Lesson Notes By Weeks and Term v3 - Senior Secondary 1
Functions
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Functions
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Subject: Further Mathematics
Class: Senior Secondary 1
Term: 2nd Term
Week: 10
Theme: Pure Mathematics
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Define function Distinguish the types of functions Solve problems which in volves functions and its in verses
Functions can be classified based on how their elements are mapped: One-to-one Function (Injective Function): Definition: Each distinct element in the domain maps to a distinct element in the range. No two different inputs produce the same output. If `f(x1) = f(x2)`, then `x1 = x2`.
Graphically: Passes the Horizontal Line Test (any horizontal line intersects the graph at most once).
Example:* `f(x) = 2x + 1`. If `f(x1) = f(x2)`, then `2x1 + 1 = 2x2 + 1`, implying `2x1 = 2x2`, so `x1 = x2`. This is a one-to-one function.
Onto Function (Surjective Function): Definition: Every element in the codomain is the image of at least one element in the domain. The range of the function is equal to its codomain. For every `y` in the codomain, there exists an `x` in the domain such that `f(x) = y`.
Example:* `f(x) = x^2` from `R` to `[0, ∞)` (real numbers to non-negative real numbers). Every non-negative real number `y` has a square root `x` (or `-x`) such that `f(x) = y`. This is an onto function. If the codomain was `R`, it would not be onto because negative numbers are not in the range. One-to-one Correspondence (Bijective Function): Definition: A function that is both one-to-one and onto. Each element in the domain maps to exactly one unique element in the codomain, and every element in the codomain is mapped from an element in the domain. Bijective functions are important because they are the only functions that have an inverse function.
Example:* `f(x) = 3x - 2` from `R` to `R`. This is both one-to-one and onto.
Many-to-one Function: Definition: Two or more distinct elements in the domain map to the same element in the range.
Example:* `f(x) = x^2` from `R` to `R`. Both `f(2) = 4` and `f(-2) = 4`. This is a many-to-one function.
Example:* `f(x) = |x|`. `f(3) = 3` and `f(-3) = 3`.
Constant Function: Definition: A function where every element in the domain maps to the same single value in the codomain. `f(x) = c`, where `c` is a constant.
Example:* `f(x) = 5`. No matter what `x` is, the output is always
5. This is a many-to-one function (unless the domain has only one element).
Identity Function: Definition: A function that maps every element to itself. `f(x) = x`. This is a one-to-one and onto function (bijective). An inverse function (denoted as `f−1(x)`) "undoes" the action of the original function `f(x)`. If `f` maps `x` to `y`, then `f−1` maps `y` back to `x`.
Condition for existence: An inverse function exists if and only if the original function `f(x)` is a one-to-one (bijective) function. If a function is many-to-one, its inverse would not be a function (because one input in the inverse would lead to multiple outputs).
Properties of Inverse Functions: `f(f−1(x)) = x` for all `x` in the domain of `f−1`. `f−1(f(x)) = x` for all `x` in the domain of `f`. The domain of `f` is the range of `f−1`. The range of `f` is the domain of `f−1`. The graph of `f−1(x)` is a reflection of the graph of `f(x)` about the line `y = x`. Steps to find the inverse of a function `y = f(x)`: Replace `f(x)` with `y`. Interchange `x` and `y` in the equation. Solve the new equation for `y`. Replace `y` with `f−1(x)`.
Example 2: Finding an Inverse Function Find the inverse of the function `f(x) = 3x - 5`.
Solution: Let `y = f(x)`. So, `y = 3x - 5`. Interchange `x` and `y`: `x = 3y - 5`. Solve for `y`: `x + 5 = 3y` `y = (x + 5) / 3` Replace `y` with `f−1(x)`: `f−1(x) = (x + 5) / 3`.
Verification: `f(f−1(x)) = f((x+5)/3) = 3((x+5)/3) - 5 = (x+5) - 5 = x`. `f−1(f(x)) = f−1(3x-5) = ((3x-5)+5)/3 = (3x)/3 = x`.
