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MATH 110
GENERAL EXAMPLES
Example: Find the domain of the given graph of the function
๐‘ฆ
420
5
-2-
Example: Find the range of the given graph of the function
1
๐‘ฅ
Example: If ๐‘“(๐‘ฅ ) = 2๐‘ฅ 2 − 4 , compute ๐‘“(๐‘Ž + 1)
2
Example: Determine whether the below curves are the graph of
a function of x
By the vertical line test:
A curve in xy-plane is a graph of a function of x if and only if no
vertical line intersects the curve more than once.
3
Even and Odd functions:
Let ๐‘“(๐‘ฅ ) be a defined function, then
๏‚ท ๐‘“ (๐‘ฅ ) is an even function if
๐‘“(−๐‘ฅ ) = ๐‘“(๐‘ฅ ) , ∀ ๐‘ฅ ∈ ๐ท๐‘“
๏‚ท ๐‘“ (๐‘ฅ ) is an odd function if
๐‘“(−๐‘ฅ ) = −๐‘“ (๐‘ฅ ) , ∀ ๐‘ฅ ∈ ๐ท๐‘“
Useful Rules:
๐‘œ๐‘‘๐‘‘
1) ๐‘’๐‘ฃ๐‘’๐‘› = ๐‘œ๐‘‘๐‘‘
๐‘’๐‘ฃ๐‘’๐‘›
3) ๐‘’๐‘ฃ๐‘’๐‘› = ๐‘’๐‘ฃ๐‘’๐‘›
5) (−1)0๐‘‘๐‘‘ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ = −1
4
2)
๐‘’๐‘ฃ๐‘’๐‘›
๐‘œ๐‘‘๐‘‘
= ๐‘œ๐‘‘๐‘‘
๐‘œ๐‘‘๐‘‘
4) ๐‘œ๐‘‘๐‘‘ = ๐‘’๐‘ฃ๐‘’๐‘›
6) (−1)๐‘’๐‘ฃ๐‘’๐‘› ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ = 1
Example: Determine whether ๐‘“ (๐‘ฅ ) is even, odd, or neither even
nor. odd functions.
1) ๐‘“(๐‘ฅ ) = ๐‘ฅ 5 + 2๐‘ฅ
2) ๐‘“(๐‘ฅ ) = 4๐‘ฅ 2 − ๐‘ฅ 4 + 4
๐‘ฅ 2 +1
3) ๐‘“(๐‘ฅ ) = ๐‘ฅ 7 −3๐‘ฅ 2
๐‘ฅ 3 −4๐‘ฅ
4) ๐‘“(๐‘ฅ ) = ๐‘ฅ 5 +2๐‘ฅ 9
5
Type of functions:
Polynomials:
General formula:
๐‘“(๐‘ฅ ) = ๐‘Ž๐‘› ๐‘ฅ ๐‘› + ๐‘Ž๐‘›−1 ๐‘ฅ ๐‘›−1 + โ‹ฏ + ๐‘Ž2 ๐‘ฅ 2 + ๐‘Ž1 ๐‘ฅ + ๐‘Ž0 Where
๐‘› is a nonnegative integer
๐‘Ž0 , ๐‘Ž1 , ๐‘Ž2 , … ๐‘Ž๐‘› ∈ R (are constants called coefficients)
The degree of the polynomial is ๐‘›
Domain of polynomials: is ๐‘… = (−∞, ∞)
Examples of polynomials:
1
๐‘“(๐‘ฅ ) = 4๐‘ฅ 3 − 2๐‘ฅ 2 + 2 ๐‘ฅ − √2
5
๐‘“(๐‘ฅ ) = √3๐‘ฅ 2 + 3 ๐‘ฅ + 1
Examples of not polynomials:
๐‘“ (๐‘ฅ ) = ๐‘ฅ −2 + 4๐‘ฅ − 1
1
2
๐‘“ (๐‘ฅ ) = ๐‘ฅ + 4๐‘ฅ + √3
6
Example: Specify the degree of the polynomial
๐‘“(๐‘ฅ ) = 4๐‘ฅ 2 + 7๐‘ฅ 5 − 1
Four types of polynomials:
7
1) ๐‘“(๐‘ฅ ) = ๐‘Ž0
โŸน Constant function
2) ๐‘“(๐‘ฅ ) = ๐‘Ž1 ๐‘ฅ + ๐‘Ž0
โŸน Linear function
3) ๐‘“(๐‘ฅ ) = ๐‘Ž2 ๐‘ฅ 2 + ๐‘Ž1 ๐‘ฅ + ๐‘Ž0
โŸน Quadratic function
4) ๐‘“(๐‘ฅ ) = ๐‘Ž3 ๐‘ฅ 3 + ๐‘Ž2 ๐‘ฅ 2 + ๐‘Ž1 ๐‘ฅ + ๐‘Ž0
โŸน Cubic function
Example: State the name of polynomials for the following
functions and specify its degree.
