How to Determine the Domain of a Function
When the domain of a function is not specified, then assume that it is the set of all possible real numbers for which the function makes sense.
ပေးထားသော function တစ်ခုအတွက် domain သတ်မှတ်ပေးထားခြင်း မရှိလျှင် function ၏ သတ်မှတ်ချက်ကို ပြေလည်စေသော အစုဝင်အားလုံးပါသည် ကိန်းစစ်အစုသည် အဆိုပါ function ၏ domain ဖြစ်သည်ဟု မှတ်ယူရမည်။
Equality of Functions
Two functions $f$ and $g$ are equal (and we write $f=g$) if and only if
1. $f$ and $g$ have the same domain, and
2. $f(c)$ = $g(c)$ for each element $c$ in the domain.
Function နှစ်ခု $f$ နှင့် $g$ ၏ domain များတူညီကြပြီး domain ထဲရှိ အစုဝင် $c$ တိုင်းအတွက် $f(c)= g(c)$ [$c$ ၏ image တူညီလျှင်] ဖြစ်လျှင် $f$ နှင့် $g$ သည် တူညီသော function များ ဖြစ်ကြသည်။
1. Determine whether each relation is a function or not. If it is a function, state the domain and range.
Show/Hide Solution
$\begin{array}{ll} \text{(a)} & \text{The relation}\ R\ \text{is a function.}\\\\ & \text{dom}\ (R ) = \{-1, 0, 1\}\\\\ & \text{ran}\ (R) = \{0, 1\}\\\\ \text{(b)} & \text{The relation}\ R\ \text{is not a function.}\\\\ \text{(c)} & \text{The relation}\ R\ \text{is not a function.} \end{array}$
2. Consider the following relations. Determine whether each relation is a function or not. If it is a function, write down the domain and range.
$\begin{array}{l} \text{(a)}\ \ \{(1,3),(2,5),(3,7),(4,9)\} \\\\ \text{(b)}\ \ \{(-2,5),(-1,3),(0,1),(-1,1)\}\\\\ \text{(c)}\ \ \{(1,3),(2,3),(3,2),(2,1)\} \\\\ \text{(d)}\ \ \{(2,4),(3,6),(4,6),(7,14)\} \\\\ \text{(e)}\ \ \{(0,0),(1,1),(3,3),(4,4)\} \\\\ \text{(f)}\ \ \{(2, a),(4, c),(5, a),(4, e)\} \end{array}$
Show/Hide Solution
$\begin{array}{l}\text{(a)}\;\;\{(1,3),(2,5),(3,7),(4,9)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{1,\ 2,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{3,\ 5,\ 7,\ 9\}\\\\\text{(b)}\;\;\{(-2,5),(-1,3),(0,1),(-1,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }-\text{1 of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(c)}\;\;\{(1,3),(2,3),(3,2),(2,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }2\text{ of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(d)}\;\;\{(2,4),(3,6),(4,6),(7,14)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{2,\ 3,\ 4,\ 7\}\\\ \ \ \ \ \text{Range}=\{4,\ 6,\ 14\}\\\\\text{(e)}\;\;\{(0,0),(1,1),(3,3),(4,4)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{0,\ 1,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{0,\ 1,\ 3,\ 4\}\\\\\text{(f)}\;\;\{(2,a),(4,c),(5,a),(4,e)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }4\text{ of domain is related to two}\\\ \ \ \ \ \text{elements }c\text{ and }e\text{.}\end{array}$
3. Let $f$ be a function from $\mathbb{R} \rightarrow \mathbb{R}$. Which of the following statements are true?
(a) If $f(x)=5-x,$ the image of -3 under $f$ is 8.
(b) If $f(x)=x^{2}+9,$ the image of -3 under $f$ is zero.
(c) If $f(x)=3 x+4,$ then $f(a)=a$ implies that $a=-2$.
(d) If $f(x)=x+3,$ there is only one value $a \in \mathbb{R}$ such that $f(a)=0$.