Example 3: Inverse of a Rational Function Find the inverse of `f(x) = (2x + 1) / (x - 3)`, for `x ≠ 3`.
Solution: Let `y = (2x + 1) / (x - 3)`. Interchange `x` and `y`: `x = (2y + 1) / (y - 3)`. Solve for `y`: `x(y - 3) = 2y + 1` `xy - 3x = 2y + 1` `xy - 2y = 3x + 1` `y(x - 2) = 3x + 1` `y = (3x + 1) / (x - 2)` Replace `y` with `f−1(x)`: `f−1(x) = (3x + 1) / (x - 2)`, for `x ≠ 2`.
Phase 1: Introduction (10 minutes)
Teacher Activity: Begins by asking students to recall what a "relation" is in mathematics (e.g., pairing students with their states of origin, or numbers with their squares). Uses simple mapping diagrams or set-builder notation. Introduces the idea that some relations are "special".
Student Activity: Students respond to questions about relations. Observe mapping diagrams and discuss examples of relations.
Phase 2: Lesson Development - Defining Functions and Types (30 minutes)
Teacher Activity: Introduces the formal definition of a function, emphasizing "each input has exactly one output". Uses the concept of a vending machine (insert N100, get one specific drink, not two or none) to illustrate. Explains domain, codomain, and range with clear examples, possibly using sets of students and their assigned school houses. Presents various mapping diagrams and asks students to identify whether they represent functions or not. Introduces and explains each type of function (one-to-one, onto, bijective, many-to-one, constant, identity) using graphical representations, set mappings, and algebraic examples. Demonstrates the Vertical and Horizontal Line Tests.
Student Activity: Students actively participate in discussions, identify functions from non-functions using mapping diagrams. Work in pairs to classify given relations as functions or non-functions, and then further classify functions by type. Attempt simple examples of finding domain and range for given relations.
Phase 3: Lesson Development - Inverse Functions (35 minutes)
Teacher Activity: Introduces the concept of an inverse function as "undoing" the original function. Explains the condition for its existence (one-to-one or bijective functions only). Demonstrates the step-by-step process of finding an inverse function algebraically using linear and rational examples. Emphasizes the interchange of `x` and `y` and solving for `y`. Explains the graphical relationship between a function and its inverse (reflection about `y=x`). Provides several worked examples, ensuring clear algebraic steps.
Student Activity: Students ask questions for clarification regarding the conditions for an inverse to exist. Students work through examples with the teacher, noting down the steps. In groups, students attempt to find the inverse of functions provided by the teacher, then share their methods and results. Discuss the domain and range changes when finding inverses.
Phase 4: Guided Practice and Consolidation (10 minutes)
Teacher Activity: Provides 2-3 scaffolded questions covering definition, types, and finding inverses. Walks students through the thought process and problem-solving steps.
Student Activity: Students attempt the questions individually or in pairs, with the teacher providing support and feedback. Solutions are discussed collectively.
Phase 5: Conclusion and Homework (5 minutes)
Teacher Activity: Summarizes the key concepts covered: definition of a function, different types of functions, and the process of finding an inverse function. Assigns independent practice questions as homework.
Student Activity: Students take notes on the summary and copy down homework assignments.
Question 1 (Target: Define function, Distinguish types): Given the set of ordered pairs: `A = {(Lagos, Ikeja), (Abuja, Garki), (Kano, Dala), (Port Harcourt, Diobu), (Benin City, Oredo)}` `B = {(Mango, Yellow), (Orange, Orange), (Apple, Red), (Banana, Yellow)}` `C = {(Student A, 1st Term Score), (Student A, 2nd Term Score), (Student B, 1st Term Score)}` For each set, determine if it represents a function. If it is a function, state its type (one-to-one, many-to-one, constant).
