IB MYP 4-5 Maths- Function notation- Study Notes - New Syllabus
IB MYP 4-5 Maths- Function notation – Study Notes
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- Function notation
IB MYP 4-5 Maths- Function notation – Study Notes – All topics
Function Notation
Function Notation
A function is a rule that assigns each input exactly one output. Instead of writing “y in terms of x”, we use a special notation:
General Form: \( f(x) \) This is read as “f of x” and means the value of function f when the input is x.
Example: If \( f(x) = 2x + 3 \), then for x = 4, \( f(4) = 2(4) + 3 = 11 \).
Why use Function Notation?
- Makes writing and calculating outputs easier.
- Helps represent functions clearly (especially when we have multiple functions).
- Used for advanced operations like composite and inverse functions.
Steps to Evaluate a Function:
- Write the given function rule.
- Replace x with the given value.
- Simplify the expression to find the output.
- Domain: The set of all possible input values (x-values).
- Range: The set of all possible output values (y-values).
- Codomain: All values that the function could theoretically output.
Example :
If \( f(x) = 3x – 7 \), find \( f(5) \).
▶️ Answer/Explanation
Substitute x = 5:
\( f(5) = 3(5) – 7 = 15 – 7 = \boxed{8} \)
Example :
If \( g(x) = x^2 – 4x + 6 \), find \( g(-3) \).
▶️ Answer/Explanation
Substitute x = -3:
\( g(-3) = (-3)^2 – 4(-3) + 6 = 9 + 12 + 6 = \boxed{27} \)
Functions and Their Representations
A function is a rule that assigns exactly one output for each input. In mathematics, we write a function as:
\( y = f(x) \) or in mapping notation: \( f: x \mapsto 2x + 3 \).
Functions vs. Relations
- A function gives exactly one output for each input.
- A relation can give more than one output for the same input.
Example:
- \( f(x) = x^2 \) is a function (every x has one value of y).
- \( x^2 + y^2 = 25 \) is a relation (a circle: each x corresponds to two y values).
Types of Functions
- Linear: \( y = mx + c \) (straight line)
- Quadratic: \( y = ax^2 + bx + c \) (parabola)
- Exponential: \( y = a \cdot b^x \) (curved growth/decay)
- Piecewise: Defined by different rules for different intervals
Real-Life Applications
- Cost function: \( C(x) = 50x + 200 \) (production cost)
- Speed function: \( s(t) = 20t \) (distance vs. time)
- Growth function: \( P(t) = P_0 e^{kt} \) (population growth)
Example :
Write the function \( f(x) = 3x – 2 \) in mapping notation and compute f(5).
▶️ Answer/Explanation
Mapping notation: \( f: x \mapsto 3x – 2 \).
Compute f(5):
\( f(5) = 3(5) – 2 = 15 – 2 = \boxed{13} \)
Example :
The relation is given by pairs: {(1, 2), (2, 4), (3, 6), (2, 5)}. Is this a function?
▶️ Answer/Explanation
For x = 2, there are two outputs (4 and 5), so it does not satisfy the definition of a function.
Answer: It is a relation but not a function.
Example :
A taxi company charges a fixed fee of $10 plus $2 per km. Write the cost function and find the cost of 15 km.
▶️ Answer/Explanation
Function: \( C(x) = 10 + 2x \).
For 15 km: \( C(15) = 10 + 2(15) = 10 + 30 = \boxed{\$40} \).
Composite Functions
Composite Functions
A composite function is formed when the output of one function becomes the input of another function. It is written as \( (f \circ g)(x) = f(g(x)) \) and read as “f of g of x”.
Idea: Apply the inside function first (g), then apply the outside function (f).
Example: If \( f(x) = 2x + 1 \) and \( g(x) = x^2 \), then:
- \( f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1 \)
- \( g(f(x)) = g(2x + 1) = (2x + 1)^2 \)
Steps to find a composite function:
- Identify the two functions \( f(x) \) and \( g(x) \).
- Start with the inner function \( g(x) \) and compute its output for the given input.
