Lesson Objective:

Dear learner,

By the end of this lesson, you will be able to:

• Define a function
• Discuss about absolute value functions
• Identify power functions from exponential functions
• Solve problems given on greatest value functions
• Compare domain and range of signum functions with other piece wisely defined functions
• Draw the graph of each type of functions

Key Terms

• Function
• Vertical line test
• Power function
• Absolute Value
• Absolute value function
• Signum function
• Floor function

Brainstorming Activity

Dear learner, discuss the difference that you see in the following relations $R_{1}$ and $R_{2}$?

3.1. Functions

Definition :Let A and B be nonempty sets. A function from  set A to set B  is a relation from A to B with the condition that for every element in the domain, there exists a unique(only one) image in the codomain ( set B). 

• If the function is called  𝑓, we write  𝑓:𝐴→𝐵. Given 𝑥∈𝐴, its associated element in B is called its image under 𝑓.
• We denote it 𝑓(𝑥), which is pronounced as “ 𝑓 of 𝑥.” or ” the functional value of x”
• Domain of $f(x)$ = A and Range of $f(x)$ is thwe subset of B

Example 1

Identify whether the following relations are functions or not?

A. $R= \ \{(1,2),(3,4),(5,4),(9,3),(6,5)\}$

B. $R =\ \{(0,3),(1,4),(2,5),(6,7),(0,8)\}$

C. $R = \ \{(x,y): x is an age of y \}$

D. $R = \ \{(x,y): x,y \in \mathbb{R}, y = x^{2}\}$

Solution:

A. $R= \ \{(1,2),(3,4),(5,4),(9,3),(6,5)\}$ is a function because every member in the domain of R is mapped on to a single element in its range. or no two elements from the second set are paired with one element in the domain set.

B. $R =\ \{(0,3),(1,4),(2,5),(6,7),(0,8)\}$ is not a function because lets see $0 \in Domain of R$ is is exactely paired with two elements 3 and 8 in the second set.

C. $R = \ \{(x,y): x is an age of y \}$ . Lets see the expression x is an age of y, take y be any two children with the same age say 10 years . according to the expression x = 10 is paired with two y values child 1 and child 2. n

Therefore the above relation R is not a function.

D.$R = \ \{(x,y): x,y \in \mathbb{R}, y = x^{2}\}$ , according to the definition on the above let (x,y) and (x,z) be two ordered pairs in R then

$y = x^{2}$ and $z = x^{2}$ implies $y = z$. therefore R is a function.

Vertical line test: A relation R is a function if and only if any vertical line intersects graph of R at most once.

Example 2

Identify the following are function graphs or not?

A.

B.

C.

3.2. Power Function and Their Graphs

Definition:  a function which can be written in the form of$f(x) = ax^{r}$, where $ a, r \in \mathbb{R}$.

$f(x) = ax^{r}$ has a variable base and constant exponenet

Example 3:

which of the following functions are power functions

a. $f(x) = 3x^{2}$

b. $g(x) = 2^{x}$

c. h(x) = $x^{\frac{4}{3}}$

d. $n(x) = \frac{2}{3^{-\frac{3}{2}}}$

Solution:

$f(x)$ , $h(x)$ and $n(x)$ are power functions but $g(x)$ is not, it is an exponenetial function

Basic Properties of Power Functions

Example 4: Find the Domain , range and draw graph of the following power functions

• a). $f(x) = x^{\frac{1}{3}}$
• b). $f(x) = x^{-\frac{3}{2}}$
• c). $f(x) = x^{\frac{2}{3}}$

Solution:

A. lets see some properties of $f(x) = x^{\frac{1}{3}}$: for any real number x the function is defined this is shown by doing the following table:

x	-3	-1	-1/2	0	1	2/5	2
$f(x) = x^{\frac{1}{3}}$	$\sqrt[3]{-3}$	-1	$\sqrt[3]{-0.5}$	0	1	$\sqrt[3]{0.4}$	$\sqrt[3]{2}$

