Section 3.1 Functions

Section 3.1 Functions

Relation – a set of ordered pair of points ( x, y ).

Function – a Relation where every x coordinate in the Domain corresponds to exactly one y coordinate in the Range.

Domain – a set of all x coordinates.

Range – a set of all y coordinates.

Dependent Variable – the variable that is isolated in an equation. Traditionally the y variable.

Independent Variable – the variable that is not isolated in an equation. Traditionally the x variable.


“Y is a function of X.” – means that y is the Dep. Var. and x is the Indep. Var. Noted as y = f(x).

Given that C is a function of t.

What is the Dep. Var.?

What is the Indep. Var.?

How would you write the notation?

C

t

Noted as C = f(t).

Given that x is a function of y.

What is the Dep. Var.?

What is the Indep. Var.?

How would you write the notation?

x

y

Noted as x = f(y).


Determine if the correspondence is a Relation or a Function.

DomainRange

12345

876910

DomainRange

12345

876910

DomainRange

1234

8769

10

Relation

x-coordinate repeated with different y-coordinate.

Function

NeitherThe y-coordinate of 9 is not matched with a x-coordinate.


Determine if the correspondence is a Relation or a Function.

{( 2, 5 ), ( 3, 15 ), ( 4, -3 ), ( -2, 1 ), ( 1, 6 )}

{( 1, 3 ), ( 2, 1 ), ( 3, 5 ), ( -2, 7 ), ( 1, -3 )}

How to write equations as a function?

Relation

Function

Solve for y!

12 yx

xxfy 21)( – 2x = – 2x

No x – coordinates repeated with a different y – coordinate.

Repeating x – coordinates with different y – coordinates.

Implicit form

Explicit form

x = – 5 , y = 1 – 2(– 5)

y = 1 + 10 = 11

y = – 7 , – 7 = 1 – 2x – 1 = – 1– 8 = – 2x – 2 – 2

x = 4

1. y can not be raised to an even power!2. y can not be in absolute value bars!

Find the value of y if x = -5.Find the value of x if y = -7

= 2(3)2 + 3(3)

xxxf 32 2 For the function f denoted by , evaluate

3f= 2(9) + 9

Substitute 3 in for x and always use ( )’s

Simplify by the right side by the Order of Operations Rules.

a.

273 f

23 yx– x3 = – x3

Replace y with g(x).

y is to an odd power … solve for y.

422 yx

32 xy

32 xxg

y is to an even power … NOT A FUNCTION!

= 18 + 9This means that if the x-coordinate is 3, then the y-coordinate is 27 or the point (3, 27).

= [2(3)2 + 3(3)] +


xff 3

= 27 +

Each function notation must be placed in [ ]’s because functions are at the “P” in PEMDAS.

Combine Like Terms…CLT

b. xx 32 2

xx 32 2

27323 2 xxxff

xf3

= 3(2x2) + 3(3x) Distribute the 3 into the [ ]’s

c. xx 323 2

xxxf 963 2

= 2(- x)2 + 3(- x)


xf

Multiply the function by -1.

Square the –x and multiply 3 and –x.

d.

xxxf 32 2

xf

Notice that this changes the sign on every term.

e.

xxxf 32 2

Substitute –x in for every x using ( )’s.

Substitute 3x in for every x using ( )’s. xf 3 = 2(3x)2 + 3(3x)

Multiply 2 and 9x2.

f.

xxxf 9183 2

= 2(9x2) + 9x Square the (3x) and multiply 3 and (3x).

= -1[2x2 + 3x]

= 2(x + 3)2 + 3(x + 3)


3xfSubstitute x + 3 in for every x using ( )’s.

The binomial being squared should be written out twice and FOILed. Distribute the 3 into (x + 3).

g.

271523 2 xxxf

F

= 2(x + 3)(x + 3) + 3(x + 3)

= 2(x2 + 3x + 3x + 9) + 3x + 9IO L

= 2x2 + 6x + 6x + 18 + 3x + 9Distribute the 2 and CLT.

= [2(x + h)2 + 3(x + h)] – [2x2 + 3x]


0;

h

h

xfhxf This formula is known as the DIFFERENCE QUOTIENT. Test ?!

Simplify the top first.We will save space by leaving the bottom out.

h.

h

hxhh 342 2

= 2(x + h)(x + h) + 3(x + h) – 2x2 – 3x

= 2(x2 + 2hx + h2) + 3x + 3h – 2x2 – 3x

= 2x2 + 4hx + 2h2 + 3x + 3h – 2x2 – 3x Distribute the 2 and that all non-h-terms cancel.

The binomial being squared should be written out twice and FOILed. Distribute the 3 into (x + h) the minus sign into [ ]’s

Factor the h as the GCF.

h

xhh 342

Cancel the h’s.

342 xh

Find the DOMAIN of a function defined by an equation.

No restrictions, D : All Real Nbrs.

Ask the question, if you can have negative values, zero, or positive values in the function?

a.

Set the denominator not equal to zero to find the restrictions.

c.

352 xxxf

b. 4

12

x

xxg STOP! We can’t have zero in the bottom.

042 x42 x

2x

D : x = + 2 This is called an implied domain, which means All Real Nbrs. except for + 2.

tth 3 STOP! We can’t take the square root of a negative value.Set the radicand, the expression in the radical symbol, greater than or equal to zero to make sure our variable generates a positive value.