- Use this output as input for the outer function \( f(x) \).
- Simplify the result to get the composite function.
Note:
- Understand notation: \( f \circ g \) means f after g (apply g first).
- Order matters: \( f(g(x)) \neq g(f(x)) \) in most cases.
- Be able to compute:
- \( (f \circ g)(x) \) (f after g)
- \( (g \circ f)(x) \) (g after f)
- Evaluate composite functions for specific values and as expressions.
- Understand domain restrictions:
- The domain of the composite function depends on the inner function.
Example :
If \( f(x) = 3x – 2 \) and \( g(x) = x^2 \), find \( (f \circ g)(x) \).
▶️ Answer/Explanation
\( f(g(x)) = f(x^2) = 3(x^2) – 2 = 3x^2 – 2 \)
So: \( (f \circ g)(x) = 3x^2 – 2 \)
Example :
If \( f(x) = 2x + 5 \) and \( g(x) = x – 3 \), find \( (g \circ f)(4) \).
▶️ Answer/Explanation
Compute f(4):
\( f(4) = 2(4) + 5 = 13 \)
Now compute g(f(4)) = g(13):
\( g(13) = 13 – 3 = 10 \)
Final Answer: \( (g \circ f)(4) = 10 \)
Example :
The cost of producing x units is given by \( g(x) = 50x + 200 \). The tax applied to the cost is \( f(C) = 1.1C \). Find a formula for the total cost after tax in terms of x, and compute it for 10 units.
▶️ Answer/Explanation
Composite function: \( f(g(x)) = 1.1(50x + 200) = 55x + 220 \)
For x = 10:
\( f(g(10)) = 55(10) + 220 = 550 + 220 = \boxed{\$770} \)
Inverse Functions
Inverse Functions
An inverse function reverses the effect of the original function. If a function \( f \) maps \( x \) to \( y \), then its inverse \( f^{-1} \) maps \( y \) back to \( x \).
If \( f(x) = y \), then \( f^{-1}(y) = x \).
Properties:
- The domain of \( f \) becomes the range of \( f^{-1} \), and vice versa.
- A function must be one-to-one (each input gives a unique output) to have an inverse function.
- The graph of \( f \) and \( f^{-1} \) are reflections of each other across the line \( y = x \).
Steps to Find the Inverse of a Function:
- Replace \( f(x) \) with \( y \).
- Swap \( x \) and \( y \).
- Solve for \( y \) in terms of \( x \).
- Replace \( y \) with \( f^{-1}(x) \).
Example :
Find the inverse of \( f(x) = 2x + 5 \).
▶️ Answer/Explanation
Step 1: Write as \( y = 2x + 5 \).
Step 2: Swap x and y: \( x = 2y + 5 \).
Step 3: Solve for y: \( y = \dfrac{x – 5}{2} \).
Final Answer: \( f^{-1}(x) = \dfrac{x – 5}{2} \).
Example :
A function converts temperature from Celsius to Fahrenheit: \( F = \dfrac{9}{5}C + 32 \). Find the inverse function to convert Fahrenheit to Celsius.
▶️ Answer/Explanation
Step 1: Start with \( F = \dfrac{9}{5}C + 32 \).
Step 2: Swap C and F: \( C = \dfrac{5}{9}(F – 32) \).
Inverse Function:
\( f^{-1}(F) = \dfrac{5}{9}(F – 32) \).
Final Answer: \( f^{-1}(x) = \dfrac{5}{9}(x – 32) \).
Example :
Find the inverse of \( f(x) = x^2 \) when \( x \ge 0 \).
▶️ Answer/Explanation
Step 1: Write as \( y = x^2 \).
Step 2: Swap x and y: \( x = y^2 \).
Step 3: Solve for y: \( y = \sqrt{x} \).
Step 4: Domain and Range: Since \( x \ge 0 \), the range of the original function becomes the domain of the inverse.
Final Answer: \( f^{-1}(x) = \sqrt{x}, \; x \ge 0 \).