(e) If $f(x)=x^{2}-1,$ then there are exactly two values $a \in \mathbb{R}$ such that $f(a)=0$.
Show/Hide Solution
$\begin{array}{l}\ \ \ \ \ \ \ \ f:\mathbb{R}\to \mathbb{R}\\\\\text{(a)}\ \ f(x)=5-x\\\ \ \ \ \ f(-3)=5-(-3)=8\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(b)}\ \ f(x)={{x}^{2}}+9\\\ \ \ \ \ f(-3)={{(-3)}^{2}}+9=18\\\ \ \ \ \ \therefore \ \text{ The statements is false}\text{.}\\\\\text{(c)}\ \ f(x)=3x+4\\\ \ \ \ \ f(a)=a\\\ \ \ \ \ 3a+4=a\\\ \ \ \ \ 2a=-4\\\ \ \ \ \ a=-2\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(d)}\ \ f(x)=x+3\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ a+3=0\\\ \ \ \ \ a=-3\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(e)}\ \ f(x)={{x}^{2}}-1\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ {{a}^{2}}-1=0\\\ \ \ \ \ {{a}^{2}}=1\\\ \ \ \ \ a=\pm 1\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\end{array}$
4. Illustrate the function $f: x \mapsto x+2$ with an arrow diagram for the domain $\{3,5,7,9,10\} .$ Write down the range of $f$.
Show/Hide Solution
$\begin{array}{l}\ \ \ f:x\mapsto x+2\\\\\ \ \ \therefore \ f(x)=x+2\\\\\ \ \ \operatorname{dom}(f)=\{3,5,7,9,10\}\\\\\ \ \ f(3)=3+2=5\\\\\ \ \ f(5)=5+2=7\\\\\ \ \ f(7)=7+2=9\\\\\ \ \ f(9)=9+2=11\\\\\ \ \ f(10)=10+2=12\\\\\ \ \ \operatorname{ran}(f)=\{5,7,9,11,12\}\end{array}$
5. Let the domain of function $h: x \mapsto 0$ be $\{2,4,6,7\} .$ What is the range of $h ?$ Draw an arrow diagram for $h$.
Show/Hide Solution
$\begin{array}{l}h:x\mapsto 0\\\\\operatorname{dom}(h)=\{2,4,6,7\}\\\\\therefore \ \ h:2\mapsto 0\\\\\ \ \ h:4\mapsto 0\\\\\ \ \ h:6\mapsto 0\\\\\ \ \ h:7\mapsto 0\\\\\operatorname{ran}(h)=\{0\}\end{array}$
6. Let the domain of function $f: x \mapsto 3 x$ be the set of natural numbers less than $5 .$ State the domain and range.
Show/Hide Solution
$\begin{array}{l} f:x\mapsto 3x\\\\ \operatorname{dom}(f)=\text{the set of natural numbers less}\ \text{than 5}\text{.}\\\\ \operatorname{dom}(f)=\{1,\ 2,\ 3,\ 4\}\\\\ f:1\mapsto 3(1)=3\\f:2\mapsto 3(2)=6\\\\ f:3\mapsto 3(3)=9\\f:4\mapsto 3(4)=12\\\\ \operatorname{ran}(f)=\{3,\ 6,\ 9,\ 12\}\ end{array}$
7. Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be given by
(a) $g(x)=3-4 x$ .Find $g(1), \quad g(3), \quad g(-2), \quad g(x+3), \quad g\left(\displaystyle \frac{1}{2}\right). $
(b) $g(x)=2 x-5 .$ Find $g(3), g\left(\displaystyle\frac{1}{2}\right), g(0), g(-4), g(4) .$ If $g(a)=99,$ find $a$.
(c) $g(x)=\displaystyle\frac{x+5}{2} .$ Find the images of $3,0,-3 .$ Find $x$ if $g(x)=0$.
(d) $g(x)=3 x-1 .$ Find $x$ such that $g(x)=20$.