Solution 1: Set A: `{(Lagos, Ikeja), (Abuja, Garki), (Kano, Dala), (Port Harcourt, Diobu), (Benin City, Oredo)}` This represents a function. Each input (Nigerian city) maps to exactly one output (its local government area or suburb).
Type: One-to-one function. Each city maps to a distinct LGA/suburb.
Set B: `{(Mango, Yellow), (Orange, Orange), (Apple, Red), (Banana, Yellow)}` This represents a function. Each input (fruit) maps to exactly one output (colour).
Type: Many-to-one function. Both 'Mango' and 'Banana' map to 'Yellow'.
Set C: `{(Student A, 1st Term Score), (Student A, 2nd Term Score), (Student B, 1st Term Score)}` This does NOT represent a function. The input 'Student A' maps to two different outputs (1st Term Score and 2nd Term Score). For a function, each input must have only one output.
Question 2 (Target: Solve problems involving functions): A local mobile data provider charges a fixed daily fee of ₦50 and then ₦10 for every 100MB of data used. Let `x` be the number of 100MB units used. a) Write a function `C(x)` that represents the total cost. b) Calculate the cost if a customer uses 5 units of 100MB data. c) If a customer pays ₦180, how many 100MB units of data did they use?
Solution 2: a) The fixed daily fee is ₦
5
0. The cost per 100MB unit is ₦
1
0. So, for `x` units, the variable cost is `10x`. The total cost function is `C(x) = 10x + 50`. b) To calculate the cost for 5 units of 100MB data, substitute `x = 5` into `C(x)`: `C(5) = 10(5) + 50` `C(5) = 50 + 50` `C(5) = ₦100` The cost is ₦100. c) If a customer pays ₦180, we set `C(x) = 180` and solve for `x`: `180 = 10x + 50` `180 - 50 = 10x` `130 = 10x` `x = 130 / 10` `x = 13` The customer used 13 units of 100MB data.
Question 3 (Target: Solve problems involving functions and its inverses): Given the function `f(x) = (x - 4) / 2`. a) Evaluate `f(10)`. b) Find the inverse function, `f−1(x)`. c) Evaluate `f−1(3)`.
Solution 3: a) Evaluate `f(10)`: `f(10) = (10 - 4) / 2` `f(10) = 6 / 2` `f(10) = 3` b) Find the inverse function, `f−1(x)`: Let `y = f(x)`: `y = (x - 4) / 2` Interchange `x` and `y`: `x = (y - 4) / 2` Solve for `y`: `2x = y - 4` `2x + 4 = y` Replace `y` with `f−1(x)`: `f−1(x) = 2x + 4` c) Evaluate `f−1(3)`: Using the inverse function `f−1(x) = 2x + 4`: `f−1(3) = 2(3) + 4` `f−1(3) = 6 + 4` `f−1(3) = 10`
Commentary: Note that `f(10) = 3` and `f−1(3) = 10`, demonstrating the inverse property where the input and output are swapped.
Market Analysis and Sales Forecasting (Economics): Businesses in Nigeria, from roadside vendors to large corporations, implicitly use functions. For instance, the demand for a product (e.g., sachet water, bread) is a function of its price. As price increases, demand often decreases. Businesses use this functional relationship to set optimal prices and forecast sales volumes. Conversely, supply is a function of price. Population Growth Modelling (Demography/Public Health): The population of a town or the number of people affected by a disease (e.g., malaria, Lassa fever) can be modelled as a function of time. This helps government agencies (like the National Population Commission or NCDC) predict future population sizes, assess resource needs (schools, hospitals), or track the spread of epidemics to implement timely interventions. Fuel Consumption and Travel Planning (Transportation): The amount of fuel consumed by a vehicle (e.g., a commercial bus or motorcycle taxi in Lagos) is a function of the distance travelled. Knowing this functional relationship helps drivers estimate fuel costs for a journey and plan stops for refuelling, especially on long inter-state trips. Similarly, travel time can be a function of distance and average speed.