Types of Functions Constant, Linear and Non-Linear

Types of Functions

Constant, Linear and

Non-Linear

7/9/2013 Types of Functions22

Types of Functions Constant Function

Each member of a set (domain) maps to the same member of another set (range)

A = domain B = range

ab k

{ }

c

, …( , ) c k,,( , ) b k( , ) a kF =


Types of Functions Constant Function Examples

f = { (-4, 3), (-1, 3), (2, 3), (4, 3), (7, 3) }

-4 3

-1 3

2 3

4 3

7 3

x y

x

y

Domain = { -4, -1, 2, 4, 7 } Range = { 3 }


Types of Functions

Constant Function Examples

f = { (x, 2) x R }

x

yy = 2

y = f (x) = 2Note:

No table Why ?

Domain = RRange = { 2 }


Types of Functions

Linear Functions

Always of form

for constants a and bf(x) = ax + b

Question: What if a = 0 ?

What properties does f(x) have ?


Types of Functions

Linear Functions

Changes in f(x) = ax + b

Always proportional to changes in x

f(x) changes by a per unit change in x


Types of Functions

Linear Functions

Rate of change of f(x) = ax + b

Rate is a, the slope of the graph

f is linear if and only if

rate of change is constant


Types of Functions Linear Function Examples

f (x) = 2x + 1 , for x A where

A = { -2, -1, 1, 3, 4 }

-2

x

y = f(x) x

y

-1 3 41

Discrete Graph

Question: Is rate of change constant ?

-3 -1 3 7 9


Types of Functions Linear Function Examples

f (x) = 2x + 1 , for all x R

Domain = R

Range = R

x

y = f(x)

y = 2x + 1

Question: Is rate of change constant ?

Continuous Graph


Rate of Change and Slope Rate of Change Example

f(x) = 2x + 1 for x R How does f(x) change

as x changes ?

Rate of change of f(x) relative to x is

x

y = f(x)

y = 2x + 1

x1 x2

y1

y2

∆x

∆y

∆f

∆x

∆y

∆x=


Rate of Change and Slope Rate of Change Example

x

y = f(x)

y = 2x + 1

x1 x2

y1

y2

∆x

∆y

y2 – y1

x2 – x1 =

∆f

∆x

∆y

∆x=

(2x2 + 1) – (2x1 + 1)

x2 – x1 =

2x2 – 2x1 + 1 – 1 x2 – x1

=

= 2 f(x) = 2x + 1

f(x) = 2x + 1 for x R


Rate of Change

We now have

∆f = ∆y = 2 ∆x Change in f(x)

is twice the change in x …

Rate of Change and Slope

∆f

∆x

∆y

∆x= = 2

x

y = f(x)

y = 2x + 1

x1 x2

y1

y2

∆x

∆y

f(x) = 2x + 1 for x R

… at any x


Intercepts from Formulas Suppose we have a line L with non-zero

slope m and slope-intercept formula

y = mx + b

Linear Functions in General

x

y

Vertical Intercept

(0, b)Slope

Horizontal Intercept

L

( ) bm–

0,

( ) bm–

0,


x

y

Find intercepts and graph for

y = x + 2

Note: y = mx + b

m = 1 and b = 2

y-intercept :

(0,b) = (0,2)

Linear Function Example

(0, 2)


y-intercept :

(0,b) = (0,2)

Slope : m = 1

x-intercept :

x

y

Linear Function Example

(0, 2)

(–2, 0)

( ) ,bm–

0 =( )

21

–

, 0

=( )

– 2 0,

m = 1 and b = 2


x

y Find the equation

Intercepts

(0,b) = (0, 3)

Linear Functions Example

= ( )

4 , 0 (4, 0)

b = 3 b

m–

= 4and

General equation

y = mx + b y = x + 334

–

becomes

(0, 3)

( )

bm–

, 0

=m b4

–

34

–

=


Average Rate of Change

For function f(x) average rate of change

of f(x) over any interval [ x1 , x2 ] is

f(x2) – f(x1) x2 – x1

=∆f ∆x y

x

y = x2 + 1

x1 x2

∆f

∆x

NOTE:

Rate not constant …

depends x1 and x2

So f not linear !



Example:

f(x) = x2 + 1y

x

y = x2 + 1

x1 x2

∆f

∆x

∆f

∆x

(x22 + 1) – (x1

2 + 1)

x2 – x1 =

x22 – x1

2 x2 – x1

=

(x2 + x1)(x2 – x1) x2 – x1

=

Rate of change, i.e. slope, is not constant

NOTE: x1 + x2 =

∆f

∆x

… depends on x1 and x2



Example:

f(x) = x2 + 1y

x

y = x2 + 1

x1 x2

∆f

∆x

∆x = 5 5 ∆f =

x1 + x2 =∆f

∆x

=∆f

∆x 5 5 = 1

On interval [-2, 3]and

∆x = 5 25 ∆f = =

∆f

∆x 5 25 = 5

On interval [0, 5]and Rate of change, i.e.

slope, is not constant… so f is not linear

NOTE:


Any function f(x) NOT of form

Example 1:

f(x) = x2 + 1

Nonlinear Functions

f(x) = ax + b

x

y

Average rate of

change (slope) not constant … P1

P2

P3 P4

so function not linear


P1

P2


Example 2:

Nonlinear Functions

f(x) = ax + b

x

y

Average rate of change (slope) not constant … P3

P4


g(x)

| |x=



Example 3:

Nonlinear Functions

f(x) = ax + b

x

y

Average rate of change (slope) not constant …

P1

P3

P4


h(x)

=1

1 – x, x < 1

P2

Asymptote


The Difference Quotient

Consider a nonlinear function f

Secant line through (x, f(x)) and (x + h, f(x + h)) has slope m

y = f(x)

x x

(x, f(x))

h

x + h

(x + h, f(x + h))

Slope m

m

=f(x + h) – f(x)

(x + h) – x

=f(x + h) – f(x)

h

=fx

f

= x

x



Consider a nonlinear function f

Secant line through (x, f(x)) and (x + h, f(x + h)) has slope m

y = f(x)

x x

(x, f(x))

h

x + h

(x + h, f(x + h))

Slope m

m =f(x + h) – f(x)

h

f

= x

x

Difference quotient for f on [ x, x+h ] , i.e. theaverage rate of change of f from x to x + h


More Secants What happens to m as h gets

smaller and smaller?

What is the slope of the tangent line at x ?

What do the secants approach as


y = f(x)

x x x+h

m =

f(x + h) – f(x) h

x+h

x+h

As h

0 m ?,

Slope m

Tangent Line

h 0 ?



Example: f(x) = x2 – 4

x

y

=((x + h)2 – 4) – (x2 – 4)

(x + h) – x

= x2 + 2xh + h2 – 4 – x2 + 4

h

=

2xh + h2 h

=

h(2x + h)

h

f(x + h) – f(x) (x + h) – x

m =

=

2x + h mThus

m



Example: f(x) = x2 – 4

x

y

m 2x as h 0Clearly,

Slope m depends only on

the value of x chosen

At x = 1 , 2x = 2

At x = 3 , 2x = 6

2x + h

= m

Conclusion ?(1, –3)

(3, 5)

6

12

1


Think about it !

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Types of Functions Constant, Linear and Non-Linear