Problem 22 - College Algebra

Question

Given the functions \(f(x)=\frac{1-x}{x}\) and \(g(x)=\frac{1}{1+x^{2}}\), fi d the following: a. \((g \circ f)(x)\) b. \((g \circ f)(2)\)

Step-by-Step Solution

Answer

(g \circ f)(x) = \frac{x^2}{2x^2 - 2x + 1}; (g \circ f)(2) = \frac{4}{5}.

Step 1: Understand function composition

Function composition involves creating a new function by applying one function to the results of another function. The notation \((g \circ f)(x)\) represents the composition of functions, where function \(f\) is applied first and then function \(g\) is applied to the result of \(f\).

Step 2: Substitute \(f(x)\) into \(g(x)\)

Given \(f(x) = \frac{1-x}{x}\), substitute this expression into the function \(g(x) = \frac{1}{1+x^2}\). This gives us:\(g(f(x)) = g\left(\frac{1-x}{x}\right) = \frac{1}{1 + \left(\frac{1-x}{x}\right)^2}\).

Step 3: Simplify the expression for \((g \circ f)(x)\)

First, simplify \(\left(\frac{1-x}{x}\right)^2 = \frac{(1-x)^2}{x^2}\). Substitute this back into \(g(f(x))\):\[g(f(x)) = \frac{1}{1 + \frac{(1-x)^2}{x^2}}\].Combine the terms in the denominator into a single fraction:\[g(f(x)) = \frac{1}{\frac{x^2 + (1-x)^2}{x^2}}\].Invert this fraction to simplify:\[g(f(x)) = \frac{x^2}{x^2 + (1-x)^2}\].Finally, expand \((1-x)^2 = 1 - 2x + x^2\), thus:\[g(f(x)) = \frac{x^2}{x^2 + 1 - 2x + x^2} = \frac{x^2}{2x^2 - 2x + 1}\].

Step 4: Evaluate \((g \circ f)(2)\) using the simplified expression

Use the expression from the previous step \(\frac{x^2}{2x^2 - 2x + 1}\) and plug in \(x = 2\):\[g(f(2)) = \frac{2^2}{2(2^2) - 2(2) + 1}\].This simplifies to:\[g(f(2)) = \frac{4}{2 \cdot 4 - 4 + 1} = \frac{4}{8 - 4 + 1} = \frac{4}{5}\].

Key Concepts

Composite Functions

Composite functions are a fundamental concept in mathematics, especially in calculus and algebra. They involve combining two functions in a specific way. This idea is central when dealing with complex mathematical models or solving equations.

To understand composite functions, consider two functions, say, \(f(x)\) and \(g(x)\). The composition of these functions is denoted as \((g \circ f)(x)\). More intuitively, this means you first apply \(f\) to \(x\) and then apply \(g\) to the result from \(f\).

Let's see how this looks in practice. If \(f(x) = \frac{1-x}{x}\) and \(g(x) = \frac{1}{1+x^{2}}\), the composite function is:\[(g \circ f)(x) = g(f(x)) = g\left(\frac{1-x}{x}\right)\].

This layered approach to function composition is useful when tackling problems that require a sequence of transformations or computations.

Function Evaluation

Function evaluation is about substituting a given value into a function to determine its output. This process is straightforward but critical in mathematics.

In the context of composite functions, function evaluation first requires substituting into the inner function before moving to the outer function. In our example, to evaluate \((g \circ f)(2)\), we first substitute \(x = 2\) into the inner function \(f(x)\), and then use that result in the outer function \(g(x)\).

Start by calculating \(f(2)\) for \(f(x) = \frac{1-x}{x}\):

• \(f(2) = \frac{1-2}{2} = \frac{-1}{2}\)

Next, substitute this into \(g\):

• \(g(f(2)) = g\left(\frac{-1}{2}\right)\)

Finally, we use the derived expression from the composition and plug in the value.

This multi-step substitution process highlights the depth of understanding required when evaluating composite functions.

Algebraic Simplification

Algebraic simplification involves reducing mathematical expressions to their simplest form. This process makes computations more manageable and results easier to interpret.

When working with complex functions or composite functions, simplifying expressions is a crucial step. For instance, in the given composite function \((g \circ f)(x)\), simplify the expression \(\frac{1}{1 + \left(\frac{1-x}{x}\right)^2}\) by working through the algebra:

• First, simplify \(\left(\frac{1-x}{x}\right)^2 = \frac{(1-x)^2}{x^2}\).
• Combine this back into the function \(g\), which yields a single fraction.
• Further simplify by inverting and expanding terms, such as \((1-x)^2 = 1 - 2x + x^2\).

This results in the expression \(\frac{x^2}{2x^2 - 2x + 1}\), a much more manageable form. Algebraic simplification is instrumental for efficient function evaluation and understanding the behavior of complex expressions.