Problem 91 - Precalculus : Building Concepts and Connections

Question

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=\frac{1}{2 x+5}$$

Step-by-Step Solution

Answer

Thus, the two functions \(f\) and \(g\) that make up the function \(h(x)\) are \(f(x) = \frac{1}{x}\) and \(g(x) = 2x+5\).

Step 1: Identify potential functions

We first identify two potential functions that when composed will produce \(h(x)\). A possible choice could be \(f(x) = \frac{1}{x}\) and \(g(x) = 2x+5\). The function \(g(x)\) goes inside the function \(f(x)\) because when two functions are composed, the inside function is processed first.

Step 2: Check the composition

We now need to check if our functions are correct. We can evaluate the composition \(f(g(x))\), using the functions we chose in step 1. Substituting \(g(x)\) into \(f(x)\), we get \(f(g(x)) = f(2x+5) = \frac{1}{2x+5}\).

Step 3: Confirm the result

Since the resulting function \(\frac{1}{2x+5}\) matches the given function \(h(x)\), we can confirm that our choice of functions \(f\) and \(g\) was correct.

Key Concepts

Composite Functions

Understanding composite functions is a fundamental aspect of precalculus, opening the door to more complex mathematical analysis. A composite function involves chaining two functions together, where the output of one becomes the input for the other. In essence, you are 'nesting' one function inside another. This operation is denoted as \(f \circ g\) which is read as 'f composed with g', and it is defined as \(f(g(x))\).

Let's break this down using a practical example from our textbook exercise. Suppose \(h(x) = \frac{1}{2x+5}\), and we want to express \(h\) as a composition of two functions, \(f\) and \(g\). A sensible approach is to pick \(g(x) = 2x+5\) as the inner function because this expression directly matches the denominator in \(h(x)\). Consequently, we let \(f(x) = \frac{1}{x}\) because it represents the reciprocal operation that we need to apply to the result of \(g(x)\).

By inserting \(g(x)\) into \(f(x)\), we complete the composition: \(f(g(x)) = f(2x+5) = \frac{1}{2x+5}\), hence \(f \circ g\) indeed equals \(h(x)\). It's crucial to approach composite functions methodically; identify the inner and outer functions first before attempting to combine them. Through this process, it becomes easier to dissect complex functions into simpler, more manageable pieces.

Function Operations

Function operations include various ways in which functions can be combined or manipulated to create new functions. In addition to composition, there are other operations such as addition, subtraction, multiplication, and division of functions. These are performed element-wise, meaning for any two functions \(f(x)\) and \(g(x)\), the operations produce new functions as follows:

• Addition: \( (f + g)(x) = f(x) + g(x) \)
• Subtraction: \( (f - g)(x) = f(x) - g(x) \)
• Multiplication: \( (f * g)(x) = f(x) * g(x) \)
• Division: \( (f / g)(x) = f(x) / g(x), \; provided \; g(x) \eq 0 \)

Understanding these operations allows students to simplify complex expressions, solve equations involving functions, and analyze the behaviors of new functions formed from familiar ones. Practical exercises, like the one provided in our textbook solution, encourage a hands-on approach to mastery, showing that these operations can be straightforward, as long as the basic rules are methodically applied.

Inverse Functions

The notion of inverse functions is another core concept in precalculus and is intertwined with the idea of function composition. In functional terms, an inverse function essentially undoes what the original function does. Mathematically, if you have a function \(f(x)\), its inverse, denoted as \(f^{-1}(x)\), will satisfy the condition \(f(f^{-1}(x)) = x\) and vice versa, \(f^{-1}(f(x)) = x\).

Hence, an inverse function is like a two-way street, allowing you to go back and forth between inputs and outputs. However, not all functions have inverses that are also functions. For a function to have an inverse, it must be bijective — both injective (one-to-one) and surjective (onto).

Taking our example of \(h(x) = \frac{1}{2x+5}\), finding an inverse would involve swapping the \(x\)'s and \(y\)'s and solving for the new \(y\). This can get a bit complex, but in simpler cases, the process gives insight into how a function behaves and lends itself to solving equations. Discovering the inverse of a function often reveals a symmetry, reflective across the line \(y = x\), which can be profoundly useful in graphical interpretations and problem-solving scenarios.