Problem 70 - Precalculus : Building Concepts and Connections

Question

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=2 x+3 ; g(x)=\frac{x-3}{2}$$

Step-by-Step Solution

Answer

The expressions for \(f \circ g(x)\) and \(g \circ f(x)\) are both \(x\). The domain of both functions is all real numbers (\(-\infty < x < +\infty\)).

Step 1: Determine \(f \circ g(x)\)

Here, the first operation is \(g(x)\), hence start by substituting \(g(x)\) into \(f(x)\) to find \(f \circ g(x)\). Wherever \(x\) is seen in \(f(x)=2x+3\), replace it with \(g(x)=\frac{x-3}{2}\). Hence, \(f \circ g(x) = 2(\frac{x-3}{2}) + 3 = x-3+3 = x\). Hence, \(f \circ g(x) = x\).

Step 2: Determine the domain of \(f \circ g(x)\)

Given that \(f \circ g(x) = x\), the domain will be all real numbers, which can be represented as \(-\infty < x < +\infty\).

Step 3: Determine \(g \circ f(x)\)

Here, the first operation is \(f(x)\), hence start by substituting \(f(x)\) into \(g(x)\) to find \(g \circ f(x)\). So, wherever \(x\) is seen in \(g(x) = \frac{x-3}{2}\), replace it with \(f(x) = 2x+3\). Hence we have \(g \circ f(x) = \frac{2x+3-3}{2} = x\). Hence, \(g \circ f(x) = x\).

Step 4: Determine the domain of \(g \circ f(x)\)

From \(g \circ f(x) = x\), the domain will be all real numbers, which again is represented as \(-\infty < x < +\infty\).

Key Concepts

Domain of a Function

The domain of a function is crucial because it tells us which input values are allowed. It's like a gatekeeper that lets certain numbers in and keeps others out. For any function, the domain is simply the set of all possible inputs, or "x" values, for which the function is defined.

For example, in the function composition problem given, we have two functions, where

• \( f(x) = 2x + 3 \)
• \( g(x) = \frac{x-3}{2} \)

When composing functions like \( f \circ g(x) \) or \( g \circ f(x) \), you'd need to find inputs that are valid for each inner function and the final result. This becomes especially important if the function has restrictions like divisions by zero or square roots of negative numbers.

In the given exercise, both composed functions \( f \circ g(x) = x \) and \( g \circ f(x) = x \) have domains of all real numbers. This means you can input any real number into these composed functions without restrictions.

Injective Functions

Injective functions, also known as "one-to-one" functions, have a unique quality: each input maps to a distinct output. This means there are no two different inputs that produce the same output. Identifying injective functions is important for understanding if a function can have an inverse.

For instance, if you have an injective function \( f(x) \), it's guaranteed that its inverse function will exist. However, in our exercise, when evaluating the compositions \( f \circ g(x) = x \) and \( g \circ f(x) = x \), we notice that these compositions reduce to the identity function \( x \). The identity function is inherently injective, as each input \( x \) produces a unique output \( x \).

Thus, both compositions maintain injectiveness, making it possible to discuss their inverses, even though the main focus in this step was checking the compositions.

Real Numbers

Real numbers form the basis of measurement and continuous quantities in mathematics. They include all the numbers on the number line, encompassing both rational numbers, like 3.5 or -2, and irrational numbers, such as \( \pi \) or \( \sqrt{2} \).

Understanding real numbers is fundamental in calculus, algebra, and most areas of mathematics since they represent all possible numbers used in real-world scenarios.

• Positive numbers, like 5
• Negative numbers, like -3
• Zero, which is neutral
• Decimal and fractional numbers, like 0.75 or \( \frac{3}{4} \)

In the domain context of our exercise, when we say that the domain is all real numbers, it translates to the fact that nothing restricts the input; every value you pick from the real number set works seamlessly in both compositions \( f \circ g(x) = x \) and \( g \circ f(x) = x \). This wide domain often simplifies problem-solving because it eliminates concerns about undefined inputs.