Problem 44 - Intermediate Algebra

Question

Let \(f(x)=2 x+1\) and \(g(x)=x^{2}-1 .\) Find each of the following. See Example 3 . $$ (f \circ g)(2 x) $$

Step-by-Step Solution

Answer

\((f \circ g)(2x) = 8x^2 - 1\).

Step 1: Write the Expression

The given expression is \[(f \circ g)(2x) = f(g(2x)).\]This means we have to first apply \(g\) to \(2x\), then apply \(f\) to the result.

Step 2: Apply the Function \(g(x)\)

Substitute \(2x\) into \(g(x)\). \[g(2x) = (2x)^2 - 1.\]Simplify the expression:\[g(2x) = 4x^2 - 1.\]

Step 3: Apply the Function \(f(x)\)

Now substitute \(g(2x) = 4x^2 - 1\) into \(f(x)\). \[f(4x^2 - 1) = 2(4x^2 - 1) + 1.\]

Step 4: Simplify the Expression

Distribute and simplify the expression:\[f(4x^2 - 1) = 2 \cdot 4x^2 - 2 \cdot 1 + 1 = 8x^2 - 2 + 1.\]Combine like terms:\[f(4x^2 - 1) = 8x^2 - 1.\]

Step 5: Final Answer

The final result of \((f \circ g)(2x)\) is \(8x^2 - 1\).

Key Concepts

Intermediate Algebra

Intermediate Algebra is a branch of mathematics that focuses on the manipulation and combination of algebraic expressions and equations. It helps students to understand and perform various operations on algebraic expressions, including addition, subtraction, multiplication, division, and composition of functions. These operations allow students to solve complex mathematical problems and understand their real-world applications.

Function composition is a key concept in intermediate algebra. It involves creating a new function by combining two or more functions. For example, in the problem, we have two functions, \(f(x)=2x+1\) and \(g(x)=x^{2}-1\). To find \((f \circ g)(2x)\), we must first find \(g(2x)\), and then use that result as the input for \(f(x)\). This step-by-step approach to solving composition problems helps in visualizing and simplifying complex expressions.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables and coefficients. These functions are characterized by the use of non-negative integer exponents on the variables. They form a crucial part of algebra due to their wide-ranging applications.

In the given exercise, \(g(x)=x^2-1\) is a polynomial function. The degree of a polynomial is determined by the highest power of the variable. For \(g(x)\), the degree is 2 since the highest power of \(x\) is 2. Polynomial functions like \(g(x)\) can be combined with other functions, such as \(f(x)\), to form composite functions.\(f(x)\) is a linear function, as it has the form \(ax + b\). The process of function composition allows for intricate transformations and simplifications of expressions, such as \(g(2x) = 4x^2 - 1\).

Algebraic Expressions

Algebraic expressions are mathematical phrases that can consist of numbers, variables, and operation symbols. They are the building blocks of algebra and are used to represent relationships and perform computations.

In our exercise, when composing functions, expressions from \(f(x)\) and \(g(x)\) combine to form new expressions. Initially, \(g(2x)\) is calculated by substituting \(2x\) into \(g(x)\), yielding \(4x^2 - 1\). This new expression serves as input for \(f(x)\), leading to \(f(4x^2 - 1)\). By understanding how to manipulate and simplify algebraic expressions, students can break down complex problems into more manageable pieces.

• Substitution: Inserting one expression into another is a key operation.
• Simplification: Combining terms and using algebraic identities help to reduce expressions to their simplest form.
• Distribution: Key in simplifying expressions like \(2(4x^2 - 1) + 1\), resulting in the final simplification to \(8x^2 - 1\).

Mastering these concepts is essential for understanding more complex algebraic operations and equations.