Unit 5: The Theory of Functions
Transcription
Unit 5: The Theory of Functions
MHF4U7 Unit 5 β 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 βeach x can only have 1 yβ How to identify a function fromβ¦ 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 = ππ β π? 1. Consider the functions π(π₯) = π₯ 2 β 3π₯ and π(π₯) = 1 β 2π₯ a) Show that π(2) > π(2) b) Determine π(3π) c) Determine π(π + 2) β π(π + 2) Page 1 MHF4U7 Unit 5 β 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) π(π₯) = 2π₯ β 3 b) π(π₯) = β3(π₯ + 1)2 + 6 c) β(π₯) = β2 β 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 Page 2 MHF4U7 Unit 5 β 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. π(π(π₯)) or ππ(π₯) or π o π(π₯) Example: Given π(π) = ππ + π and π(π) = πππ , determine π π¨ π(π) π₯ 1. Given π(π₯) = π₯+1 , π₯ β β1 and π(π₯) = π₯ 2 , determine a) π o π(π₯) b) π o π(π₯) Note: π o π(π₯) does not necessarily equal to π o π(π₯) 1 π₯ 2. Given π(π₯) = π₯ 2 β 1 and π(π₯) = , π₯ β 0, determine π o π(π₯) Page 3 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok 3. Given π(π₯ + 1) = π₯ 2 + 4π₯ + 5, find π(π₯) 4. Given β(π₯) = βπ₯ + 2 and π(π₯) = 5π₯ 2 + 2 a) determine the domain and range of β(π₯) and π(π₯) h(x) g(x) Domain: Range: b) determine the domain of β(π(π₯)) Page 4 MHF4U7 Unit 5 β 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 π, we define it as inverse function π β1 (π₯) From a table of values and graphs, the inverse function can be found by exchanging DOMAIN(π) and RANGE(π). 1 2 3 Example: π β1 (π₯) π(π₯) Example: a b c a b c 1 2 3 Draw the inverse function of π(π₯) You need to understand that the graph of π¦ = π β1 (π₯) is the reflection of the graph of π¦ = π(π₯) in the line π¦ = π₯ From an equation, the inverse function can be found by following: Step 1. 2. 3. Example: Swap position of π₯ and π¦ Solve for π¦ (isolate π¦) Replace y with π β1 (π₯) Determine the inverse function of π(π₯) = 3π₯ + 5 Page 5 MHF4U7 1. If π(π₯) = Unit 5 β The Theory of Functions (Ch.4) 3π₯+7 π₯+1 Mr. Kwok and π(π₯) is the inverse of π, find the value of π(2). 2. Consider the functions a) Find πβ1 (β1) π: π₯ β 2π₯ + 5 and π: π₯ β 8βπ₯ , 2 b) solve the equation (π o πβ1 )(π₯) = 9 3. Consider the functions π: π₯ β 5π₯ and π: π₯ β βπ₯, solve the equation (πβ1 o π)(π₯) = 25 Page 6 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok 4. Given the functions π: π₯ β 2π₯ and π: π₯ β 4π₯ β 3, Show that (π β1 o πβ1 )(π₯) = (π π π)β1 (π₯) Identity Function: π β1 (π(π₯)) = π(π β1 (π₯)) = π₯ This means π β1 π π and π π π β1 are both equal to the identity function. The graph of π¦ = β(π₯) is shown. a) Sketch the graph of π¦ = ββ1 (π₯) and π¦ = β π ββ1 (π₯) Page 7 MHF4U7 Unit 5 β 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) π(π₯) = π₯ + 2 5 π(π₯) πβ1 (π₯) Domain: Domain: Range: Range: b) π(π₯) = βπ₯ + 1, π₯ β₯ 0 π(π₯) π β1 (π₯) Domain: Domain: 6. Find the inverse of β(π₯) = 1ββπ₯ ,π₯ βπ₯ Range: Range: >0 Page 8 MHF4U7 Unit 5 β 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: π(π₯) = 2π₯ + 3 and π(π₯) = π₯ 2 πΉπππ (π + π)(π₯). 1. πΉπππ (π β π)(π₯) ππ π(π₯) = 5π₯ + 1 and π(π₯) = 3π₯ β 2 2. πΉπππ (π + π)(3) ππ π(π₯) = 5π₯ + 12 and 8 π(π₯) = π₯+1 3. Graph the following onto the grid to the right. a) π(π₯) = 2π₯ 2 + 6π₯ b) π(π₯) = β4π₯ β 4 c) (π + π)(π₯) Page 9 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok 4. Complete the table of values. x π(π₯) x π(π₯) x -3 -3 -3 -2 -2 -2 -1 -1 -1 0 0 0 1 1 1 5. Graph Page 10 MHF4U7 4E Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok Reciprocal Functions π A reciprocal function has the form of π(π) = π Graph of a reciprocal function is a __________________. 1 The simplest reciprocal function is π(π₯) = π₯ x y Horizontal Asymptote: Vertical Asymptote: π Why is the reciprocal function (π¦ = π₯ ) a self-inverse function? Page 11 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok Review of Rational Functions 1. Simplify and state any restrictions on the variables 3π3 β3π2 8π3 β12π2 +4π Multiplying and Dividing Rational Functions 2. Simplify and state restrictions a) 6π₯ 2 15π₯π¦ 3 ( ) 5π₯π¦ 8π₯π¦ 4 Adding and Subtracting Rational Functions 3. Simplify and state restrictions 3 1 5 a) 8π₯ 2 + 4π₯ β 6π₯ 3 b) 21πβ3π2 16π+4π2 b) 3π 4 + πβ3 2π+1 ÷ 14β9π+π2 12+7π+π2 Page 12 MHF4U7 Unit 5 β 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: π(π₯) = π₯ 2 β1 π₯β3 Vertical Asymptote: Example: π(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 numeratorβ¦ the horizontal asymptote is the x-axis. Example: ο· π₯ 6 β1 π₯ 2 +5π₯+3 Horizontal Asymptote: π¦ = 0 If the degree of the numerator is bigger than the denominatorβ¦ there is no horizontal asymptote. Example: ο· π(π₯) = π(π₯) = π₯ 3 +27 π₯β3 Horizontal Asymptote: ππππ If the degrees of the numerator and denominator are the sameβ¦ 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: ππππ πππππππππππ‘ ππ ππ’πππππ‘ππ ) ππππ πππππππππππ‘ ππ πππππππππ‘ππ Example: π(π₯) = 3π₯ 2 β5π₯β2 π₯ 2 +6π₯+9 3 Horizontal Asymptote: π¦ = 1 π¦=3 HINT: Use mnemonic device β BOB0 BOTN EATS DC Bigger on Bottom Zero Bigger on Top None Exponents are the Same Divide Coefficients Page 13 MHF4U7 Unit 5 β 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 π(π₯) = 2π₯+1 π₯β1 Step 1: Vertical Asymptote Step 2: Horizontal Asymptote Step 3: x-intercept Step 4: y-intercept Page 14 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) 2. Sketch π(π₯) = Mr. Kwok π₯ 2 β3π₯+2 π₯ 2 β1 Step 1: Vertical Asymptote Step 2: Horizontal Asymptote Step 3: x-intercept Step 4: y-intercept Page 15 MHF4U7 Unit 5 β The Theory of Functions (Ch.4) Mr. Kwok π₯β10 3. Sketch π(π₯) = 2π₯+5 Step 1: Vertical Asymptote Step 2: Horizontal Asymptote Step 3: x-intercept Step 4: y-intercept Page 16