Week 6

1. Take out the writing utensils

2. Put your backpack away

3. Work on the warm up

4. Check the agenda on the board

5. Put your homework on the desk

6. Unit 1 Project due on Friday 9/13:

Set Theory Flipbook

Evaluating Functions and Difference Quotient

Objectives

Standard 24.0

•I will evaluate the value of any functions, including piecewise functions.

Calculus Standard 4.0

•I will evaluate the difference quotient of any given function.

When we evaluate a function we are finding the function value fora specific input. To do this we replace the function variable inthe function’s formula with the specific input and proceed from there.

The “specific input” can be a constant, another variable or an algebraic expression. The important thing to remember is it replacesthe function variable everywhere in the function’s formula

For example:

1539639232

9393

9909030

69159535

93

ccccf

aaaf

f

f

xxf

Another example:

12131)1(

13)(

31214137)7(

13)(

f

ababf

f

xxf

Because 2 is an imaginary number 1 is not in the domainof f (x)

Another example:

aa

aaa

aaag

g

tttg

2

2

2

2

2

23312

21311

302214927377

23

Another example:

0

5

1616

5

416

54

916

5

316

53

16

5

016

50

16

5

2

22

2

2

h

bbbh

h

xxh

Because 5/0 is undefined 4 is not in the domain of h(x).

Problems - 1

Given 22 5 3f x x x find

4f

3f

2f h

22(4) 5(4) 3 2(16) 20 3 15

22( 3) 5( 3) 3 2(9) 15 3 36

2

2 2

2

2

2(2 ) 5(2 ) 3

2(2 2(2 ) ) 5(2) 5 3

8 8 2 10 5 3

1 3 2

h h

h h h

h h h

h h

Problems - 2

2Given 4 find (click on mouse to see answer)

3

6

0

1

g x x

g

g

g

g a

33 4 9 4 5

26 5 36 5 31

20 5 0 5 5

because 5 is not a real number 0 is not in the domain of g

2 2 21 5 2 1 5 2 4a a a a a

Sometimes a function has different rules or formulas dependingon what the input value is. These functions are known as piece-wise defined functions.

31412)2(

110)0(

615)5(

2,1

2,1)(

2

2

2

f

f

f

xx

xxxf

Problems - 3

2

2

Given 2 1

4 1

evaluate

3

2

1

f x x x

x x

f

f

f

23 2 9 2 11

24 2 4 4 0

21 2 1 2 3

The Difference Quotient

The difference quotient of a function f (x) is defined as follows:

h

xfhxf

This is used in calculus when finding derivatives so it isworthwhile to become familiar with it in precalculus.

As we take Q closer to P, the accuracy with which the slope of the secant line approximates the slope of the tangent line increases.

h

xfhxf

Find the difference quotients for the following functions:

3

2

)(

2)(

126)(

xxf

xxxf

xxf

The difference quotient for 126)( xxf

6

6

1261266

12612)(6

)()(

h

hh

xhxh

xhxh

xfhxf

The difference quotient for xxxf 2)( 2

22

22

2222

2)(2)(

)()(

2

222

22

hxh

hhxh

h

xxhxhxhx

h

xxhxhx

h

xfhxf

The difference quotient for 3)( xxf

22

322

33223

33

33

33

33

)(

)()(

hxhx

h

hxhhx

h

xhxhhxx

h

xhx

h

xfhxf

Problems - 4

Find the difference quotient for the following function (click on mouse to see answer).

5 2f x x

5 2 5 2

5 2 2 5 2 5 2 2 5 2

22

x h xf x h f x

h hx h x x h x

h hh

h

Problems - 5

Find the difference quotient for the following function (click on mouse to see answer).

2 1g x x

2 2

2 2 2 2 2 2

2

1 1

2 1 1 2 1 1

22

x h xg x h g x

h h

x xh h x x xh h x

h h

xh hx h

h

Practice

Textbook P.69 Q.73-80

Answers

Problems - 6

Find the difference quotient for the following function (click on mouse to see answer).

h x x

1

to simplify this further we need to

rationalize the numerator

1

k x k x x h x

h h

x h x x h x

h x h xx h x h

h x h x h x h x

x h x

Average Rate of Change

ab

afbf

change of rate average

A special kind of difference quotient is the average rate of change. We can use the function values at two different points, a and b to find the average rate of change of a function over the interval [ a, b ]. This is given by:

Notice, this is equal to the slope of the line connecting the two points ( a, f(a) ) and ( b, f(b) ).

The average rate of change for the function f (x) = x2 over the interval [2,4] is:

62

12

2

416

24

24 change of rate average

22

Find the average rate of change for each of the following functions over the given intervals (click on mouse to see answer).

8,3 126 xxf

6

5

30

5

636

5

121812485

12361286

38

38

ff

1,3 22 xxxg

6

2

12

2

153

2

6921

31

323121

31

31 22

gg