Show/Hide Solution
$\begin{array}{ll} \text{(a)} & g(x) = 3-4x\\ & g(1) = 3-4(1) = -1\\ & g(3) = 3-4(3) = -9\\ & g(-2) = 3-4(-2) = 11\\ & g(x+3) = 3-4(x+3) = -4x-9\\ & g \left({\displaystyle\frac{1}{2}} \right) = 3- 4\left(\displaystyle\frac{1}{2}\right) = 1\\\\ \text{(b)} & g(x) = 2x-5\\ & g(3) = 2(3)-5 = 1\\ & g \left(\displaystyle\frac{1}{2}\right) = 2 \left(\displaystyle\frac{1}{2}\right)-5 = -4\\ & g(0) = 2(0)-5 = -5\\ & g(-4) = 2(-4)-5 = -13\\ & g(4) = 2(4)-5 = 3\\ & g(a) = 99\\ & 2a-5 = 99\\ & 2a = 104\\ & a = 52 \\\\ \text{(c)} & g(x) = \displaystyle\frac{x+5}{2}\\ & g(3) = \displaystyle\frac{3+5}{2} = 4\\ & g(0) = \displaystyle\frac{0+5}{2} = \displaystyle\frac{5}{2}\\ & g(-3) = \displaystyle\frac{-3+5}{2} = 1\\ & g(x)=0\\ & \displaystyle\frac{x+5}{2} = 0\\ & x=-5\\\\ \text{(d)} & g(x) = 3x-1\\ & g(x) = 20\\ & 3x-1 = 20\\ & x = 7 \end{array}$
8. A function $f$ from $A$ to $A$, where $A$ is the set of positive integers, is given by $f(x)=$ the sum of all possible divisors of $x$.
For example $f(6)=1+2+3+6=12$
(a) Find the values of $f(2), f(5), f(13), f(18)$.
(b) Show that $f(14)=f(15)$ and $f(3) \cdot f(5)=f(15)$.
Show/Hide Solution
$\begin{array}{l}f(x)=\text{the sum of all possible divisors of }x\\\\f(2)=1+2=3\\\\f(5)=1+5=6\\\\f(13)=1+13=14\\\\f(15)=1+3+5+15=24\\\\f(14)=1+2+7+14=24\\\\\therefore \ f(15)=f(14)\\\\f(3)=1+3=4\\\\f(3)\cdot f(5)=4\times 6=24\\\\\therefore \ f(3)\cdot f(5)=f(15)\end{array}$
9. Let $A=$ the set of positive integers greater than 3 and $B=$ the set of all positive integers. Let $d: A \rightarrow B$ be a function given by $d(n)=\frac{1}{2} n(n-3)$ the number of diagonals of a polygon of $n$ sides.
(a) Find $d(6), d(8), d(10), d(12)$.
(b) How many diagonals will a polygon of 20 sides have?
Show/Hide Solution
$ \begin{array}{l}\text{(a)}\ \ \ \ A=\text{ the set of positive integers greater than 3}\\\\\ \ \ \ \ \ \ A=\{4,\ 5,\ 6,\ ...\}\\\\\ \ \ \ \ \ \ B=\text{ the set of positive all integers }\\\\\ \ \ \ \ \ \ B=\{1,\ 2,\ 3,\ ...\}\\\\\ \ \ \ \ \ \ d:A\to B\ \\\\\ \ \ \ \ \ d(n)=\displaystyle \frac{1}{2}n(n-3)\\\\\ \ \ \ \ \ d(6)=\displaystyle \frac{1}{2}\cdot 6\cdot (6-3)=9\\\\\ \ \ \ \ \ d(8)=\displaystyle \frac{1}{2}\cdot 8\cdot (8-3)=20\\\\\ \ \ \ \ \ d(10)=\displaystyle \frac{1}{2}\cdot 10\cdot (10-3)=35\\\\\ \ \ \ \ \ d(12)=\displaystyle \frac{1}{2}\cdot 12\cdot (12-3)=54\\\\\text{(b)}\ \ \ d(20)=\displaystyle \frac{1}{2}\cdot 20\cdot (20-3)=170\end{array}$
10. Determine whether $f$ and $g$ are equal functions or not. Give reason:
(a) $f(x)=x^{2}+2,\ g(x)=(x+2)^{2}$.
