace x \in \mathbb{R}-\lbrace 0 \rbrace|\dfrac{1}{x}>1 \rbrace=(0,1)$
$D_{g \circ f} = \lbrace x \in D_f|f(x) \in D_g \rbrace= \lbrace x \in (1,+\infty)| \dfrac{1}{x}\in \mathbb{R}-\lbrace 0 \rbrace \rbrace=(1,+\infty)$
On the other hand
$(f\circ g)_{(x)} = f(g(x))=f(\dfrac{1}{x})=10^{-10^{\log(\log(\dfrac{1}{x}))}} \rightarrow y=(f \circ g)_{(x)}=$
$10^{-\log \dfrac{1}{x}}=10^{\log x} \rightarrow y=(f \circ g)_{(x)}\,\,\, x \in (0,1) \rightarrow 0 < y < 1 \rightarrow R_{f \circ g}=(0,1)$
Now for finding the range of $g \circ f$, notice that
$Z=(g\circ f)_{(x)}=x \rightarrow x=Z\in (1,+\infty) \rightarrow Z>1 \rightarrow R_{g \circ f}=(1,+\infty)$
Graph of $f$ is
Graph of $g$ is
Graph of $f \circ g$ is
Graph of $g \circ f$ is

Example:
If $f(x)=x-1$ and $(f \circ g)_{(x)}=\dfrac{1}{x-1}$, then determine domain and range of $g \circ f$.
Solution:
First let's define $g \circ f$
$f(x)=x-1 \rightarrow f(g(x))=g(x)-1 \rightarrow (f \circ g)_{(x)}=g(x)-1 \rightarrow \\ \dfrac{1}{x-1}=g(x)-1 \rightarrow g(x)=\dfrac{x}{x+1}$
So
$y=(g \circ f)_{(x)}=g(f(x))=\dfrac{f(x)}{f(x)-1}=\dfrac{x-1}{x+1}$
Therefore
$D_{g \circ f}=\lbrace x|x \in \mathbb{R}, x \neq 2 \rbrace \rightarrow D_{g \circ f}=\mathbb{R}-\lbrace 2 \rbrace$
Also
$y=\dfrac{x-1}{x-2} \rightarrow x=\dfrac{2y-1}{y-1}$
$R_{g \circ f}=\lbrace y | y \in \mathbb{R}, y \neq 1 \rbrace \rightarrow$
$R_{g \circ f}=\mathbb{R}-\lbrace 1 \rbrace$
Graph of $f$ is
Graph of $f \circ g$ is
Graph of $g$ is
Graph of $g \circ f$ is

Exercises

1) If $f(x)=2^{\log_2 x}$ and $g(x)=\dfrac{x-1}{x^2-x}$, then determine domain and range of $f \circ g$.
2) If $(g \circ f)_{(x)}=\sqrt{x}$ and $f(x)=\dfrac{4}{x}$, then determine domain and range of $f \circ g$.
3) If $f(x)=2x$ and $(f \circ g)_{(x)}=\dfrac{1}{x-1}$, then determine domain and range of $g \circ f$.
4) If $3f(-x)+4f(x)=\dfrac{1}{x}$, then determine domain and range of $f$.

Important inequalities

In this part some important inequalities will be discussed which can make the process of finding range easier.

Inequality 1:
If $u>0$, then
$u+\dfrac{1}{u} \geq 2$
and $u+\dfrac{1}{u}=2$ if and only if $u=1$.

If $u<0$, then
$u+\dfrac{1}{u} \leq -2$
and $u+\dfrac{1}{u}=-2$ if and only if $u=-1$.
Example:
Determine domain and range of $f(x)=\dfrac{7x^6}{x^{12}+1}$.
Solution:
It is clear that $D_f=\mathbb{R}$. Now notice that $f(0)=0$. So let's suppose $x\neq 0$, then
$f(x)=\dfrac{7x^6}{x^{12}+1}=\dfrac{7}{\dfrac{x^{12}+1}{x^6}}=\dfrac{7}{x^6+\dfrac{1}{x^6}}$
Becuase
$f(0)=0$ for all $x \in \mathbb{R}, \dfrac{7x^6}{x^{12}+1} \geq 0, \forall x \in \mathbb{R}, x^6+\dfrac{1}{x^6} \geq 2$
So
$0 \leq f(x) \leq \dfrac{7}{2} \rightarrow R_f=[0,\dfrac{7}{2}]$
Graph of $f$ is

Inequality 2:
$\forall a,b,k,x \in \mathbb{R} \,\,\, -\sqrt{a^2+b^2} \leq a \sin kx+b\cos kx \leq \sqrt{a^2+b^2}$

Example:
Determine domain and range of $y=\sin ^6 kx+\cos ^6 kx$.
Solution:
It is clear that $D_f=\mathbb{R}$ and $y>0$. On the other hand
$\sin^6 kx \leq \sin^2 kx$

$\cos ^6 kx \leq \cos ^2 kx$
Therefore
$\sin^6 kx+\cos^6 kx \leq \sin^2 kx+\cos ^2 kx=1 \rightarrow \sin ^6 kx+\cos ^6 kx \leq 1 \rightarrow$
$y \leq 1, y > 0 \rightarrow 0 < y \leq 1$
Also notice
$y=\sin ^6 kx+\cos^6 kx=1-\dfrac{3}{4}\sin ^2 2kx \,\,\, -1 \leq \sin 2kx \leq 1$
$\rightarrow \sin 2kx= \pm 1 \rightarrow y=\dfrac{1}{4} , \sin 2x=0 \rightarrow y=1$
$\rightarrow \dfrac{1}{4} \leq y \leq 1 \rightarrow R_f=[\dfrac{1}{4},1]$

Part 1

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