1
1) ๐‘“(๐‘ฅ ) = 2
3
2) ๐‘“(๐‘ฅ ) = 2 ๐‘ฅ 2 + ๐‘ฅ + 3
3) ๐‘“(๐‘ฅ ) = 4๐‘ฅ 2 + 1
4) ๐‘“(๐‘ฅ ) = ๐‘ฅ 3 − 3๐‘ฅ 2
8
5) ๐‘“(๐‘ฅ ) = 5๐‘ฅ − 4
6) ๐‘“(๐‘ฅ ) = 3๐‘ฅ − √3๐‘ฅ 2
9
Power function:
General formula:
๐‘“ (๐‘ฅ ) = ๐‘ฅ ๐‘Ž , ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘Ž ∈ ๐‘…
Examples:
๐‘“(๐‘ฅ ) = ๐‘ฅ
3
, ๐‘“ (๐‘ฅ ) = ๐‘ฅ
๐‘“(๐‘ฅ ) = ๐‘ฅ 1.3
1
2
, ๐‘“ (๐‘ฅ ) = ๐‘ฅ −2
, ๐‘“(๐‘ฅ ) = ๐‘ฅ √3
Root (radical) Functions:
General formula:
๐‘“ (๐‘ฅ ) = ๐‘›√๐‘ƒ(๐‘ฅ)
, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘ƒ(๐‘ฅ ) ๐‘–๐‘  ๐‘๐‘œ๐‘™๐‘ฆ๐‘›๐‘œ๐‘š๐‘–๐‘Ž๐‘™
and ๐‘› ≥ 2
10
Example: Determine the root functions from the following
functions
3
1) ๐‘“(๐‘ฅ ) = √2๐‘ฅ 2 + 4๐‘ฅ − 1
2) ๐‘“(๐‘ฅ ) = √๐‘ฅ −3 + 4๐‘ฅ
1
4
3) ๐‘“(๐‘ฅ ) = √๐‘ฅ 2 + ๐‘ฅ 5 − 2 ๐‘ฅ + 3
4
4) ๐‘“(๐‘ฅ ) = √√2๐‘ฅ 3 + 3 ๐‘ฅ
6
1
5) ๐‘“(๐‘ฅ ) = √4๐‘ฅ 2 − ๐‘ฅ 3
1
6) ๐‘“(๐‘ฅ ) = √6๐‘ฅ 2 − 4๐‘ฅ + 3
11
Domain of the Root functions:
๐‘›
๐‘“(๐‘ฅ ) = √๐‘ƒ(๐‘ฅ) =
๐‘… = (−∞, ∞)
๐‘–๐‘“ ๐’ ๐’Š๐’” ๐’๐’…๐’…
๐‘ƒ(๐‘ฅ ) ≥ 0
๐‘–๐‘“ ๐’ ๐’Š๐’” ๐’†๐’—๐’†๐’
Example: Find the domain of the below functions:
1) ๐‘“(๐‘ฅ ) = √๐‘ฅ − 1
3
2) ๐‘“(๐‘ฅ ) = √6๐‘ฅ 2 − 1
5
3) ๐‘“(๐‘ฅ ) = √๐‘ฅ − 1
12
4) ๐‘“(๐‘ฅ ) = √๐‘ฅ 2 − 4
13
5) ๐‘“(๐‘ฅ ) = √๐‘ฅ 2 − 5๐‘ฅ + 6
14
Rational Function:
General formula:
๐‘ƒ(๐‘ฅ)
๐‘“(๐‘ฅ ) = ๐‘„(๐‘ฅ) , where ๐‘ƒ(๐‘ฅ )๐‘Ž๐‘›๐‘‘ ๐‘„ (๐‘ฅ ) are polynomials
๐‘„ (๐‘ฅ ) ≠ 0.
