2.3 Functions
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Learning Objectives
- Understand the concept of a function and its notation
- Identify the domain and range of a function
- Evaluate composite functions
- Find and use inverse functions
- Recognize and use the modulus function
Key Concepts
Introduction to Functions
A function is a relation that assigns to each element of a set (the domain) exactly one element of another set (the codomain).
Functions are denoted using function notation, such as f(x) = x² + 3, where f is the name of the function and x is the input variable.
Functions are fundamental in mathematics and are used to model relationships between quantities in various fields.
Function Notation and Evaluation
Function notation f(x) represents the output of the function f when the input is x.
To evaluate a function at a specific value, substitute that value for x in the function expression.
For example, if f(x) = 2x² - 3x + 1, then f(2) = 2(2)² - 3(2) + 1 = 2(4) - 6 + 1 = 8 - 6 + 1 = 3.
Example 1
Given f(x) = x³ - 4x + 7, find f(-2).
Solution:
f(-2) = (-2)³ - 4(-2) + 7 f(-2) = -8 + 8 + 7 f(-2) = 7
Example 2
Given g(x) = (2x - 1)/(x + 3), find g(0).
Solution:
g(0) = (2(0) - 1)/(0 + 3) g(0) = -1/3
Domain and Range
The domain of a function is the set of all possible input values (x-values) for which the function is defined.
The range of a function is the set of all possible output values (y-values) that the function can produce.
Restrictions on the domain can arise from mathematical operations like division by zero or taking the square root of a negative number.
Example 1
Find the domain of f(x) = 1/(x - 2).
Solution:
The function f(x) = 1/(x - 2) is undefined when the denominator equals zero. Setting x - 2 = 0, we get x = 2. Therefore, the domain of f is all real numbers except 2, which can be written as {x ∈ ℝ | x ≠ 2} or ℝ\{2}.
Example 2
Find the domain and range of g(x) = √(4 - x²).
Solution:
For the domain, we need 4 - x² ≥ 0 (since we can't take the square root of a negative number). Solving this inequality: 4 - x² ≥ 0 x² ≤ 4 -2 ≤ x ≤ 2 So the domain is {x ∈ ℝ | -2 ≤ x ≤ 2} or [-2, 2]. For the range, when x is in the domain, g(x) = √(4 - x²) takes values from 0 (when x = ±2) to 2 (when x = 0). So the range is {y ∈ ℝ | 0 ≤ y ≤ 2} or [0, 2].
Composite Functions
A composite function is formed by applying one function to the result of another function.
If f and g are functions, the composite function (f ∘ g)(x) is defined as f(g(x)), which means 'apply g first, then apply f to the result'.
The domain of f ∘ g consists of all x in the domain of g such that g(x) is in the domain of f.
Example 1
Given f(x) = x² + 1 and g(x) = 2x - 3, find (f ∘ g)(x) and (g ∘ f)(x).
Solution:
(f ∘ g)(x) = f(g(x)) = f(2x - 3) = (2x - 3)² + 1 = 4x² - 12x + 9 + 1 = 4x² - 12x + 10 (g ∘ f)(x) = g(f(x)) = g(x² + 1) = 2(x² + 1) - 3 = 2x² + 2 - 3 = 2x² - 1
Example 2
Given f(x) = √x and g(x) = x + 4, find the domain of (f ∘ g)(x).
Solution:
(f ∘ g)(x) = f(g(x)) = f(x + 4) = √(x + 4) For this to be defined, we need x + 4 ≥ 0 (since we can't take the square root of a negative number). Solving this inequality: x + 4 ≥ 0 x ≥ -4 So the domain of (f ∘ g)(x) is {x ∈ ℝ | x ≥ -4} or [-4, ∞).
Inverse Functions
The inverse of a function f, denoted by f⁻¹, is a function that 'undoes' the effect of f.
If f maps x to y, then f⁻¹ maps y back to x. In other words, f⁻¹(f(x)) = x for all x in the domain of f.
A function has an inverse if and only if it is one-to-one (injective) and onto (surjective), i.e., it is bijective.
Example 1
Find the inverse of f(x) = 3x - 5.
Solution:
Step 1: Replace f(x) with y: y = 3x - 5 Step 2: Swap x and y: x = 3y - 5 Step 3: Solve for y: x + 5 = 3y y = (x + 5)/3 Step 4: Replace y with f⁻¹(x): f⁻¹(x) = (x + 5)/3 To verify: f⁻¹(f(x)) = f⁻¹(3x - 5) = ((3x - 5) + 5)/3 = 3x/3 = x
Example 2
Find the inverse of g(x) = (x - 2)³ + 1, and state its domain and range.
Solution:
Step 1: Replace g(x) with y: y = (x - 2)³ + 1 Step 2: Swap x and y: x = (y - 2)³ + 1 Step 3: Solve for y: x - 1 = (y - 2)³ ∛(x - 1) = y - 2 y = ∛(x - 1) + 2 Step 4: Replace y with g⁻¹(x): g⁻¹(x) = ∛(x - 1) + 2 The domain of g⁻¹ is the range of g, which is ℝ (all real numbers). The range of g⁻¹ is the domain of g, which is also ℝ (all real numbers).
The Modulus Function
The modulus function, denoted by |x|, gives the absolute value or magnitude of a real number x.
Formally, |x| = x if x ≥ 0, and |x| = -x if x < 0.
The modulus function is useful in defining distances and solving equations and inequalities involving absolute values.
Example 1
Solve the equation |2x - 3| = 5.
Solution:
The modulus equation |2x - 3| = 5 has two cases: Case 1: 2x - 3 = 5 (when 2x - 3 ≥ 0) 2x = 8 x = 4 This is valid since 2(4) - 3 = 5 > 0. Case 2: 2x - 3 = -5 (when 2x - 3 < 0) 2x = -2 x = -1 This is valid since 2(-1) - 3 = -5 < 0. So the solutions are x = 4 and x = -1.
Example 2
Sketch the graph of f(x) = |x - 2| + 1.
Solution:
The graph of f(x) = |x - 2| + 1 is the graph of |x - 2| shifted up by 1 unit. The graph of |x - 2| is the graph of |x| shifted right by 2 units. So f(x) = |x - 2| + 1 has a V-shape with its vertex at (2, 1), and it increases linearly as x moves away from 2 in either direction.
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