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Topic 2/3
15 Flashcards in this deck.
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Formally, a function \( f \) from a set \( X \) to a set \( Y \) is denoted as \( f: X \rightarrow Y \), where \( f(x) = y \) for \( x \in X \) and \( y \in Y \).
Composition of functions involves applying one function to the result of another function. If \( f: X \rightarrow Y \) and \( g: Y \rightarrow Z \), then the composition \( g \circ f \) is a function from \( X \) to \( Z \) defined by:
$$ (g \circ f)(x) = g(f(x)) $$**Example:**
Let \( f(x) = 2x + 3 \) and \( g(x) = x^2 \). Then:
$$ (g \circ f)(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 $$An inverse function reverses the effect of a function. Formally, if \( f: X \rightarrow Y \) is a bijection (one-to-one and onto), then its inverse \( f^{-1}: Y \rightarrow X \) satisfies:
$$ f^{-1}(f(x)) = x \quad \text{and} \quad f(f^{-1}(y)) = y $$**Example:**
Let \( f(x) = 2x + 3 \). To find \( f^{-1}(x) \), solve for \( x \):
$$ y = 2x + 3 \\ y - 3 = 2x \\ x = \frac{y - 3}{2} \\ \Rightarrow f^{-1}(x) = \frac{x - 3}{2} $$To find the inverse of a function \( f(x) \), follow these steps:
**Example:**
Find the inverse of \( f(x) = \frac{x + 5}{3} \):
To verify that \( f^{-1} \) is indeed the inverse of \( f \), check that:
$$ f^{-1}(f(x)) = x \quad \text{and} \quad f(f^{-1}(x)) = x $$**Example:**
Given \( f(x) = 2x + 3 \) and \( f^{-1}(x) = \frac{x - 3}{2} \):
Verify \( f^{-1}(f(x)) = x \):
$$ f^{-1}(f(x)) = f^{-1}(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x $$Verify \( f(f^{-1}(x)) = x \):
$$ f(f^{-1}(x)) = f\left(\frac{x - 3}{2}\right) = 2\left(\frac{x - 3}{2}\right) + 3 = (x - 3) + 3 = x $$The composition of a function and its inverse yields the identity function:
$$ f \circ f^{-1} = f^{-1} \circ f = I $$Where \( I(x) = x \).
The Inverse Function Theorem provides conditions under which a function has a continuously differentiable inverse. Specifically, if \( f: \mathbb{R}^n \rightarrow \mathbb{R}^n \) is a continuously differentiable function and its Jacobian matrix at a point \( a \) is invertible, then \( f \) has a locally defined inverse function near \( a \).
**Formal Statement:**
If \( f \) is continuously differentiable at \( a \) and the Jacobian matrix \( J_f(a) \) is invertible, then there exists an open set around \( a \) where \( f \) is invertible and its inverse is also continuously differentiable.
**Implications:** This theorem is crucial in multivariable calculus and differential geometry, providing a foundation for understanding local behaviors of functions and mappings.
When composing more than two functions, associativity allows for flexibility in the order of composition. For functions \( f, g, h \) where the compositions are defined:
$$ h \circ g \circ f = h \circ (g \circ f) = (h \circ g) \circ f $$>**Example:**
Let \( f(x) = x + 1 \), \( g(x) = 2x \), and \( h(x) = x^2 \).
Both yield \( (2x + 2)^2 = 4x^2 + 8x + 4 \).
The inverse of a composite function follows the reverse order of the composition:
$$ (f \circ g)^{-1} = g^{-1} \circ f^{-1} $$>**Proof:** Given \( h = f \circ g \), then \( h^{-1} = g^{-1} \circ f^{-1} \) because: $$ h^{-1}(h(x)) = h^{-1}(f(g(x))) = g^{-1}(f^{-1}(f(g(x)))) = g^{-1}(g(x)) = x $$>
**Example:**
Let \( f(x) = \frac{x}{2} + 1 \) and \( g(x) = 3x - 4 \). Then:
$$ (f \circ g)(x) = f(g(x)) = f(3x - 4) = \frac{3x - 4}{2} + 1 = \frac{3x - 4 + 2}{2} = \frac{3x - 2}{2} $$>Finding the inverse:
The inverse \( h^{-1} = g^{-1} \circ f^{-1} \).
First, find \( f^{-1}(x) \):
$$ y = \frac{x}{2} + 1 \\ y - 1 = \frac{x}{2} \\ x = 2(y - 1) = 2y - 2 \\ \Rightarrow f^{-1}(x) = 2x - 2 $$>Next, find \( g^{-1}(x) \):
$$ y = 3x - 4 \\ y + 4 = 3x \\ x = \frac{y + 4}{3} \\ \Rightarrow g^{-1}(x) = \frac{x + 4}{3} $$>Thus:
$$ h^{-1}(x) = g^{-1}(f^{-1}(x)) = g^{-1}(2x - 2) = \frac{(2x - 2) + 4}{3} = \frac{2x + 2}{3} = \frac{2x}{3} + \frac{2}{3} $$>Function iteration involves composing a function with itself multiple times. If \( f \) is a function, then its \( n \)-th iterate is defined as:
$$ f^{(n)}(x) = \underbrace{f \circ f \circ \cdots \circ f}_{n \text{ times}}(x) $$>For invertible functions, iteration interacts with inversion as follows:
Inverse trigonometric functions are the inverses of the basic trigonometric functions, restricted to specific domains to ensure bijectivity:
**Example:**
If \( f(x) = \sin(x) \) with \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), then \( f^{-1}(x) = \arcsin(x) \).
Inverse functions are pivotal in calculus, particularly in:
Exploring the composition and inversion of functions with complex behaviors, such as polynomial, exponential, and logarithmic functions, requires a deep understanding of their properties:
Extending the concept of inverse functions to higher dimensions involves mappings between spaces, where functions become transformations:
Understanding how these properties apply to function composition and inversion is essential:
Applying composition and inverse functions to solve real-world problems enhances mathematical proficiency:
Aspect | Composition of Functions | Inverse of Functions |
---|---|---|
Definition | Combining two functions where the output of one is the input of another: \( g \circ f \) | A function that reverses another function: \( f^{-1}(f(x)) = x \) |
Notation | \( g \circ f \) | \( f^{-1} \) |
Existence | Always exists if the codomain of \( f \) matches the domain of \( g \) | Exists only if the function is bijective |
Properties | Associative but not commutative | Inverse functions compose to the identity function |
Applications | Modeling sequential processes, function chaining | Solving equations, reversing transformations |
To master composition and inverse functions, always verify if a function is bijective before seeking its inverse. A helpful mnemonic for remembering the steps to find an inverse is "Replace, Swap, Solve, Substitute" (RSSS). Practice by composing multiple functions and finding their inverses to build confidence. Additionally, drawing function graphs can provide visual insight into how composition and inversion affect function behavior, aiding in better retention and understanding for exam success.
Did you know that the concept of inverse functions was pivotal in developing early cryptographic systems? By reversing function processes, cryptographers created methods to encode and decode secret messages effectively. Additionally, inverse functions play a crucial role in computer graphics, enabling transformations that render complex images by reversing scaling and rotation operations.
Students often mistake composition for simple multiplication of functions, leading to incorrect results. For example, confusing \( (f \circ g)(x) \) with \( f(x) \times g(x) \) is a frequent error. Another common mistake is neglecting to restrict the domain when finding inverse functions, which can result in functions that aren't truly inverses. Ensuring functions are bijective before attempting to find their inverses can help avoid these pitfalls.