Example: Determine the rational functions from the below
functions
1) ๐‘“(๐‘ฅ ) =
2) ๐‘“(๐‘ฅ ) =
3) ๐‘“(๐‘ฅ ) =
4) ๐‘“(๐‘ฅ ) =
5) ๐‘“(๐‘ฅ ) =
15
๐‘ฅ 2 −7๐‘ฅ −1 +4
3๐‘ฅ 2 −4๐‘ฅ
1 2
๐‘ฅ −4๐‘ฅ
2
2๐‘ฅ 3 −4๐‘ฅ 2 +1
√3๐‘ฅ 3 −4๐‘ฅ−7
1
2๐‘ฅ 2 +4๐‘ฅ 3
5๐‘ฅ 4 +3๐‘ฅ 2 −1
1
4
7๐‘ฅ 3 − ๐‘ฅ
2√๐‘ฅ+4๐‘ฅ 2
3๐‘ฅ 4 −2๐‘ฅ
โŸน
โŸน
โŸน
โŸน
โŸน
Domain of Rational Functions:
๐ท = ๐‘… − {๐‘ง๐‘’๐‘Ÿ๐‘œ๐‘  โ€ซ}ู…ุงุนุฏุง ุงุตูุงุฑ ุงู„ู…ู‚ุงู…โ€ฌ
Example: Find the domain of the below functions:
1) ๐‘“(๐‘ฅ ) =
2) ๐‘“(๐‘ฅ ) =
16
2๐‘ฅ 2 +7๐‘ฅ−4
๐‘ฅ−1
7๐‘ฅ 2 −2๐‘ฅ 3 +1
๐‘ฅ 2 −1
3) ๐‘“(๐‘ฅ ) =
17
5๐‘ฅ 2 −4๐‘ฅ+3
๐‘ฅ 2 +๐‘ฅ−6
Algebraic Functions:
All previous functions are algebraic functions.
Example: Find the domain of the below functions
1
1) ๐‘“(๐‘ฅ ) = ๐‘ฅ−3 + √๐‘ฅ − 1
18
2) ๐‘“(๐‘ฅ ) =
19
√๐‘ฅ−4
√7−๐‘ฅ
Trigonometric Functions:
Examples:
๐‘“(๐‘ฅ ) = sin(2๐‘ฅ − 1)
๐‘“ (๐‘ฅ ) = cos(3๐‘ฅ 2 − 1)
๐‘“(๐‘ฅ ) = sec(2๐‘ฅ 3 + 4)
๐‘“(๐‘ฅ ) = csc(4๐‘ฅ)
๐‘“(๐‘ฅ ) = cot(4๐‘ฅ − 1)
๐‘“(๐‘ฅ ) = tan(2๐‘ฅ − 1)
Exponential Functions:
General formula:
๐‘“ (๐‘ฅ ) = ๐‘Ž ๐‘ฅ , ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘Ž ≠ 1 ๐‘Ž๐‘›๐‘‘ ๐‘Ž > 0
Examples:
๐‘“(๐‘ฅ ) = 4
๐‘ฅ
;
1 ๐‘ฅ
๐‘“(๐‘ฅ ) = (2)
Logarithmic Functions:
General formula:
๐‘“ (๐‘ฅ ) = log ๐‘Ž ๐‘ฅ
, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘Ž ≠ 1 ๐‘Ž๐‘›๐‘‘ ๐‘Ž > 0
Examples:
๐‘“(๐‘ฅ ) = log 3 ๐‘ฅ
20
;
๐‘“ (๐‘ฅ ) = log 4 (2๐‘ฅ 2 − 1)
Transformation of functions:
Vertical and horizontal shifts:
Assume that โ„Ž > 0. If ๐‘ฆ = ๐‘“(๐‘ฅ) is the graph of a given function,
then:
1) The graph of ๐‘ฆ = ๐‘“ (๐‘ฅ + โ„Ž) is shifted to the left h units.
