Composite Functions Functions of Functions. 10/11/2013 Composite Functions 2 2 Fencing a Square Lot...

Composite Functions

Functions of Functions

10/11/2013 Composite Functions 22

Fencing a Square Lot A square 2-acre yard is to be fenced How many feet of fencing is needed for each side? (1 acre = 43,560 sq.ft.) Solution:

Converting acres to square feet: f(x) = 43560x sq.ft.

f(2) = 43560(2) = 87120 sq.ft. = y sq.ft.

Composition of Functions

Let x = no. of acres y = no. of square ft.


Fencing a Square Lot Solution:

f(2) = 43560(2) = 87120 sq.ft. = y sq.ft. Converting sq.ft. to side length in feet:

g(y) = y1/2 ft.

g(f(2)) = g(43560(2))

= (43560(2))1/2

= (87120)1/2

= 295.16 ft.



Fencing a Square Lot Solution:

f(2) = y sq.ft.

g(f(2)) = 295.16 ft

4

y Square Feet

x Acres


● ● ●2 87120 295.16

f g

?

Feetg(f(x))


Functions of Functions If y = f(x) and y is in the domain of g, then g(y) = g(f(x)) is called a composite function of x

g is a function of a function

5



Functions of Functions Definition

For functions f and g, with g defined on the range of f , the function

is the composite function of g composed with f

6


(g f)(x) = g(f(x))°

Note: g f° is NOT the product of g and f


Functions of Functions Domain of

{ x | x is in domain f and f(x) is in domain g }

{ x | x dom f ; f(x) dom g }

Domain of

7


is a subset of domain f

(g f)(x) = g(f(x))° is

(g f)(x)°

OR


Stopping a Vehicle Find the reaction distance r(x) = tx traveled in estimated reaction time t = 2.5 secs at velocity x = 60 mph

Converting mph to ft/hr:

h(x) = 5280x ft/hr

h(60) = 5280(60)

= 316800 ft/hr

= y ft/hr



Stopping a Vehicle h(60) = 316800 ft/hr = y ft/hr Converting ft/hr to ft/sec: g(y) = y/3600 ft/sec

g(h(60)) = g(31600)

= 316800/3600

= 88 ft/sec

= z ft/sec



Stopping a Vehicle h(60) = 316800 ft/hr = y ft/hr g(y) = g(h(60)) = 88 ft/sec = z ft/sec Converting ft/sec to feet:

f(z) = (5/2)z ft Thus reaction distance at 60 mph is

r(60) = f(g(h(60)))

= f(g(316800))

= f(88)

= (5/2)88 ft = 220 ft



Stopping a Vehicle h(60) = 316800 ft/hr = y ft/hr g(y) = g(h(60)) = 88 ft/sec = z ft/sec f(z) = (5/2)z ft = 220 ft r(60) = f(g(h(60))) = 220 ft


z ft/sec y ft/hr x mi/hr

● ● ●60 316800 88

h g

r(x)

●f

220

feet


Composite Diagram:

12

Dom f


Range f

f g

Dom g

Range ga f(a)

bg(b)

Remember:

Question:Is Dom g = Range f ? … not necessarily ! b Dom g but b Range ff(x) ≠ b for any x but g(b) Range ga Dom f but f(a) Dom g

(version 1)g f°

Range is a subset of Range gg f °

Dom is a subset of Dom fg f °

So ( )(a) = g(f(a)) does not exist g f °

x f(x) g(f(x))● ● ●

°Dom g f °Range g f°g f

Where are Dom g f and Range g f ?° °


Domain f

Composite Diagram


Range f

● ● ●x f(x) g(f(x))

f gRange ga f(a)

b g(b)

(version 2)g f°

°Dom g f Range g f°

Domain g

Range is a subset of Range gg f °

Dom is a subset of Dom fg f °

° g f


Example:

