Problem 28 - Intermediate Algebra

Question

Determine the indicated functional values. (Objective 2 ) If \(f(x)=-2 x-6\) and \(g(x)=3 x+10\), find \((f \circ g)(5)\) and \((g \circ f)(-3)\).

Step-by-Step Solution

Answer

\((f \circ g)(5) = -56\) and \((g \circ f)(-3) = 10\).

Step 1: Understand Composition of Functions

To solve this exercise, we need to understand that \((f \circ g)(x)\) means \(f(g(x))\), which is plugging the result of \(g(x)\) into \(f(x)\). Similarly, \((g \circ f)(x)\) means \(g(f(x))\).

Step 2: Evaluate \(g(5)\)

Substitute \(x = 5\) into \(g(x) = 3x + 10\): \[ g(5) = 3(5) + 10 = 15 + 10 = 25 \]

Step 3: Evaluate \(f(g(5))\) or \(f(25)\)

Substitute \(x = 25\) into \(f(x) = -2x - 6\): \[ f(25) = -2(25) - 6 = -50 - 6 = -56 \] This gives us \((f \circ g)(5) = -56\).

Step 4: Evaluate \(f(-3)\)

Substitute \(x = -3\) into \(f(x) = -2x - 6\): \[ f(-3) = -2(-3) - 6 = 6 - 6 = 0 \]

Step 5: Evaluate \(g(f(-3))\) or \(g(0)\)

Substitute \(x = 0\) into \(g(x) = 3x + 10\): \[ g(0) = 3(0) + 10 = 0 + 10 = 10 \] This gives us \((g \circ f)(-3) = 10\).

Key Concepts

Function Evaluation

Function evaluation is the process of determining the output of a function for a specified input value. In simpler terms, we take an equation, such as a function, and substitute a known value for its variable.
For example, for the function \(f(x) = -2x - 6\), evaluating \(f(3)\) means substituting \(x = 3\) into the equation. This gives us:

• Replace \(x\) with \(3\): \(-2(3) - 6 = -6 - 6\).
• Then, simplify the result to find \(f(3) = -12\).

When evaluating functions as part of a composition, like \((f \circ g)(x)\), we first find \(g(x)\) and use its output as the input for \(f(x)\). The evaluation of this process plays a crucial role in solving composition of functions problems.

Substitution Method

The substitution method is a key technique when dealing with function compositions such as \((f \circ g)(x)\) and \((g \circ f)(x)\). This involves taking the output of one function and using it as the input for another function.
Let's break down the process with an example:

• First, evaluate \(g(x)\). For an input \(x = 5\) in \(g(x) = 3x + 10\), substitute \(5\) for \(x\) and solve: \(g(5) = 3(5) + 10 = 25\).
• Next, take this output, \(g(5) = 25\), and substitute it into \(f(x) = -2x - 6\) to find \(f(g(5))\). Thus, \(f(25) = -2(25) - 6 = -56\).

The substitution method helps in unraveling the layers of composition by handling one function at a time.
Thus, by efficiently using substitution, students can accurately solve and understand composite functions.

Algebraic Operations

Algebraic operations are at the heart of function compositions, where we perform tasks such as addition, subtraction, multiplication, or division on function inputs and outputs. When you evaluate a function like \(f(x) = -2x - 6\), you engage in algebraic operations.
For instance, suppose you want to find \(f(-3)\). The steps would include:

• Substitution: Substitute \(-3\) into \(f(x)\), getting \(-2(-3) - 6\).
• Multiplication: Calculate \(-2\times -3 = 6\).
• Subtraction: Perform the subtraction \(6 - 6\).

These operations allow us to simplify expressions and solve for specific values.
Understanding how these operations work is essential for smoothly solving questions that involve function evaluation and composition.
Thus, as functions are evaluated and composed, applied algebraic operations streamline the process, providing clarity in solution methods.