(b) $f(x)=x^{2},\ g(x)=|x|^{2}$.
(c) $f(x)=\displaystyle\frac{x^{2}-1}{x+1},\ g(x)=x-1$.
(d) $f(x)=\displaystyle\frac{x+2}{x^{2}-4},\ g(x)=\frac{1}{x-2}$.
Show/Hide Solution
$\begin{array}{l} \text{(a)}\ \ f(x)={{x}^{2}}+2,\;g(x)={{(x+2)}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ f(1)={{1}^{2}}+2=3\\\ \ \ \ \ g(1)={{(1+2)}^{2}}=9\\ \ \ \ \ \ \therefore \ \operatorname{ran}(f)\ne \operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(b)}\ \ f(x)={{x}^{2}},\;g(x)=|x{{|}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ \operatorname{ran}(f)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \,\,\,\ \ \ \operatorname{ran}(g)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)=\operatorname{dom}(g)\ \ \text{and }\operatorname{ran}(f)=\operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f=g\\\\ \text{(c)}\ \ f(x)=\displaystyle \frac{{{{x}^{2}}-1}}{{x+1}},\;g(x)=x-1\\ \ \ \ \ f(x)\ \text{exists only when }x+1\ne 0,\ \text{i}\text{.e,}\ x\ne -1\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{-1\},\ \operatorname{dom}(g)=\mathbb{R}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(d)}\ \ f(x)=\displaystyle \frac{{x+2}}{{{{x}^{2}}-4}},\;g(x)=\displaystyle \frac{1}{{x-2}}\\ \ \ \ \ \ f(x)\ \text{exists only when }\\ \,\ \ \ \ {{x}^{2}}-4\ne 0\\ \ \ \ \ \ {{x}^{2}}\ne 4\\ \ \ \ \ \ \therefore x\ne \pm \ 2\\ \ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{\pm \ 2\}\\ \ \ \ \ \ g(x)\ \text{exists only when }\\ \,\ \ \ \ x-2\ne 0\\\ \ \ \ \ x\ne 2\\ \ \ \ \ \ \therefore \ \operatorname{dom}(g)=\mathbb{R}\smallsetminus \{2\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\end{array}$
11. State the domain of the following functions.
(a) $f(x)=\sqrt{x-2}$.
(b) $f(x)=\displaystyle\frac{1}{2 x-1}$.
(c) $f(x)=\displaystyle\frac{4}{x-3}$.
(d) $f(x)=\displaystyle\frac{2}{x^{2}-1}$.
Show/Hide Solution
$\begin{array}{l}\text{(a)}\ \ f(x)=\sqrt{{x-2}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-2\ge 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ge 2\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ge 2,\ x\in \mathbb{R}\}\\\\\text{(b)}\ \ f(x)=\displaystyle \frac{1}{{2x-1}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ 2x-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne \displaystyle \frac{1}{2}\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \displaystyle \frac{1}{2},\ x\in \mathbb{R}\}\\\\\text{(c)}\ \ f(x)=\displaystyle \frac{4}{{x-3}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-3\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne 3\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne 3,\ x\in \mathbb{R}\}\\\\\text{(d)}\ \ f(x)=\displaystyle \frac{2}{{{{x}^{2}}-1}}\\\ \ \ \ \ f(x)\ \text{exists only when }\\\,\ \ \ \ {{x}^{2}}-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ {{x}^{2}}\ne 1\\\ \ \ \ \ \therefore x\ne \pm \ 1\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \pm \ 1,\ x\in \mathbb{R}\}\end{array}$
