Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-34: Find (a) \((f \circ g)(x)\) and the domain of \(f \circ g\) and (b) \((g \circ f)(x)\) and the domain of \(g \circ f\). $$ f(x)=\frac{3 x+5}{2}, \quad g(x)=\frac{2 x-5}{3} $$

Step-by-Step Solution

Answer

Both \((f \circ g)(x)\) and \((g \circ f)(x)\) simplify to \(x\), with domains \((-\infty, \infty)\).

Step 1: Understand the Composition of Functions

To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). Similarly, to find \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\).

Step 2: Calculate \((f \circ g)(x)\)

First, compute \((f \circ g)(x)\). Substitute \(g(x)\) into \(f(x)\):\[ f(g(x)) = f\left(\frac{2x-5}{3}\right) = \frac{3\left(\frac{2x-5}{3}\right) + 5}{2} \]Simplify the expression:\[ = \frac{2x-5 + 5}{2} = \frac{2x}{2} = x \]

Step 3: Determine the Domain of \(f \circ g\)

The domain of \(f \circ g\) is determined by the domain of \(g(x)\) since \(f(x)\) is defined for all real numbers. Therefore, the domain of \(g(x) = \frac{2x-5}{3}\) is all real numbers since there are no restrictions (the denominator must not be zero, but in this case, it never happens). Thus, the domain of \(f \circ g\) is all real numbers \(( -\infty, \infty )\).

Step 4: Calculate \((g \circ f)(x)\)

Now, compute \((g \circ f)(x)\). Substitute \(f(x)\) into \(g(x)\):\[ g(f(x)) = g\left(\frac{3x+5}{2}\right) = \frac{2\left(\frac{3x+5}{2}\right) - 5}{3} \]Simplify the expression:\[ = \frac{3x+5 - 5}{3} = \frac{3x}{3} = x \]

Step 5: Determine the Domain of \(g \circ f\)

The domain of \(g \circ f\) is determined by the domain of \(f(x)\), which is all real numbers since there are no restrictions on \(f(x)\). Similarly, \(g(x)\) must also be defined for the output of \(f(x)\), which is true for all real numbers. Thus, the domain of \(g \circ f\) is all real numbers \(( -\infty, \infty )\).

Key Concepts

Domain of a Function

When dealing with functions, it's important to understand that the domain of a function is the set of all possible input values (usually represented as 'x') for which the function is defined. Knowing the domain helps us avoid undefined expressions, such as division by zero or taking the square root of a negative number.
To find the domain of a composite function like \(f \circ g\), we first need to consider the domain of \(g(x)\) and ensure that its outputs are valid inputs for \(f(x)\). In our example:

• The function \g(x) = \frac{2x - 5}{3}\ is defined for all real numbers since its denominator never becomes zero.
• Thus, every output of \g(x)\ can be used in \f(x)\ without causing any undefined expressions.

Therefore, the domain of \(f \circ g\) is simply all real numbers, denoted as \(-\infty, \infty\).
Likewise, the domain for \(g \circ f\) follows the same consideration. Here, \(f(x) = \frac{3x + 5}{2}\) is also defined for all real numbers, leading to the composite function \(g(f(x))\) having a domain of all real numbers as well.

Simplification of Expressions

Simplifying expressions makes them easier to understand and work with. It's a crucial skill in mathematics, especially when dealing with composite functions.
Consider the simplification steps for both \(f \circ g\) and \(g \circ f\):

• For \f(g(x))\, we substituted \(g(x) = \frac{2x - 5}{3}\) into \(f(x)\). The expression was simplified as follows: \ f\left(\frac{2x-5}{3}\right) = \frac{3\left(\frac{2x-5}{3}\right) + 5}{2}\. After simplification, this reduces to \x\.
• Similarly, \g(f(x))\ involves plugging \(f(x) = \frac{3x + 5}{2}\) into \(g(x)\) and simplifying the result: \g\left(\frac{3x+5}{2}\right) = \frac{2\left(\frac{3x+5}{2}\right) - 5}{3}\, simplifying it to \x\.

The key to simplifying these expressions is performing basic arithmetic operations (addition, subtraction) and canceling terms where possible to achieve the simplest form. Here, both expressions simplified perfectly back to \x\, showcasing the "undoing" nature of certain compositions.

Real Numbers

Real numbers are a fundamental concept in mathematics, representing a continuous range of values without gaps. They include both rational numbers (like fractions and integers) and irrational numbers (like \(\pi\) and \(\sqrt{2}\)).
This rich set of numbers is crucial when discussing the domain of functions, as we often want to determine the broadest possible range of inputs a function can handle. In the case of function composition, it is vital to ensure that both functions involved can accept real number inputs from start to finish.
For example, in the given exercise:

• Both \f(x) = \frac{3x + 5}{2}\ and \g(x) = \frac{2x - 5}{3}\ define outputs for all real numbers given their simple linear forms.
• This makes them inherently compatible and allows their composite functions \(f \circ g\) and \(g \circ f\) to maintain the domain of all real numbers.

Understanding this ensures that we can predict the values of these functions reliably in virtually any application, reaffirming the significance of the real number set.