There fore Domain = $\mathbb{R}$ and Range = $\mathbb{R}$

The graph of the function is given below where it can easily show the nature of the graph as the values of x is expanding:

fig 3.4 graph of $f(x) = x^{\frac{1}{3}}$

b. $f(x) = x^{-\frac{3}{2}}$

Domain of f(x) = $\{x : x \in \mathbb{R}, x > 0\}$ = $(0, \infty )$

Range od f(x) = $\{x : x \in \mathbb{R}, x > 0\}$ = $(0, \infty )$

The graph of the function is given as follows:

fig.3.5. graph of $f(x) = x^{-\frac{3}{2}}$

c. $f(x) = x^{\frac{2}{3}}$

Domain of f(x) = $\mathbb{R}$

Range of f(x) = $ [0, \infty)$ 

The graph of the function looks like given below

fig.3.6. graph of $f(x) = x^{\frac{2}{3}}$

In general the graph of some power functions and their properties are given below

i. $f(x) = x^{\frac{m}{n}}$, m = even and n=odd, (m <  n)  

Domain = $\mathbb{R}$ and Range =  $ [0, \infty)$    

The graph is given below

fig. 3.7. $f(x) = x^{\frac{m}{n}}$, m = even and n=odd, (m <  n)  

ii.$f(x) = x^{\frac{m}{n}}$, m = odd and n=even, (m < n) or (m > n)

Domain =  $ [0, \infty)$ and Range =  $ [0, \infty)$

Graph:

fig . 3.8. $f(x) = x^{\frac{m}{n}}$, m = odd and n=even, (m < n) or (m > n)

iii. $f(x)=x^{-\frac{1}{n}}$, where n = odd

Domain = $\mathbb{R}_/{0}$

Range = $\mathbb{R}_/{0}$

Graph

fig. 3.9. graph of$f(x)=x^{-\frac{1}{n}}$, where n = odd

iv. $f(x)= x^{\frac{1}{n}}$, where n = even

Domain =  $ (0, \infty)$ and Range =  $(0, \infty)$

Graph:

fig. 3.10. graph of $f(x)= x^{\frac{1}{n}}$, where n = even

v. $f(x) = x^{\frac{m}{n}}$, m = even and n=odd,

Domain = $\mathbb{R}/ \{0\}$

Range = $(0, \infty)$

Graph:

fig.3.11. graph of $f(x) = x^{\frac{m}{n}}$, m = even and n=odd,

https://www.geogebra.org/m/tk2abnev

https://www.geogebra.org/m/VSxSDAty

3.3. Modulus ( Absolute value ) Function

Definition: the absolute value of a real number a denoted by $|a|$ is anon negative real number defined by:

$|a| =
\begin{cases}
a & \text{if } a > 0 \\
0 & \text{if } a = 0 \\
-a & \text{if } a < 0
\end{cases}$

Example 5:

Evaluate the following absolute value problems

A. $|-8| $ b). $|0.25|$ c). $|0|$

Solution:

A. $|-8|$ = -(-8)= 8, since -8 is less than zero

B. $|0.25|$ = 0.25, since 0.25 is greater than zero

C. $|0|$ = 0

Properties of Absolute Value

For any real numbers A and B:

i. |a| = |-a|

ii.$ |a| \ge 0$

iii. |ab| = |a||b|

iv. $|\frac{a}{b}| = \frac{|a|}{|b|}$, where $b\ne 0$

v. $|a + b| \le |a| +|b|$ ( triangular property)

vi. $|a| \ge a$

vii. $|x|$ = a if and only if x = a or x = -a

Example 6 :

Solve the following questions

a. |x| = 9

b. |x+2| = 3

c. |x| +5 = 3

Solution:

A. |a| = 9, if a >0 then x = 9 and if a < 0, then x = -9 but a cannot be zero.

solution={-9,9}

B. |x + 2| = 3,

if x +2 > 0, x +2 = 3 implies x = 3-2 = 1

if x+2 cannot be zero since $0 \ne3$

if x+2 < 0, then -(x+2) = 3, implies x+2 = -3, then x = -3-2 = -5

solution = {-5,1}

C. |x| +5 = 3, lets re write the equation as |x| = 3-5 implies |x| = -3, here it shows that this question has no solution by property ii, hence $|x| \ge 0$

Modulus Function

Definition: A real function $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)$ = $\begin{cases}\ x & \text{if } x > 0, \\ 0 & \text{if} x = 0,\\-x & \text{if } x < 0. \end{cases}$, for every $x \in \mathbb{R}$ is the modulus function.