14


f(x) = x + 2 g(x) = 5x + x

= g(x + 2)= 5(x + 2) + x + 2

Dom f = R Dom g = { x x ≥ 0 }Range f = R Range g = { x x ≥ 0 }

(g f)(x) = g(f(x))°


Example:

15


f(x) = x + 2 g(x) = 5x + x

= [ –2 , )= [ 0 , )

= 5(x + 2) + x + 2 (g f)(x) °

Dom g f = { x x ≥ –2 } °Range g f = { x x ≥ 0} °


Example:

16


f(x) = x + 2 g(x) = 5x + x

= f ( 5x + x )= 5x + x + 2

(f g)(x) = f(g(x))°

Question:

(g f)(x)° = 5(x + 2) + x + 2

°How do g f and differ ?f g°


Example:

17


f(x) = x + 2 g(x) = 5x + x

= 5x + x + 2

(f g)(x) °

= [ 2 , )Range f g

°

(g f)(x)° = 5(x + 2) + x + 2

Dom f g ° = [ 0 , )

= [ –2 , )Dom g f

° = [ 0 , )Range g f °

What if f(x) = x2 – 4x – 2 = x + 2 , for x ≠ 2 ?


Example:

18


f(x) = x + 2 g(x) = 5x + x

f(x) = x2 – 4x – 2 = x + 2 , for x ≠ 2 ?

Dom f g ° = { x | x ≥ 0 , x ≠ .7350 }

= { x | x ≥ –2 , x ≠ 2 }Dom g f

°= { x | x ≥ 0 , x ≠ 22 }Range g f °

Range f g

° = { x | x ≥ 2, x ≠ 13.414 }


Domain f

Example:

Composite Diagram


g(x) = 5x + x= g(x + 2) = 5(x + 2) + x + 2

Range f

● ● ●x f(x) g(f(x))

f g

Domain g

Range g–5 –3

4 22

f(x) = x2 – 4x – 2 = x + 2 , for x ≠ 2, f(x) ≠ 4

f

(g f)(x) = g(f(x))°

g f°

Dom g f = { x x ≥ –2 , x ≠ 2 }

° Range g f ={ x x ≥ 0 , x ≠ 22 }

°


Composition Versus Multiplication NOTE: Example : Let f(x) = x2 and g(x) = x1/2


= g(x2)= (x2)1/2 = |x|

... then

(gf)(x) = g(x) • f(x) = x1/2 • x2 = x5/2 ... which is NOT |x| !!

, for all x

(g f)(x) ≠ (gf)(x)°

But …

(g f)(x) = g(f(x)) °

But …


Composition Versus Multiplication Is it possible that


... well ... = f(g(x)) = f(x1/2)= (x1/2)2

Question:

Only if their domains are equal ! Are they ?

, for x ≥ 0

?(g f)(x) = (f g)(x)

° °

Are and the same function ?

g f ° f g °

= (g f)(x)

°

(f g)(x)

°

= x


Find

22

Combining Functions

Note:Output of the first function is input to the second

= f(1) = 3 = -2= g(-1)

and

graphically (g f)(–3) °(f g)(2)

°

= g(f(-3))(g f)(-3)

°= f(g(2))(f g)(2) °

f(1) = 3

-3-1

-2

3

1

1 2 3x

y

●●

●

● g(2) = 1

y = f(x) y =g(x)

-3-1

-2

3

1

1 2 3x

y

y = f(x) y =g(x)

●

●

●●f(-3) = -1


Combining Functions

Find

= f(g(2))

= g(f(-3))

Note: Output of the first function is input to the second

= f(1) = 3

= -2= g(-1)

and in tabular form

(g f)(–3) °(f g)(2) °

x –3 –2 –1 0 1 2 3

f(x) –1 0 1 2 3 4 5 g(x) 5 1 –2 –3 –2 1 5 –2

–1

–1 3

1

–3 2 1

°(f g)(2)

°(g f)(-3)


Think about it !

Composite Functions Functions of Functions. 10/11/2013 Composite Functions 2 2 Fencing a Square Lot...

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