Unit 5: The Theory of Functions

Transcription

Unit 5: The Theory of Functions
MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4A What is a Function?
Functions are relations where an independent value correspond with only one dependent value.
In other words β€œeach x can only have 1 y”
How to identify a function from…
A table?
x
y
3
2
1
1
5
-2
3
8
x
3
2
1
0
y
5
-2
3
8
A graph?
Function Notation:
What are other ways to write π’š = πŸ‘π’™ βˆ’ 𝟐?
1. Consider the functions 𝑓(π‘₯) = π‘₯ 2 βˆ’ 3π‘₯ and 𝑔(π‘₯) = 1 βˆ’ 2π‘₯
a) Show that 𝑓(2) > 𝑔(2)
b) Determine 𝑔(3𝑏)
c) Determine 𝑓(𝑐 + 2) βˆ’ 𝑔(𝑐 + 2)
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4B Domain and Range
Domain
Set of all possible x values
(independent variable)
Range
Set of all possible y values
(dependent variable)
1. Determine the domain and range of each function
a) 𝑓(π‘₯) = 2π‘₯ βˆ’ 3
b) 𝑔(π‘₯) = βˆ’3(π‘₯ + 1)2 + 6
c) β„Ž(π‘₯) = √2 βˆ’ π‘₯
When working with real numbers, the three most common reasons to restrict the domain are:
ο‚· You cannot divide by zero
ο‚· You cannot take the square root of a negative number
ο‚· You cannot take the logarithm of a negative number or zero
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4C Composition of Functions
You get a composite function when you apply one function to the result of another.
𝑓(𝑔(π‘₯))
or
𝑓𝑔(π‘₯)
or
𝑓 o 𝑔(π‘₯)
Example:
Given 𝒇(𝒙) = πŸ‘π’™ + 𝟐 and π’ˆ(𝒙) = πŸπ’™πŸ , determine 𝒇 𝐨 π’ˆ(𝒙)
π‘₯
1. Given 𝑓(π‘₯) = π‘₯+1 , π‘₯ β‰  βˆ’1 and 𝑔(π‘₯) = π‘₯ 2 , determine
a) 𝑓 o 𝑔(π‘₯)
b) 𝑔 o 𝑓(π‘₯)
Note:
𝑓 o 𝑔(π‘₯) does not necessarily equal to 𝑔 o 𝑓(π‘₯)
1
π‘₯
2. Given 𝑓(π‘₯) = π‘₯ 2 βˆ’ 1 and 𝑔(π‘₯) = , π‘₯ β‰  0, determine 𝑔 o 𝑓(π‘₯)
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
3. Given 𝑓(π‘₯ + 1) = π‘₯ 2 + 4π‘₯ + 5, find 𝑓(π‘₯)
4. Given β„Ž(π‘₯) = √π‘₯ + 2 and 𝑔(π‘₯) = 5π‘₯ 2 + 2
a) determine the domain and range of β„Ž(π‘₯) and 𝑔(π‘₯)
h(x)
g(x)
Domain:
Range:
b) determine the domain of β„Ž(𝑔(π‘₯))
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4D Inverse Functions
Functions transforms an input into an output.
There are situations where we are interested in finding out what input produced a particular output.
When we reverse the actions of a function 𝑓, we define it as inverse function 𝑓 βˆ’1 (π‘₯)
From a table of values and graphs, the inverse function can be found by exchanging DOMAIN(𝒙) and RANGE(π’š).
1
2
3
Example:
𝑓 βˆ’1 (π‘₯)
𝑓(π‘₯)
Example:
a
b
c
a
b
c
1
2
3
Draw the inverse function of 𝑓(π‘₯)
You need to understand that the graph of 𝑦 = 𝑓 βˆ’1 (π‘₯) is the reflection of the graph of
𝑦 = 𝑓(π‘₯) in the line 𝑦 = π‘₯
From an equation, the inverse function can be found by following:
Step
1.
2.
3.
Example:
Swap position of π‘₯ and 𝑦
Solve for 𝑦 (isolate 𝑦)
Replace y with 𝑓 βˆ’1 (π‘₯)
Determine the inverse function of 𝑓(π‘₯) = 3π‘₯ + 5
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1. If 𝑓(π‘₯) =
Unit 5 – The Theory of Functions (Ch.4)
3π‘₯+7
π‘₯+1
Mr. Kwok
and 𝑔(π‘₯) is the inverse of 𝑓, find the value of 𝑔(2).
