Problem 6
Question
Find functions \(f\) and \(g\) such that the given function is the composition \(f(g(x))\). $$ \frac{\sqrt{x}-1}{\sqrt{x}+1} $$
Step-by-Step Solution
Verified Answer
Functions are \(f(x) = \frac{x-1}{x+1}\) and \(g(x) = \sqrt{x}\).
1Step 1: Identify the Inner Function g(x)
The first step in finding the composition of functions is to identify the potential inner function, \(g(x)\). A natural choice here is to work with the expression under the radical. Set \(g(x) = \sqrt{x}\). This function maps \(x\) to \(\sqrt{x}\), which is exactly what we see in the components of the outer function.
2Step 2: Substitute g(x) into the Given Function
With \(g(x) = \sqrt{x}\), substitute this into the original expression \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\) to achieve a new expression purely in terms of \(g(x)\). This yields the expression \(\frac{g(x)-1}{g(x)+1}\).
3Step 3: Define the Outer Function f(x)
Determine the outer function, \(f(x)\), that when applied to \(g(x)\) will result in the original function. Notice that \(\frac{g(x)-1}{g(x)+1}\) can be defined as \(f(x) = \frac{x-1}{x+1}\). This function, \(f(x)\), takes the form \(\frac{x-1}{x+1}\) when \(x = g(x)\).
4Step 4: Verify the Composition
Check the composition \(f(g(x)) = \frac{g(x) - 1}{g(x) + 1}\) to ensure it matches the original function \(\frac{\sqrt{x}-1}{\sqrt{x}+1}\). Substituting \(g(x) = \sqrt{x}\) into \(f(g(x))\) results in \(\frac{\sqrt{x} - 1}{\sqrt{x} + 1}\), confirming correctness.
Key Concepts
Inner Function IdentificationOuter Function DefinitionComposition Verification
Inner Function Identification
When you first encounter a function composition problem, your task is to break down a complex function into two simpler functions: an inner function and an outer function. The inner function, often notated as \( g(x) \), is a simpler component within the more complex original function. The key here is to look for parts of the function that can be isolated, such as a base operation or transformation.
In this exercise, we started with the function \( \frac{\sqrt{x}-1}{\sqrt{x}+1} \). The presence of the square root of \( x \) is a clear indicator of a potential candidate for the inner function. By setting the inner function as \( g(x) = \sqrt{x} \), we simplify the expression. This choice leverages the operation within the square root, which is present in both the numerator and the denominator.
Identifying the inner function correctly is crucial because it influences the rest of the process, including the definition of the outer function. Always look for transformations or specific operations as cues.
In this exercise, we started with the function \( \frac{\sqrt{x}-1}{\sqrt{x}+1} \). The presence of the square root of \( x \) is a clear indicator of a potential candidate for the inner function. By setting the inner function as \( g(x) = \sqrt{x} \), we simplify the expression. This choice leverages the operation within the square root, which is present in both the numerator and the denominator.
Identifying the inner function correctly is crucial because it influences the rest of the process, including the definition of the outer function. Always look for transformations or specific operations as cues.
Outer Function Definition
Once you have identified the inner function, the next step is to define the outer function, denoted as \( f(x) \). The purpose of the outer function is to work with the output of the inner function to recreate the original complex expression. Essentially, \( f(x) \) "wraps around" \( g(x) \).
In our example, after setting \( g(x) = \sqrt{x} \), the expression becomes \( \frac{g(x)-1}{g(x)+1} \). This transformation allows us to consider what the outer function must be to arrive back at the original expression using \( g(x) \).
Thus, we define the outer function as \( f(x) = \frac{x-1}{x+1} \). This formulation uses simple algebraic manipulation, operating directly on the variable \( x \), which represents \( g(x) \). The outer function captures the structure of \( g(x) \) as it appears in the original expression.
In our example, after setting \( g(x) = \sqrt{x} \), the expression becomes \( \frac{g(x)-1}{g(x)+1} \). This transformation allows us to consider what the outer function must be to arrive back at the original expression using \( g(x) \).
Thus, we define the outer function as \( f(x) = \frac{x-1}{x+1} \). This formulation uses simple algebraic manipulation, operating directly on the variable \( x \), which represents \( g(x) \). The outer function captures the structure of \( g(x) \) as it appears in the original expression.
Composition Verification
Verifying the composition of the two functions ensures that they indeed recreate the original function. This is where you test your choices for \( f(x) \) and \( g(x) \) by performing the composition \( f(g(x)) \) and checking if it matches the given function.
For our problem, substituting \( g(x) = \sqrt{x} \) into \( f(x) = \frac{x-1}{x+1} \) gives us \( f(g(x)) = \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \). Comparing this with the original function \( \frac{\sqrt{x}-1}{\sqrt{x}+1} \), they perfectly match.
For our problem, substituting \( g(x) = \sqrt{x} \) into \( f(x) = \frac{x-1}{x+1} \) gives us \( f(g(x)) = \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \). Comparing this with the original function \( \frac{\sqrt{x}-1}{\sqrt{x}+1} \), they perfectly match.
- First, ensure \( g(x) \) correctly represents the parts of the function under transformation.
- Second, check that \( f(x) \) operates on those parts to reproduce the original function.
Other exercises in this chapter
Problem 5
Use the definition of the derivative to show that the following functions are not differentiable at \(x=0\). \(f(x)=|2 x|\)
View solution Problem 5
Find the derivative of each function. $$ f(x)=x^{1 / 2} $$
View solution Problem 6
Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(x)=x^{3}\left(x^{2}+1\right) $$
View solution Problem 6
Use the definition of the derivative to show that the following functions are not differentiable at \(x=0\). \(f(x)=|3 x|\)
View solution