Problem 53 - Elementary and Intermediate Algebra

Question

Find \(f(x)\) and \(g(x)\) such that \(h(x)=(f \circ g)(x) .\) Answers may vary. $$ h(x)=\sqrt{16 x-1} $$

Step-by-Step Solution

Answer

Choose \(f(x) = \sqrt{x}\) and \(g(x) = 16x - 1\). Then \((f \circ g)(x) = \sqrt{16x - 1} = h(x)\).

Step 1: Understanding Composition of Functions

The operation \( (f \circ g)(x) \) means \( f(g(x)) \). This implies that \( g(x) \) is first applied to \( x \), and then \( f \) is applied to the result of \( g(x) \). We need to find two functions \( f(x) \) and \( g(x) \) such that applying them in succession results in \( h(x) = \sqrt{16x-1} \).

Step 2: Identifying Possible Function for g(x)

Given the expression under the square root \( 16x - 1 \), a natural choice is to set \( g(x) = 16x - 1 \). This transformation simplifies what's inside the square root, which is necessary for the composite function.

Step 3: Determining the Function for f(x)

Now that we have \( g(x) = 16x - 1 \), we can express the function \( h(x) \) as \( \sqrt{g(x)} \). Therefore, \( f(x) = \sqrt{x} \) should be chosen because \( f(g(x)) = \sqrt{16x - 1} \).

Step 4: Verifying the Composition

Let's verify if \( h(x) = (f \circ g)(x) \) holds true: Start with \( g(x) = 16x - 1 \), then apply \( f(x) = \sqrt{x} \) on \( g(x) \). Substituting gives \( f(g(x)) = \sqrt{g(x)} = \sqrt{16x - 1} \), which matches \( h(x) \). Therefore, our choices of \( f(x) \) and \( g(x) \) are correct.

Key Concepts

Square Root Function

The square root function, mathematically represented as \(f(x) = \sqrt{x}\), is crucial in various mathematical operations and applications. At its core, this function returns the positive square root of a non-negative real number.

• The expression \(\sqrt{16x - 1}\) means finding the value that, when multiplied by itself, yields \(16x - 1\).
• In the context of our problem, where \(h(x) = \sqrt{16x - 1}\), it plays the role of an outer function, taking the result of another operation as its input.
• Understanding how the square root affects algebraic expressions is key to decomposing complex functions into compositions of simpler ones.

When working with square roots, remember that the operation assumes the principal (non-negative) square root when working with real numbers.

Inner Function

In the context of function composition, the inner function is the one that is applied first. With the problem at hand, the inner function is identified as \(g(x) = 16x - 1\).

• This function prepares the input for the outer function, setting the stage for further operations.
• By transforming \(x\) into \(16x - 1\), \(g(x)\) simplifies the expression so that the subsequent application of the square root function becomes straightforward.
• Understanding the role of the inner function is crucial, as it affects the domain and the overall behavior of the composition.

Prioritizing the inner calculation ensures that every subsequent step of function composition logically fits together, leading to the correct result.

Outer Function

The outer function in a composition, denoted as \(f(x) = \sqrt{x}\), operates on the result provided by the inner function.

• It acts as the final step in a series of transformations, refining the output into its ultimate form.
• In our specific example of \(h(x) = f(g(x))\), the outer function takes the result of \(g(x) = 16x - 1\) and applies the square root operation.
• This layering of functions is a hallmark of composition, allowing us to build complex operations from simpler, more manageable pieces.

By carefully selecting and ordering these functions, we ensure the desired transformation is achieved—exemplified by rewriting \(h(x)\) as a composition of \(f\) and \(g\). This approach illuminates the versatility and power of function composition in mathematical modeling.