evaluating functions and difference quotient. objectives standard 24.0 i will evaluate the value of...
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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