2. Consider the functions
a) Find π‘”βˆ’1 (βˆ’1)
𝑓: π‘₯ β†’ 2π‘₯ + 5 and 𝑔: π‘₯ β†’
8βˆ’π‘₯
,
2
b) solve the equation (𝑓 o π‘”βˆ’1 )(π‘₯) = 9
3. Consider the functions
𝑓: π‘₯ β†’ 5π‘₯ and 𝑔: π‘₯ β†’ √π‘₯, solve the equation (π‘”βˆ’1 o 𝑓)(π‘₯) = 25
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4. Given the functions
𝑓: π‘₯ β†’ 2π‘₯ and 𝑔: π‘₯ β†’ 4π‘₯ βˆ’ 3,
Show that (𝑓 βˆ’1 o π‘”βˆ’1 )(π‘₯) = (𝑔 π‘œ 𝑓)βˆ’1 (π‘₯)
Identity Function:
𝑓 βˆ’1 (𝑓(π‘₯)) = 𝑓(𝑓 βˆ’1 (π‘₯)) = π‘₯
This means 𝑓 βˆ’1 π‘œ 𝑓 and 𝑓 π‘œ 𝑓 βˆ’1 are both equal to the identity function.
The graph of 𝑦 = β„Ž(π‘₯) is shown.
a) Sketch the graph of 𝑦 = β„Žβˆ’1 (π‘₯)
and
𝑦 = β„Ž π‘œ β„Žβˆ’1 (π‘₯)
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
5. Find and graph the inverse of the following. State domain and range for each function and its inverse.
2
a) 𝑔(π‘₯) = π‘₯ + 2
5
𝑔(π‘₯)
π‘”βˆ’1 (π‘₯)
Domain:
Domain:
Range:
Range:
b) 𝑓(π‘₯) = √π‘₯ + 1, π‘₯ β‰₯ 0
𝑓(π‘₯)
𝑓 βˆ’1 (π‘₯)
Domain:
Domain:
6. Find the inverse of β„Ž(π‘₯) =
1βˆ’βˆšπ‘₯
,π‘₯
√π‘₯
Range:
Range:
>0
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4C Adding and Subtracting Functions
You can add and subtract functions to create a new function.
Example:
𝑓(π‘₯) = 2π‘₯ + 3
and
𝑔(π‘₯) = π‘₯ 2
𝐹𝑖𝑛𝑑 (𝑓 + 𝑔)(π‘₯).
1. 𝐹𝑖𝑛𝑑 (𝑓 βˆ’ 𝑔)(π‘₯) 𝑖𝑓
𝑓(π‘₯) = 5π‘₯ + 1 and
𝑔(π‘₯) = 3π‘₯ βˆ’ 2
2. 𝐹𝑖𝑛𝑑 (𝑓 + 𝑔)(3) 𝑖𝑓
𝑓(π‘₯) = 5π‘₯ + 12
and
8
𝑔(π‘₯) = π‘₯+1
3. Graph the following onto the grid to the right.
a) 𝑓(π‘₯) = 2π‘₯ 2 + 6π‘₯
b) 𝑔(π‘₯) = βˆ’4π‘₯ βˆ’ 4
c) (𝑓 + 𝑔)(π‘₯)
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4. Complete the table of values.
x
𝑓(π‘₯)
x
𝑔(π‘₯)
x
-3
-3
-3
-2
-2
-2
-1
-1
-1
0
0
0
1
1
1
5. Graph
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MHF4U7
4E
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
Reciprocal Functions
π’Œ
A reciprocal function has the form of 𝒇(𝒙) = 𝒙
Graph of a reciprocal function is a __________________.
1
The simplest reciprocal function is 𝑓(π‘₯) = π‘₯
x
y
Horizontal Asymptote:
Vertical Asymptote:
π‘˜
Why is the reciprocal function (𝑦 = π‘₯ ) a self-inverse function?