2) The graph of ๐‘ฆ = ๐‘“ (๐‘ฅ − โ„Ž) is shifted to the right h units.
3) The graph of ๐‘ฆ = ๐‘“ (๐‘ฅ ) + โ„Ž is shifted upward h units.
4) The graph of ๐‘ฆ = ๐‘“ (๐‘ฅ ) − โ„Ž is shifted downward h units.
21
Example: Suppose that ๐‘ฆ = ๐‘“(๐‘ฅ) is a given function. Describe
the new graph of the below function in words.
1) ๐‘ฆ = ๐‘“(๐‘ฅ ) − 3 โŸบ the graph is shifted downward 3 units.
2) ๐‘ฆ = ๐‘“(๐‘ฅ ) + 3 โŸบ the graph is shifted upward 3 units.
3) ๐‘ฆ = ๐‘“(๐‘ฅ − 3) โŸบ the graph is shifted to the right 3 units
4) ๐‘ฆ = ๐‘“(๐‘ฅ + 3) โŸบ the graph is shifted to the left 3 units
22
5) ๐‘ฆ = ๐‘“(๐‘ฅ + 6) + 3 โŸบ the graph is shifted to the left 6 units
and then shifted upward 3 units.
6) ๐‘ฆ = ๐‘“(๐‘ฅ − 6) + 3 โŸบ the graph is shifted to the right 6 units
and then shifted upward 3 units.
7) ๐‘ฆ = ๐‘“(๐‘ฅ − 6) − 3 โŸบ the graph is shifted to the right 6 units
and then shifted downward 3 units.
8) ๐‘ฆ = ๐‘“(๐‘ฅ + 6) − 3 โŸบ the graph is shifted to the left 6 units
and then shifted downward 3 units.
23
Reflecting Transformation:
If ๐‘ฆ = ๐‘“(๐‘ฅ) is the graph of a given function, then
1) The graph of ๐‘ฆ = −๐‘“(๐‘ฅ) is reflected the graph about x-axis
2) The graph of ๐‘ฆ = ๐‘“(−๐‘ฅ) is reflected the graph about y-axis
24
Example: Suppose that ๐‘ฆ = ๐‘“(๐‘ฅ) is a given graph of the
function. Describe in words the following new functions
1) ๐‘ฆ = ๐‘“(−๐‘ฅ ) โŸบ the graph is reflected about y-axis
2) ๐‘ฆ = −๐‘“ (๐‘ฅ ) โŸบ the graph is reflected about x-axis
3) ๐‘ฆ = −๐‘“ (๐‘ฅ ) + 6 โŸบ the graph is reflected about x-axis and
then shifted upward 6 units.
4) ๐‘ฆ = ๐‘“(−๐‘ฅ ) − 6 โŸบ the graph is reflected about y-axis and
then shifted downward 6 units.
