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Exercises \(57-72:\) Use the given \(f(x)\) and \(g(x)\) to find each of the following. Identify its domain. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=2-x, \quad g(x)=\frac{1}{x^{2}} $$

Short Answer

Expert verified
(f ∘ g)(x) = 2 - 1/x², domain: x ≠ 0. (g ∘ f)(x) = 1/(2-x)², domain: x ≠ 2. (f ∘ f)(x) = x, domain: all x.

Step by step solution

01

Understand Function Composition

Function composition (\(f \circ g\)(x)) means applying one function to the result of another. Here it implies that \((f \circ g)(x) = f(g(x))\) which requires us to substitute \(g(x)\) into the function \(f(x)\).
02

Compute (f ∘ g)(x)

Calculate \(f(g(x))\). Substitute \(g(x) = \frac{1}{x^2}\) into \(f(x)\) to get \(f(g(x)) = 2 - \frac{1}{x^2}\). Therefore, \((f \circ g)(x) = 2 - \frac{1}{x^2}\).
03

Identify the Domain of (f ∘ g)(x)

The domain of \((f \circ g)(x)\) must satisfy the domain restrictions of both \(f\) and \(g\). Since \(g(x) = \frac{1}{x^2}\), \(x\) cannot be zero. So, the domain of \((f \circ g)(x)\) is all real numbers except \(x = 0\).
04

Compute (g ∘ f)(x)

Calculate \(g(f(x))\). Substitute \(f(x) = 2-x\) into \(g(x)\) to get \(g(f(x)) = \frac{1}{(2-x)^2}\). Therefore \((g \circ f)(x) = \frac{1}{(2-x)^2}\).
05

Identify the Domain of (g ∘ f)(x)

The domain of \((g \circ f)(x)\) requires that \((2-x) eq 0\) because the denominator can't be zero. Solving \(2-x = 0\), we find \(x = 2\) must be excluded. Thus, the domain is all real numbers except \(x = 2\).
06

Compute (f ∘ f)(x)

Calculate \(f(f(x))\). Substitute \(f(x) = 2-x\) back into itself to get \(f(f(x)) = 2 - (2-x) = x\). Thus \((f \circ f)(x) = x\).
07

Identify the Domain of (f ∘ f)(x)

The function \(f(x) = 2-x\) is defined for all real numbers. Therefore, the domain of \((f \circ f)(x)\) is also all real numbers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Domain of Functions
In mathematics, the domain of a function consists of all the input values (typically represented by "x") for which the function is defined. Understanding the domain is crucial because any input outside of this set results in an undefined or non-existent output.

The domain can be influenced by constraints such as:
  • Fractions - the denominator cannot be zero.
  • Square roots - the radicand must be non-negative for real number outputs.
  • Logarithms - the argument must be positive.
For instance, if a function involves a fraction, like the function g(x) = \( \frac{1}{x^2} \), the domain excludes zero since division by zero is undefined.

When dealing with function composition, such as \((f \circ g)(x)\), it's essential to satisfy the domain criteria of both individual functions involved. This may lead to constraints that further limit the domain of the composed function. For example, in \((f \circ g)(x) = 2 - \frac{1}{x^2}\), not only must \( x eq 0 \) due to \( g(x) \), but the result must be valid under \( f(x) \) as well.
Algebraic Functions
Algebraic functions are built using basic algebraic operations such as addition, subtraction, multiplication, division, and exponentiation. These are fundamental in expressing relationships and solving equations in algebra.

Examples include linear functions like \( f(x) = 2 - x \) and rational functions like \( g(x) = \frac{1}{x^2} \). Algebraic functions can vary from simple polynomials to more complex forms involving multiple operations.

Composition allows us to create new functions by applying one algebraic function to the output of another. In this way, more complex expressions or transformations can model real-world situations or abstract problems more accurately.

In the exercises discussed, we performed algebraic operations in function compositions like \((f \circ g)(x)\). Understanding underlying algebraic manipulation simplifies solving or rearranging these functions.
Function Operations
Function operations encompass ways to combine functions, such as addition, subtraction, multiplication, division, and composition. These operations expand what we can express and calculate using functions.

The operation of composition, denoted \((f \circ g)(x)\), essentially means plugging one function's output into another. It's like inputting \( x \) into \( g \), and then using this result as the input for \( f \). This was illustrated when \((f \circ g)(x)\) turned into \( f\left( \frac{1}{x^2} \right) = 2 - \frac{1}{x^2} \).

For operations like \((g \circ f)(x)\), it's crucial to remember order matters—\((g \circ f)(x)\) is typically not the same as \((f \circ g)(x)\). Here, \( f(x) = 2-x \) serves as an input to \( g(x) \), giving \( \frac{1}{(2-x)^2} \). Understanding this helps in predicting function behavior and ensures calculations are made correctly.

To correctly identify feasible operations, always start by considering the domains resulting from each operation, ensuring inputs remain valid for the functions involved.

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