03 tt3

3t D : t < 3 or (- , 3]

Find the DOMAIN of a function defined by an equation.Ask the question, if you can have negative values, zero, or positive values in the function?

Set the radicand greater than or equal to zero, but remember the denominator can not be equal to zero as well. This will find the restrictions.

e.

d. 1

2

x

xxF STOP! We can’t have zero in the bottom

and take the square root of a negative.

01xD : x > 1 or (1, )

12

x

xtG STOP! We can’t have zero in the bottom.

012 x D : (- , ) or All Real Nbrs.

01x1x


12 xNot possible, x2 is always a non-negative value.

Find the DOMAIN of a function defined by an equation.Ask the question, if you can have negative values, zero, or positive values in the function?

f. 25

52

x

xtM STOP! We can’t have zero in the bottom.

0252 x


252 x 5x

STOP! We can’t have negatives in the square root.

05 x 5x

5 5

,55,5:D

Find the domain of the function , which represents the area of a circle with radius r.

Ask the question, if you can have negative values, zero, or positive values in the function?

STOP! We can’t have zero or negatives for the Area or the radius.

D : r > 0 or (0, )

2rrA

The Algebra of functions. Sum, Difference, Product, and Quotient of Two Functions

SUM : xgxfxgfgf DIFFERENCE: xgxfxgfgf

PRODUCT : xgxfxfgfg QUOTIENT : _________;xg

xfx

g

f

g

f

0xg



PRODUCT : xgxfxfgfg QUOTIENT : 0)(;

xg

xg

xfx

g

f

g

f

Let and . Determine the functions. 92 xxf 53 xxg

a. 2gf b.

242 gf

= f(2) + g(2)

Substitute 2 into both functions and simplify each one separately. Next, perform the operation on the two values.

= [(2)2 + 9] + [3(2) + 5]

= [4 + 9] + [6 + 5]

= 13 + 11

5gf = f(5) – g(5)

= [(5)2 + 9] – [3(5) + 5]

= [25 + 9] – [15 + 5]

= 34 – 20

145 gf



PRODUCT : xgxfxfgfg QUOTIENT : 0)(;

xg

xg

xfx

g

f

g

f

Let and . Determine the functions. 92 xxf 53 xxg

c. 2fg d.

132 fg

= f(-2) * g(-2)

Substitute 2 into both functions and simplify each one separately. Next, perform the operation on the two values.

= [(-2)2 + 9] * [3(-2) + 5]

= [4 + 9] * [-6 + 5]

= 13 * (-1)

1

g

f f(1)

[(1)2 + 9]

[3 + 5] 10

4

51

g

f

g(1)=

=

= =

[3(1) + 5]

[1 + 9]8

Let and . Determine the functions. 92 xxf 53 xxge. xgf f.

1432 xxxgf

= f(x) + g(x) Substitute in both functions and CLT. Beware of subtraction, distribute the minus sign.

= [x2 + 9] + [3x + 5]

xfg = g(x) – f(x)

= [3x + 5] – [x2 + 9]

432 xxxfg

= 3x + 5 – x2 – 9

g. xfg h.= f(x) * g(x)

= [x2 + 9] * [3x + 5]

xg

f

f(x) [x2 + 9]g(x)

= = [3x + 5]

452753 23 xxxxfg

; x2 + 9 = 0

3

5;

53

92

x

x

xx

g

f

i. xf

g

f(x)

g(x)=

[3x + 5] = [x2 + 9]

9

532

x

xx

f

g

Not Possible to equal zero.

Let and . Determine the functions. 2

1

xxj

1

x

xxk

a. xkj b.

1,2;12

132

xxx

xxxkj

= j(x) + k(x) xkj = j(x) – k(x)

(x – 1)

Bring in both domain restrictions.Check for canceling.

Warning! Find domain restrictions.D : x = -2 D : x = 1

1(x + 2)

= + x

(x – 1)

Find LCD and add.

(x – 1)

(x + 2)

(x + 2)

= x – 1 + x2 + 2x(x + 2)(x – 1)

1,2;12

12

xxx

xxxkj

(x – 1)


1(x + 2)

= – x

(x – 1)

Find LCD and subtract.

(x – 1)

(x + 2)

(x + 2)

= x – 1 – x2 – 2x(x + 2)(x – 1)

Distribute minus sign.

CLT on top. CLT on top.

Let and . Determine the functions. 2

1

xxj

1

x

xxk

c. xjk d.

1,2;12

xxx

xxjk

= j(x) * k(x) xk

j

k(x)


D : x = -2 D : x = 1

1(x + 2)

= * x

(x – 1)

1,0,2;2

1

xxx

xx

k

j

Another look at the domain restrictions.

1(x + 2)

= x(x – 1)

=

Multiply tops and bottoms, leave denominator in factored form.

New domain restriction!

j(x)=

= 1(x + 2)

x(x – 1)

(x – 1) x

1(x + 2)

*

1,2: xD

1(x + 2)

x(x – 1)

2x

0x1x

-2 makes top denominator 0.

1 makes bottom denominator 0.0 makes the denominator of BIG fraction 0.

Section 3.1 Functions

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