Problem 30 - Algebra for College Students

Question

If \(f(x)=1 / x\) and \(g(x)=2 x+1\), find \((f \circ g)(1)\) and \((g \circ f)(2) . \quad \frac{1}{3} ; 2\)

Step-by-Step Solution

Answer

\((f \circ g)(1)=\frac{1}{3}; (g \circ f)(2)=2\).

Step 1: Understand Composition of Functions

To find \((f \circ g)(x)\), we apply the function \(g(x)\) first and then apply \(f(x)\) to the result: \((f \circ g)(x) = f(g(x))\). For \((g \circ f)(x)\), we apply \(f(x)\) first and then \(g(x)\) to the result: \((g \circ f)(x) = g(f(x))\).

Step 2: Calculate \((f \circ g)(1)\)

First, calculate \(g(1)\) using \(g(x) = 2x + 1\).\[ g(1) = 2(1) + 1 = 3 \]Now, substitute \(g(1)\) into \(f(x) = \frac{1}{x}\):\[ f(3) = \frac{1}{3} \]Thus, \((f \circ g)(1) = \frac{1}{3}\).

Step 3: Calculate \((g \circ f)(2)\)

First, calculate \(f(2)\) using \(f(x) = \frac{1}{x}\).\[ f(2) = \frac{1}{2} \]Now, use the result in \(g(x) = 2x + 1\):\[ g\left(\frac{1}{2}\right) = 2 \left(\frac{1}{2}\right) + 1 = 2 \]Thus, \((g \circ f)(2) = 2\).

Key Concepts

Function Notation

Function notation is a unique way to denote and evaluate functions easily and consistently. When we write a function, such as \(f(x)\), we specify the rule for the function and identify it with a small letter (often \(f\), \(g\), or \(h\)). Here, \(x\) denotes the input value of the function. This means that the output of the function is determined based on \(x\).

Understanding function notation is key when evaluating and manipulating functions, especially when dealing with multiple functions at a time. For example, if we have two functions \(f(x) = \frac{1}{x}\) and \(g(x) = 2x + 1\), reading \(f(g(x))\) indicates a composition where we substitute the entire function \(g(x)\) into \(f(x)\).

Function notation helps us not only write these functions concisely but also makes it clearer and formal to work with them, especially when solving problems that involve more complex operations like compositions.

Algebraic Functions

Algebraic functions are functions that involve operations such as addition, subtraction, multiplication, division, and taking roots of polynomials. These are the basic building blocks of most mathematical models and are essential in expressing equations and relationships among variables.

For instance, in the exercise provided, we have two algebraic functions: \(f(x) = \frac{1}{x}\) and \(g(x) = 2x + 1\).

• \(f(x) = \frac{1}{x}\) defines a simple rational function, which involves division.
• \(g(x) = 2x + 1\) represents a linear function, characterized by its constant rate of change.

Algebraic functions are common in textbook exercises as they lay the foundation for understanding how equations describe relationship patterns. When you compose such functions, it deepens your algebraic skills as you substitute one function into another and simplify.

Function Evaluation

Function evaluation involves determining the output of a function given an input value. This essential process involves substituting the specific input into the function's formula to find the corresponding output.

In the example, to compute \((f \circ g)(1)\), you evaluate \(g(1)\) first. Given \(g(x) = 2x + 1\), substituting 1 gives us \(g(1) = 3\). Next, substitute this result into \(f(x) = \frac{1}{x}\), finding \(f(3) = \frac{1}{3}\). Thus, \((f \circ g)(1) = \frac{1}{3}\).

Similarly, to find \((g \circ f)(2)\), start with \(f(2)\).

• With \(f(x) = \frac{1}{x}\), substituting 2 yields \(f(2) = \frac{1}{2}\).
• Then, apply this in \(g(x) = 2x + 1\) to obtain \(g\left(\frac{1}{2}\right) = 2\).

Function evaluation is crucial in applying function compositions and verifying results accurately. It showcases the dynamic interaction of multiple algebraic expressions through substitution.