Domain = $\mathbb{R}$

Range = $[0, \infty)$

Graph : the graph of $f(x) = |x| $ is a continues curve passing through (0,0) and has also a sharp corner at (0,0) too. It is a symmetric graph with the line $x = 0$ ( y-axis).

fig 3.12. Graph of $f(x) = |x|$

Example 7: Find domain, range and sketch the graph of the following functions

A. $f(x) = |x+1|$

b. $f(x) = 2- |x|$

c. $f(x) = x-|x|$

d. $f(x) = ln |x|$

Solution:

A). $f(x) = |x+1|$, as $f(x) = |x+1|$ is defined in the set of real numbers then $|x+1| \ge 0$, therefore

$f(x) = |x + 1| =
\begin{cases}
x + 1 & \text{if } x > -1 \\
0 & \text{if } x = -1 \\
-(x + 1) & \text{if } x < -1
\end{cases}$

Domain = $\mathbb{R}$

Range = $[0, \infty)$

Graph of $f(x) = |x+1|$, is a continoues curve having a sharp corner at (-1,0)

fig 3.13. Graph of $f(x) = |x+1|$

B). $f(x) = 2- |x|$, the function $f(x) = 2- |x|$ is defined in the set of all real numbers so that $f(x) = 2- |x| \le 2$

implies that :

$f(x) = 2 – |x| =
\begin{cases}
2 – x & \text{if } x \geq 0 \\
2 + x & \text{if } x < 0
\end{cases}$

Domain = $\mathbb{R}$ and Range = $(-\infty, 2]$

The graph of f(x) has a continoues curve opens down ward with sharp corner at (0,2)

fig. 3.14 Graph of $f(x) = 2- |x|$

C). $f(x) = x-|x|$, this function is defined in the set of real numbers and :

https://www.geogebra.org/m/wqq85pvt

$f(x) = x – |x| =
\begin{cases}
x – x = 0, & \text{if } x \geq 0 \\
-x – x = -2x, & \text{if } x < 0
\end{cases}$

Domain = $\mathbb{R}$ and Range = $(-\infty, 0]$

Graph of $f(x)$ is a continoues curve opens down ward with sharp corner at (0,0)

fig .3.15 Graph of $f(x) = x-|x|$

D). $f(x) = ln |x|$, the function is defined in the set of real numbers except zero like:

$f(x) = \ln|x| =
\begin{cases}
\ln x, & \text{if } x \neq 0 \\
\text{does not exist}, & \text{if } x = 0
\end{cases}$.

Domain = $\mathbb{R} – {0}$ and Range = $\mathbb{R} – {0}$.

Graph of f(x) is a curve expanding down ward indefinitely as the value of x is getting closer to zero:

fig3.16. Graph of $f(x) = |ln |x|$

https://www.geogebra.org/m/eenf6zhe

3.4. Signum Function  

Definition:

$~\text{The real function}~f:\mathbb{R}\rightarrow\mathbb{R},~\text{defined by}~f(x)=\text{sgn}x=
\begin{cases}
1,&\text{if}~x>0 \\
0,&\text{if}~x=0 \\
-1,&\text{if}~x<0
\end{cases}
~\text{is the signum function.}$

Domain of $sgn~ x = \mathbb{R}$ and Range of $sgn ~x = \{-1, 0, 1\}.$

The function $f(x) = sgn ~x$ is a piece wisely defined function so that we can define this function using Absolute value function as:

$\operatorname{sgn}(x) = \begin{cases}
\dfrac{|x|}{x} & \text{for } x \neq 0 \\
0 & \text{for } x = 0
\end{cases}$.

Graph of $sgn ~x$ are two horizontal straight lies discontinued at $x =0$ , one is marked on the first quadrant while the other is lying on the third quadrant:

Fig . 3.17. Graph of $f(x) = sgn ~x$

Examples 8:

Evaluate the following

a. $sgn (3.452)$

b. $sgn(-0.00001)$

c. $sgn (e)$

Solution:

a. sgn(3.452) = 1, since 3.452 > 0.

b. sgn(-0.000001) = -1, since -0.000001 <0

c. sgn(e) = 1, since} e = 2.713….. >0.

