Problem 25 - Algebra and Trigonometry

Question

Use \(f(x)=3 x-5\) and \(g(x)=2-x^{2}\) to evaluate the expression. $$ \begin{array}{ll}{\text { (a) }(f \circ g)(x)} & {\text { (b) }(g \circ f)(x)}\end{array} $$

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = 1 - 3x^2\) (b) \((g \circ f)(x) = -9x^2 + 30x - 23\).

Step 1: Understanding Composition of Functions

When we have two functions, say \(f(x)\) and \(g(x)\), the composition \((f \circ g)(x)\) means applying \(g(x)\) first and then applying \(f\) to the result of \(g(x)\). Conversely, \((g \circ f)(x)\) means applying \(f(x)\) first and then \(g\) to the result of \(f(x)\). Our task is to find these compositions.

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

To find \((f \circ g)(x)\), we first substitute \(g(x)\) into \(f\). Since \(f(x) = 3x - 5\) and \(g(x) = 2 - x^2\), \((f \circ g)(x)\) will be \(f(g(x)) = f(2 - x^2)\). Substitute \(2 - x^2\) into \(f(x)\) to get:\[ f(2 - x^2) = 3(2 - x^2) - 5 \]Calculate:\[ = 3 \times 2 - 3 \times x^2 - 5 \]\[ = 6 - 3x^2 - 5 \]\[ = 1 - 3x^2 \]Thus, \((f \circ g)(x) = 1 - 3x^2\).

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

To find \((g \circ f)(x)\), we first substitute \(f(x)\) into \(g\). Given \(g(x) = 2 - x^2\) and \(f(x) = 3x - 5\), \((g \circ f)(x)\) becomes \(g(f(x)) = g(3x - 5)\). Substitute \(3x - 5\) into \(g(x)\) to get:\[ g(3x - 5) = 2 - (3x - 5)^2 \]Calculate:\[ = 2 - (3x - 5)(3x - 5) \]\[ = 2 - (9x^2 - 30x + 25) \]\[ = 2 - 9x^2 + 30x - 25 \]\[ = -9x^2 + 30x - 23 \]Thus, \((g \circ f)(x) = -9x^2 + 30x - 23\).

Key Concepts

Algebra

Algebra is a branch of mathematics that uses symbols, usually letters, to represent numbers in expressions and equations. One of its primary functions is to enable the manipulation of these expressions to solve equations or to understand variable relationships. In the context of functions, algebra becomes crucial for the manipulation and transformation of mathematical expressions.

• Operations: Algebra focuses on performing operations, such as addition, subtraction, multiplication, and division, using algebraic expressions. When dealing with function composition, these operations are necessary to properly evaluate and simplify expressions.
• Substitution: It involves replacing a variable within an equation or function with another expression. In evaluating composed functions like \(f(g(x))\), substitution illustrates that one function's output becomes the input for another.

To solve exercises using algebra, a thorough understanding of the order of operations, known by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction), is vital. It ensures that you simplify expressions correctly, especially when dealing with composite functions.

Function Evaluation

Function evaluation simply means substituting a particular value for the variable in a function. In more complex terms, it often involves evaluating several layers of functions, like in function composition. For our exercise, we evaluate two kinds of compositions: \(f \circ g\) and \(g \circ f\).

• Substituting \(g(x)\) into \(f(x)\): To find \(f(g(x))\), replace \(x\) in \(f(x)\)'s formula with the entire function \(g(x)\). Given \(f(x) = 3x - 5\) and \(g(x) = 2 - x^2\), we substitute into \(f\) to get \(f(2-x^2) = 1 - 3x^2\).
• Substituting \(f(x)\) into \(g(x)\): Similarly, for \(g(f(x))\), replace \(x\) in \(g(x)\) with \(f(x)\). Following the same process gives us \(g(3x-5) = -9x^2 + 30x - 23\).

Function evaluation in composition tasks requires careful substitution and simplification to ensure accuracy. Checking your work by re-evaluating the derived function is a good practice to confirm the solution.

Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of powers of variables with coefficients. They are defined as \(a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where \(a_n, a_{n-1}, \ldots, a_0\) are constants and \(n\) is a non-negative integer. In our context of function composition, both \(f(x)\) and \(g(x)\) are polynomials.

• Linear function: \(f(x) = 3x - 5\) is a linear polynomial because its highest degree is one (x\^1).
• Quadratic function: \(g(x) = 2 - x^2\) is a quadratic polynomial as it includes \(x^2\), a term with the degree of two.

When we evaluate \(f \circ g\) or \(g \circ f\), we end up with polynomial expressions. The degree of the resulting polynomial may change depending on the combination of operations and the degree of input functions. Properly simplifying these polynomials to reach an extracted polynomial function is essential to retrieving the accurate output for any value of \(x\). Breaking down a polynomial into its terms and simplifying by grouping like terms is part of the critical understanding in algebra.