FUNCTIONS

GRAPHS OF FUNCTIONS; PIECEWISE DEFINED FUNCTIONS; ABSOLUTE VALUE

FUNCTION; GREATEST INTEGER FUNCTION OBJECTIVES:

•  sketch the graph of a function;•  determine the domain and range  of a     function from its graph; and•  identify whether a relation is a function or     not from its graph.•  define piecewise defined functions;•  evaluate piecewise defined functions;•  define absolute value function; and •  define greatest integer function

     As we mentioned in our previous lesson, a function can be represented in different ways and one of which is through a graph or its geometric representation. We also mentioned that a function may be represented as the set of ordered pairs (x, y).  That is plotting the set of ordered pairs as points on the rectangular coordinates system and joining them will determine a curve called the graph of the function. The graph of a function f consists of all points (x, y) whose coordinates satisfy y = f(x), for all x in the domain of f.  The set of ordered pairs (x, y) may also be represented by (x, f(x)) since y = f(x). 

     Knowledge of the standard forms of the special curves discussed in Analytic Geometry such as lines and  conic  sections  is  very  helpful  in  sketching  the graph  of  a  function.  Functions  other  than  these curves can be graphed by point-plotting.

To facilitate the graphing of a function, the following steps are suggested:•  Choose suitable values of x from the domain of a     function and •  Construct a table of function values y = f(x) from the     given values of x.•  Plot these points (x, y) from the table.•  Connect the plotted points with a smooth curve.

1

23)(.4

4)(.3

9)(.2

)(.1

2

2

2

+++=

+=−=

=

x

xxxh

xxG

xxG

xxf

A. Sketch the graph of the following functions and     determine the domain and range.

EXAMPLE:

23)(.6

9)(.5 2

++=−=xxg

xxh

SOLUTIONS:

(-3, 0) (3, 0)

(0, 3)(-2, 3)

(9, 0)(0, 4)

(-1, 1)

2)(.1 xxf = xxF −= 9)(.2 4)(.3 2 += xxG

1

23)(.4

2

+++=

x

xxxh

29)(.5 xxh −=23)(.6 ++= xxg

( )[ )+∞

+∞∞−,0:

,:

R

D( )[ )+∞

+∞−,0:

9,:

R

D ( )[ )+∞+

+∞∞−,4:

,:

R

D

( )

( ) 1,:

1,:

++∞∞−

−+∞∞−

exceptR

exceptD [ ][ ]3,0:R

3,3:D

++−

( )[ )+∞+

+∞∞−,3:

,:

R

D

When the graph of a function is given, one can easily determine its domain and range. Geometrically, the domain and range of a function refer to all the x-coordinate and y-coordinate for which the curve passes, respectively.

Recall that all relations are not functions. A function is one that has a unique value of the dependent variable for each value of the independent variable in its domain. Geometrically speaking, this means:

Consider the relation defined as {(x, y)|x2 + y2 = 9}. When graphed, a circle is formed with center at (0, 0) having a radius of 3 units. It is not a function because for any x in the interval (-3, 3), two ordered pairs have x as their first element. For example, both (0, 3) and (0, -3) are elements of the relation. Using the vertical line test, a vertical line when drawn within –3 ≤ x ≤ 3 intersects the curve at two points. Refer to the figure below.

A relation f is said to be a function if and only if, in its graph, each vertical line cuts or touches the curve at no more than one point. This is called the vertical line test.

(0, 3)

(3, 0)(-3, 0)

(0, -3)

DEFINITION: PIECEWISE DEFINED FUNCTION

if x<0

+=

1x

x)x(f.1

2

0x ≥if

A piecewise defined function is defined by different formulas on different parts of its domain.

Example:

−−

−=

3x

x9

x

)x(f.2 2

0x ≤

1x >3x0 ≤<

if

if

if

Sometimes a function is defined by more than one rule or by different formulas. This function is called a piecewise define function.

if x<0 f(-2), f(-1), f(0), f(1), f(2)

A. Evaluate the piecewise function at the indicated values.

+=

1x

x)x(f.1

2

0x ≥if

−+=

2)2x(

1x

x3

)x(f.2

f(-5), f(0), f(1), f(5)

0x <if

if

if

2x0 ≤≤2x >

EXAMPLE:

B. Define g(x) = |x| as a piecewise defined function and evaluate g(-2), g(0) and g(2).

EXAMPLE:

Solution:

From the definition of |x|,

<≥

−=

0x

0x

if

if

x

x)x(g

2)2(g

0)0(g

2)2()2(g

Therefore

==

=−−=−

Sketch the graph of the following functions and determine the domain and range.

EXAMPLE:

−

=

−=

2

23)(.2

3

2

4

)(.1

x

xxf

xg

if

if

if

1

21

2

−<<≤−

≥

x

x

x

if

if

1

1

≥<x

x

>+≤

=1 12

1 )(.3

2

xifx

xifxxf

DEFINITION: ABSOLUTE VALUE FUNCTION

Recall that the absolute value or magnitude of a real number is defined by

Properties of absolute value:

<−≥

=0 ,

0 ,

xifx

xifxx

yineaqualit triangleThe baba .4

valuesabsolute theof ratio theis ratio a of valueabsolute The 0b ,b

a

b

a .3

valuesabsolute theofproduct theisproduct a of valueabsolute The b aab .2

valueabsolute same thehave negative its andnumber A aa .1

+≤+

≠=

=

=−

The graph of the function can be obtained by graphing the two parts of the equation

separately. Combining the two parts produces the V-shaped graph. It may help to generate the graph of absolute value function by expressing the function without using absolute values.

xxf =)(

<−≥

=0 if ,

0 if ,

xx

xxy

Example: Sketch the graph of the following functions and determine the domain and range.

5x23)x(f.2

1x3x)x(f.1

++=

++=

DEFINITION: GREATEST INTEGER FUNCTION

greatest integer less than or equal to x

The greatest integer function is defined by

=x

Example: =0

=1.0

=3.0

=9.0

=1

=1.1

=2.1

=9.1

=2

=1.2

=4.3

=− 4.3

=− 9.0

0

0

0

0

1

1

1

1

2

2

3

-4

-1

Graph of greatest integer function.

xy =Sketch the graph of

x xy =1x2 −<≤−

0x1 <≤−1x0 <≤2x1 <≤3x2 <≤

2−1−012

x

y

o

EXERCISES:

22

2

2

4:.6

1

12:.5

3:.4

21:.3

1:.2

34:.1

xyh

x

xxyg

xyh

xyG

xyF

xyH

+=

−+−=

+=

−=

+=

+=

( )( )( )( )312

943.10

4:.9

23

211

13

:.8

312

31:.7

2

22

+−+−−+=

−=

≥

<<−

≤−

=

≥+

<−=

xxx

xxxy

xyG

xif

xif

xif

yf

xifx

xifxyF

A. Given the following functions, determine the domain and range, and sketch the graph:

EXERCISES:

B.Compute the indicated values of the given functions.

4t

4t4

4t

if

if

if

t

1t

3

)x(f

>≤≤−

−<

+=

)16(andf),4(f),6(f −−

a.

<≤≤−

−<−−

=x2

2x2

2x

if

if

if

3

1

4

)x(hb.

c.

)2(h and),e(h,2

h),2(h),3(h 2

π−−

−=−≠

−−

=3x

3x

if

if

2

4x)x(F

2

−−

32

F and),3(F),0(F),4(F

C. Define H(x) as a piecewise defined function and evaluate H(1), H(2), H(3), H(0) and H(-2) given by,

H(x) = x - |x – 2|.