Example 9 :

Determine Domain , Range and draw graph of the following questions

a. $f(x) = 3 sgn x$ b. $g(x) = 2- sgn x $ c. $h(x) = x sgn x$

Solution:

a. $f(x) = 3 sgn x$, the function is re- defined as:

$f(x)=3\text{sgn}x=
\begin{cases}
3,&\text{if}~x>0 \\
0,&\text{if}~x=0 \\
-3,&\text{if}~x<0
\end{cases}$

Domain = $\mathbb{R}$ and Range = {-3, 0, 3}

The graph of the function is a straight line discontinoues at x = 0

fig 3.18. grapg of f(x) = 3 sgn x

b). $g(x) = 2- sgn x$

$G(x)=2-\text{sgn}x=2-
\begin{cases}
1,&\text{if}~x>0 \\
0,&\text{if}~x=0 \\
-1,&\text{if}~x<0 \end{cases}= \begin{cases} 2-1,&\text{if}~x>0 \\
2-0,&\text{if}~x=0 \\
2-(-1),&\text{if}~x<0 \end{cases}= \begin{cases} 1,&\text{if}~x>0 \\
2,&\text{if}~x=0 \\
3,&\text{if}~x<0
\end{cases}$

Domain = $\mathbb{R} $ and Range = {1, 2, 3}

The graph of g(x) is shown below:

Fig 3.19. Graph of $g(x) = 2 – sgn x$

C). $h(x) = x sgn x$:

Domain = $\mathbb{R}$ and Range = $[0,\infty)$

Graph of h(x) is shown below

fig. 3.20 . Graph of h(x) = x sgnx

https://www.geogebra.org/m/umuxztdd

3.5. The Greatest Integer(floor or step ) Function

Definition: The real function $f \colon \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\lfloor x \rfloor$ is the greatest integer lesthan or equal to x, is said to be the greatest integer function.

Example 10: evaluate the following

a $\lfloor 0.04\rfloor$ b. $\lfloor -2.001 \rfloor$ c. $\lfloor 4.999 \rfloor$

Solution:

a. $\lfloor 0.04\rfloor$ = 0 since the greatest integer lessthan or equal to 0.04 is 0.

b. $\lfloor -2.001 \rfloor$ = -3, since the greatest integer lessthan or equal to -2.001 is -3

c. $\lfloor 4.999 \rfloor$ = 4, since the greatest integer less than or equal to 4.999 is 4.

Domain $f(x) = \lfloor x \rfloor$ = $\mathbb{R}$

Range $f(x) = \lfloor x \rfloor$ = $\mathbb{Z}$

Graph : the value of x is given using intervals since there are infinite real numbers between [-3,-2)

we use a chart containing intervals we can do as follows:

The graph of f(x) is a step like graph where the dash line represents the floor( integer) representing an interval

Fig .3.20. Graph of $f(x) = \lfloor x \rfloor$

Example 11: Draw the graph and determine domain and range of the following questions

A. $f(x) = \lfloor x+1 \rfloor$ b). $g(x) = 2 + \lfloor x \rfloor $ c). $h(x) = 2\lfloor x \rfloor$

Solution:

A. $f(x) = \lfloor x+1 \rfloor$

lets use chart to see how the function behaves

Therefore : Domain = $\mathbb{R}$ and Range = $\mathbb{Z}$

Graph of f(x) is drawn by moving the graph of $\lfloor x \rfloor$ one step up:

Fig 3.21. Graph of $f(x) = \lfloor x+1 \rfloor$

B). $g(x) = 2 + \lfloor x \rfloor $

The function is defined in the set of all real numbers so that using a chart help you to see in detail:

as you have seen from the chart:

Domain = $\mathbb{R}$ and Range = $\mathbb{Z}$

The graph of g(x) is drawn by moving grapg of $\lfloor x \rfloor$ two steps up.

C). $h(x) = 2\lfloor x \rfloor$

The function is defined in the set of all real numbers and using a chart makes clear about the domain and range as welas the graph of the function

hence, Domain = $\mathbb{R}$ and Range = the set of multiples of 2 = {…, -4, -2, 0, 2, 4 ,6,…..}

Example 12: solve the following equations

a). $\lfloor x+3 \rfloor = 6$ b). $\lfloor 3+\lfloor x \rfloor \rfloor = 5$

HINT: $\lfloor x \rfloor = a$ if and ony if $a \le x < a+1$

Solution:

A). $\lfloor x+3 \rfloor = 6$

$6 \le x+3 < 6+1$ implies $6-3 \le x < 7-3$,

$\iff 6 \le x+3 < 6+1$ = $3 \le x < 4$

Solution $x = [3,4).$

B). $\lfloor 3+\lfloor x \rfloor\rfloor = 5$

$5 \le 3+\lfloor x \rfloor < 5+1 \implies$ , $5-3 \le \lfloor x \rfloor < 6-3$

$\iff 2 \le \lfloor x \rfloor < 3$ here it shows that $\lfloor x \rfloor \in [2,3) = 2$

$\iff x = [2, 3)$.

https://www.geogebra.org/m/Fax2Wgsw