25
Arithmetic Combination of functions:
Suppose that ๐‘“(๐‘ฅ ) ๐‘Ž๐‘›๐‘‘ ๐‘”(๐‘ฅ) are functions, then
1) (๐‘“ + ๐‘”)(๐‘ฅ ) = ๐‘“(๐‘ฅ ) + ๐‘”(๐‘ฅ)
2) (๐‘“ − ๐‘”)(๐‘ฅ ) = ๐‘“(๐‘ฅ ) − ๐‘”(๐‘ฅ)
3) (๐‘“. ๐‘”)(๐‘ฅ ) = ๐‘“(๐‘ฅ ). ๐‘”(๐‘ฅ)
๐‘“
๐‘“(๐‘ฅ)
4) (๐‘”) (๐‘ฅ ) = ๐‘”(๐‘ฅ)
,
๐‘”(๐‘ฅ) ≠ 0
Example: If ๐‘“(๐‘ฅ ) = ๐‘ฅ 2 − 1 and ๐‘”(๐‘ฅ ) = ๐‘ฅ − 1 , find:
1) (๐‘“ + ๐‘”)(๐‘ฅ ) = ๐‘“(๐‘ฅ ) + ๐‘”(๐‘ฅ)
2) (๐‘“ − ๐‘”)(๐‘ฅ ) = ๐‘“(๐‘ฅ ) − ๐‘”(๐‘ฅ)
26
3) (๐‘“. ๐‘”)(๐‘ฅ ) = ๐‘“ (๐‘ฅ ). ๐‘”(๐‘ฅ)
๐‘“
๐‘“ (๐‘ฅ )
๐‘”
๐‘” (๐‘ฅ )
4) (๐‘”) (๐‘ฅ ) = ๐‘”(๐‘ฅ)
5) (๐‘“ ) (๐‘ฅ ) = ๐‘“(๐‘ฅ)
27
Composition of functions:
Suppose that ๐‘“(๐‘ฅ ) ๐‘Ž๐‘›๐‘‘ ๐‘”(๐‘ฅ ) are given functions. Then, the
composite functions is defined as:
(๐‘“๐œŠ๐‘”)(๐‘ฅ) = ๐‘“(๐‘”(๐‘ฅ ))
Example: If ๐‘“(๐‘ฅ ) = ๐‘ฅ 2 + 1 and ๐‘”(๐‘ฅ ) = ๐‘ฅ − 1 , find:
1) (๐‘“๐œŠ๐‘”)(๐‘ฅ) = ๐‘“(๐‘”(๐‘ฅ )) = ๐‘“(๐‘ฅ − 1)
2) (๐‘”๐œŠ๐‘“)(๐‘ฅ) = ๐‘”(๐‘“(๐‘ฅ )) = ๐‘”(๐‘ฅ 2 + 1)
28
3) (๐‘“๐œŠ๐‘“)(๐‘ฅ) = ๐‘“(๐‘“(๐‘ฅ )) = ๐‘“(๐‘ฅ 2 + 1)
4) (๐‘”๐œŠ๐‘”)(๐‘ฅ) = ๐‘”(๐‘”(๐‘ฅ )) = ๐‘”(๐‘ฅ − 1)
29
Domain of Composite functions:
๐ท{(๐‘“๐œŠ๐‘”)(๐‘ฅ)} =๐ท(๐‘“๐œŠ๐‘”)(๐‘ฅ) ∩ ๐ท๐‘”(๐‘ฅ)
๐ท{(๐‘”๐œŠ๐‘“)(๐‘ฅ)} =๐ท(๐‘”๐œŠ๐‘“)(๐‘ฅ) ∩ ๐ท๐‘“(๐‘ฅ)
30
and
Example: If ๐‘“(๐‘ฅ ) = √๐‘ฅ − 4 and ๐‘”(๐‘ฅ ) = ๐‘ฅ 2 , find domain of
(๐‘“๐œŠ๐‘”)(๐‘ฅ)
( ๐ท{(๐‘“๐œŠ๐‘”)(๐‘ฅ)} )
๐ท{(๐‘“๐œŠ๐‘”)(๐‘ฅ)} =๐ท(๐‘“๐œŠ๐‘”)(๐‘ฅ) ∩ ๐ท๐‘”(๐‘ฅ)
31
Exponential Function:
General formula:
๐‘“ (๐‘ฅ ) = ๐‘Ž ๐‘ฅ , ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘Ž ≠ 1 ๐‘Ž๐‘›๐‘‘ ๐‘Ž > 0
Example: Determine the exponential functions of the following
functions:
a) ๐‘“(๐‘ฅ ) = 2๐‘ฅ
2 ๐‘ฅ
b) ๐‘“(๐‘ฅ ) = (3)
c) ๐‘“(๐‘ฅ ) = (3.2)๐‘ฅ
d) ๐‘“(๐‘ฅ ) = (√3)
๐‘ฅ
1 ๐‘ฅ
e) ๐‘“(๐‘ฅ ) = (− 2)
e) ๐‘“(๐‘ฅ ) = (1)๐‘ฅ
1 ๐‘ฅ
f) ๐‘“(๐‘ฅ ) = − (2)
32
Domain of the exponential function:
Domain is ๐‘… = (−∞, ∞)
Example: Find the domain of the below functions
a) ๐‘“(๐‘ฅ ) = 3๐‘ฅ
4
b) ๐‘“(๐‘ฅ ) = ( )๐‘ฅ − 5
7