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
Review of Rational Functions
1. Simplify and state any restrictions on the variables
3𝑛3 βˆ’3𝑛2
8𝑛3 βˆ’12𝑛2 +4𝑛
Multiplying and Dividing Rational Functions
2. Simplify and state restrictions
a)
6π‘₯ 2 15π‘₯𝑦 3
(
)
5π‘₯𝑦 8π‘₯𝑦 4
Adding and Subtracting Rational Functions
3. Simplify and state restrictions
3
1
5
a) 8π‘₯ 2 + 4π‘₯ βˆ’ 6π‘₯ 3
b)
21π‘βˆ’3𝑝2
16𝑝+4𝑝2
b)
3𝑛
4
+ π‘›βˆ’3
2𝑛+1
÷
14βˆ’9𝑝+𝑝2
12+7𝑝+𝑝2
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
How to Find Vertical and Horizontal Asymptotes?
Vertical asymptote(s)
To find the vertical asymptote, find the value(s) of x where it will make the function undefined.
Hint:
Set the denominator equal to zero and solve for the unknown.
Example:
𝑓(π‘₯) =
π‘₯ 2 βˆ’1
π‘₯βˆ’3
Vertical Asymptote:
Example:
𝑓(π‘₯) =
8
π‘₯(π‘₯+4)
Vertical Asymptote:
Horizontal Asymptote
ο‚·
If the degree (the largest exponent) of the denominator is bigger than the degree of the numerator…
the horizontal asymptote is the x-axis.
Example:
ο‚·
π‘₯ 6 βˆ’1
π‘₯ 2 +5π‘₯+3
Horizontal Asymptote: 𝑦 = 0
If the degree of the numerator is bigger than the denominator…
there is no horizontal asymptote.
Example:
ο‚·
𝑓(π‘₯) =
𝑓(π‘₯) =
π‘₯ 3 +27
π‘₯βˆ’3
Horizontal Asymptote: π‘›π‘œπ‘›π‘’
If the degrees of the numerator and denominator are the same…
the horizontal asymptote equals the leading coefficient (the coefficient of the largest exponent) of the
numerator divided by the leading coefficient of the denominator.
(HA:
π‘™π‘’π‘Žπ‘‘ π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘›π‘’π‘šπ‘’π‘Ÿπ‘Žπ‘‘π‘œπ‘Ÿ
)
π‘™π‘’π‘Žπ‘‘ π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘–π‘’π‘›π‘‘ π‘œπ‘“ π‘‘π‘’π‘›π‘œπ‘šπ‘–π‘›π‘Žπ‘‘π‘œπ‘Ÿ
Example:
𝑓(π‘₯) =
3π‘₯ 2 βˆ’5π‘₯βˆ’2
π‘₯ 2 +6π‘₯+9
3
Horizontal Asymptote: 𝑦 = 1
𝑦=3
HINT: Use mnemonic device – BOB0 BOTN EATS DC
Bigger on Bottom Zero
Bigger on Top None
Exponents are the Same
Divide Coefficients
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
4H Rational Functions
When sketching rational functions, you need to perform these steps:
1) Vertical Asymptote
set denominator = 0 then solve
Beware of holes
2) Horizontal Asymptote
BOB0 BOTN EATS DC
3) x-intercept
set numerator = 0 then solve
4) y-intercept
set x=0
5) Perform tests on both sides of asymptote
6) Perform tests on ends
1. Sketch 𝑓(π‘₯) =
2π‘₯+1
π‘₯βˆ’1
Step 1:
Vertical Asymptote
Step 2:
Horizontal Asymptote
Step 3:
x-intercept
Step 4:
y-intercept
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
2. Sketch 𝑓(π‘₯) =
Mr. Kwok
π‘₯ 2 βˆ’3π‘₯+2
π‘₯ 2 βˆ’1
Step 1:
Vertical Asymptote
Step 2:
Horizontal Asymptote
Step 3:
x-intercept
Step 4:
y-intercept
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MHF4U7
Unit 5 – The Theory of Functions (Ch.4)
Mr. Kwok
π‘₯βˆ’10
3. Sketch 𝑓(π‘₯) = 2π‘₯+5
Step 1:
Vertical Asymptote
Step 2:
Horizontal Asymptote
Step 3:
x-intercept
Step 4:
y-intercept
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