Problem 68 - College Algebra with Modeling and Visualization

Question

Exercises \(57-72:\) Use the given \(f(x)\) and \(g(x)\) to find each of the following. Identify its domain. $$ \begin{array}{llll} \text { (a) }(f \circ g)(x) & \text { (b) }(g \circ f)(x) & \text { (c) }(f \circ f)(x) \end{array} $$ $$ f(x)=\frac{x-3}{2}, \quad g(x)=2 x+3 $$

Step-by-Step Solution

Answer

All compositions are defined for all real numbers.\((f\circ g)(x) = x, (g\circ f)(x) = x, (f \circ f)(x) = \frac{x-9}{4}.\) Domain: \((-\infty,\infty)\) for each.

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

The composition \((f \circ g)(x)\) means plugging the expression for \(g(x)\) into \(f(x)\). The expression for \(g(x)\) is \(2x + 3\). Substitute this into \(f(x) = \frac{x-3}{2}\): \[ (f \circ g)(x) = f(2x + 3) = \frac{(2x + 3) - 3}{2} = \frac{2x}{2} = x \] The function simplifies to \((f \circ g)(x) = x\). The domain of this function is all real numbers, \( (-\infty, \infty) \).

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

The composition \((g \circ f)(x)\) involves substituting \(f(x)\) into \(g(x)\). So, with \(f(x) = \frac{x-3}{2}\), plug this into \(g(x) = 2x + 3 \): \[ (g \circ f)(x) = g\left(\frac{x-3}{2}\right) = 2\left(\frac{x-3}{2}\right) + 3 = (x-3) + 3 = x \] The function simplifies to \((g \circ f)(x) = x\). Again, the domain is all real numbers, \( (-\infty, \infty) \).

Step 3: Calculating \((f \circ f)(x)\)

For \((f \circ f)(x)\), the function \(f(x)\) is applied to itself. Start by plugging \(f(x) = \frac{x-3}{2}\) back into \(f\): \[ (f \circ f)(x) = f\left(\frac{x-3}{2}\right) = \frac{\left(\frac{x-3}{2}\right) - 3}{2} \] Simplify this expression:\[ = \frac{\frac{x-3}{2} - 3}{2} = \frac{\frac{x-3-6}{2}}{2} = \frac{x-9}{4} \] The composition simplifies to \((f \circ f)(x) = \frac{x-9}{4} \). The domain is also all real numbers, \( (-\infty, \infty) \).

Key Concepts

Domain of a Function

Understanding the domain of a function is essential for comprehending the limitations and applicability of a given function. The domain is simply the set of all possible input values (usually denoted as "x") that a function can accept. For instance, in the function \( f(x) = \frac{x-3}{2} \), the domain is determined by identifying all values of \( x \) for which the function is defined.

In general, we look for values of \( x \) that do not cause any undefined operations. Common causes of undefined operations include division by zero and square roots of negative numbers in the context of real numbers.

In our example, the function \( f(x) \) can accept any real number; hence, its domain is all real numbers, represented as \( (-\infty, \infty) \). Similarly, when examining cases like composite functions, such as \((f \circ g)(x)\), you'll want to ensure both \( g(x) \) and \( f(g(x)) \) are defined for the chosen \( x \). Here, both compositions \((f \circ g)(x)\) and \((g \circ f)(x)\) simplify to \( x \), maintaining their domain as all real numbers.

Composite Functions

Composite functions form when one function is applied to the result of another. This concept involves plugging one complete function into another, denoted as \((f \circ g)(x) = f(g(x))\). The trick is to carefully replace the variable \( x \) in the outer function with the entire inner function.

For example, in the problem's Step 1, we computed \((f \circ g)(x)\) by substituting \(g(x) = 2x + 3\) into \(f(x) = \frac{x-3}{2}\). This resulted in \( f(2x + 3) \). By simplifying the resulting algebraic expression, reducing \( \frac{2x}{2} \) gives us the elegant expression \( x \).

Composite functions are powerful because they frequently simplify into new functions. These enable us to combine different transformations conveniently. In practical applications, composite functions appear in various disciplines, illustrating transformations and movements through equations.

Algebraic Expressions

Algebraic expressions represent a set of numbers, symbols, and operations (like addition, subtraction, etc.) structured as meaningful combinations. They are fundamental in building functions and evaluating compositions.

Consider the given functions, \( f(x) = \frac{x-3}{2} \) and \( g(x) = 2x + 3 \). Each of these is an algebraic expression. The act of finding \((g \circ f)(x)\) involved inserting the expression for \( f(x) \) into \( g(x) \). This substitution yields \( g\left(\frac{x-3}{2}\right) = 2\left(\frac{x-3}{2}\right) + 3 \), which simplifies to the algebraic expression \( x \).

Algebraic manipulation like this requires practicing combining expressions and simplifying step-by-step. Simplifying complex algebraic expressions is key to understanding how transformations work when analyzing functions. Practice is crucial; by working through exercises, students can sharpen their skills in manipulating expressions to reveal intuitive meanings and relationships between variables.