c) ๐‘“(๐‘ฅ ) = −(√3)๐‘ฅ + 4
The domain for any exponential function is always R
33
The range of the exponential functions:
Let ๐‘Ž๐œ–๐‘… and ๐‘ > 0 and ๐‘ ≠ 1. Then:
If ๐‘ฆ = ๐‘ ๐‘ฅ ± ๐‘Ž
โŸน ๐‘…{๐‘ฆ} = (±๐‘Ž, ∞)
If ๐‘ฆ = −๐‘ ๐‘ฅ ± ๐‘Ž โŸน ๐‘…{๐‘ฆ} = (− ∞, ±๐‘Ž)
Example: Find the range of the below functions:
1) ๐‘“(๐‘ฅ ) = 2๐‘ฅ − 3 โŸน ๐‘…{๐‘“} = (−3, ∞)
1 ๐‘ฅ
2) ๐‘“(๐‘ฅ ) = (2) + 4 โŸน ๐‘…{๐‘“} = (+4, ∞)
๐‘ฅ
3
3
3) ๐‘“(๐‘ฅ ) = −(√2) + 2 โŸน ๐‘…{๐‘“} = (−∞, 2)
3 ๐‘ฅ
2
5
7
4) ๐‘“(๐‘ฅ ) = − ( ) −
34
2
โŸน ๐‘…{๐‘“} = (−∞, − )
7
Laws of exponents: let ๐‘Ž, ๐‘ ∈ ๐‘…+ , then :
๐‘ฅ ๐‘ฆ
1) ๐‘Ž ๐‘Ž = ๐‘Ž
๐‘ฅ+๐‘ฆ
3) (๐‘Ž ๐‘ฅ )๐‘ฆ = ๐‘Ž ๐‘ฅ๐‘ฆ
๐‘Ž๐‘ฅ
2) ๐‘Ž๐‘ฆ = ๐‘Ž ๐‘ฅ−๐‘ฆ , ๐‘Ž ≠ 0
4) (๐‘Ž๐‘)๐‘ฅ = ๐‘Ž ๐‘ฅ ๐‘ ๐‘ฅ
Example: Use the Laws of Exponential to rewrite and simplify
the expression.
8−4
a) 2−10
35
b) (๐‘ฅ 2 )−3 (4๐‘ฅ −4 )2
Natural Exponential Function:
General Formula: ๐‘“ (๐‘ฅ ) = ๐‘’ ๐‘ฅ ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘’ ≈ 2.71
Example: Find the domain and range of the below functions.
1) ๐‘“(๐‘ฅ ) = ๐‘’ ๐‘ฅ − 3
2) ๐‘“(๐‘ฅ ) = ๐‘’ ๐‘ฅ + 3
3) ๐‘“(๐‘ฅ ) = −๐‘’ ๐‘ฅ − 3
4) ๐‘“(๐‘ฅ ) = −๐‘’ ๐‘ฅ + 3
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Horizontal line test:
We say that the graph of a function is one-to-one if and only if
no horizontal line intersects its graph more than once
Example: Determine whether the below graphs of the functions
are one-to one function or not.
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1)
2)
3)
4)
One-to-One functions:
๐‘“(๐‘ฅ ) = ๐‘ฅ ๐‘› =
๐‘“(๐‘ฅ ) ๐‘–๐‘  1 – 1 ๐‘“๐‘ข๐‘›๐‘๐‘ก๐‘–๐‘œ๐‘›
๐‘–๐‘“ ๐‘› ๐‘–๐‘  ๐‘œ๐‘‘๐‘‘
๐‘“(๐‘ฅ ) ๐‘–๐‘  not 1 – 1 ๐‘“๐‘ข๐‘›๐‘๐‘ก๐‘–๐‘œ๐‘› ๐‘–๐‘“ ๐‘› ๐‘–๐‘  ๐‘’๐‘ฃ๐‘’๐‘›
Example:
One-to-one functions
Not one-to-one functions
๐‘“(๐‘ฅ ) = ๐‘ฅ 3 , ๐‘“(๐‘ฅ ) = ๐‘ฅ 5 + 3
๐‘“(๐‘ฅ ) = ๐‘ฅ 2 − 5 , ๐‘“ (๐‘ฅ ) = 5 − ๐‘ฅ 4
Example: Which one of the following is a one-to-one function.
a) 5 – ๐‘ฅ 4
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b) ๐‘’ ๐‘ฅ
6
c) ๐‘ฅ 3 + 2
d) ln(๐‘ฅ 2 − 1)
Inverse Function:
Definition:
Example: If ๐‘“(2) = 5 , then ๐‘“ −1 (5) =
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Cancellation equations:
๐‘“ −1 (๐‘“ (๐‘ฅ )) = ๐‘ฅ , ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘›๐‘ฆ ๐‘ฅ ∈ ๐‘‹
๐‘“(๐‘“ −1 (๐‘ฅ )) = ๐‘ฅ , ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘›๐‘ฆ ๐‘ฅ ∈ ๐‘Œ
Example:
tan(tan−1 ๐‘ฅ) = ๐‘ฅ
tan−1 (tan ๐‘ฅ) = ๐‘ฅ
log ๐‘Ž (๐‘Ž ๐‘ฅ ) = ๐‘ฅ
๐‘Žlog๐‘Ž ๐‘ฅ = ๐‘ฅ
ln(๐‘’ ๐‘ฅ ) = ๐‘ฅ
๐‘’ ln ๐‘ฅ = ๐‘ฅ
Finding the inverse of the function:
The graph of a function and the graph of its inverse are
symmetrical about the line ๐’š = ๐’™
Ex) The graph of ๐‘“ −1 is obtained by reflecting the graph of ๐‘“ about
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a) origin
b) x-axis
c) y-axis
d) the line ๐‘ฆ = ๐‘ฅ
Example: Find the inverse function of the functions below:
1) ๐‘“(๐‘ฅ ) = ๐‘ฅ 3 + 2
2) ๐‘“(๐‘ฅ ) =
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2๐‘ฅ−4
๐‘ฅ
Logarithmic Functions:
๐‘“(๐‘ฅ ) = log ๐‘Ž ๐‘ฅ
, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘Ž ≠ 1 ๐‘Ž๐‘›๐‘‘ ๐‘Ž > 0
Note that:
๐‘“(๐‘ฅ ) = ln ๐‘ฅ is called the Natural Logarithmic Function
Laws of Logarithmic:
If ๐‘ฅ, ๐‘ฆ ∈ ๐‘…+ ๐‘Ž๐‘›๐‘‘ ๐‘Ÿ ∈ ๐‘… , ๐‘กโ„Ž๐‘’๐‘›
1) log ๐‘ (๐‘ฅ๐‘ฆ) = log ๐‘ (๐‘ฅ) + log ๐‘ (๐‘ฆ)
๐‘ฅ
2) log ๐‘ (๐‘ฆ) = log ๐‘ (๐‘ฅ) − log ๐‘ (๐‘ฆ)
3) log ๐‘ (๐‘ฅ ๐‘Ÿ ) = ๐‘Ÿ log ๐‘ ๐‘ฅ
4) log ๐‘Ž ๐‘Ž = 1
,
log ๐‘Ž 1 = 0
Example: Use the laws of logarithmic to compute
log 2 80 − log 2 5
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Example: Solve the following equations:
1) ๐‘’ 5−3๐‘ฅ = 10
2) log 3 (2๐‘ฅ + 3) = 3
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Inverse Trigonometric Functions:
Example: Evaluate
1
1) sin−1 (2)
√3
2) cos −1 ( 2 )
√3
3) sin−1 ( 2 )
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Example: Find the domain of the function ๐‘ฆ = sin−1 (3๐‘ฅ + 1)
Solution:
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