Chapter 2: Problem 8
Find functions \(f\) and \(g\) such that the given function is the composition \(f(g(x))\). $$\sqrt{\frac{x-1}{x+1}}$$
Short Answer
Expert verified
The functions are \( f(u) = \sqrt{u} \) and \( g(x) = \frac{x-1}{x+1} \).
Step by step solution
01
Identify the Inner Function
To express the given function as a composition, recognize that the most inner operation in the expression should correspond to the inner function. Here, the fraction inside the square root is \( \frac{x-1}{x+1} \). Let this be our inner function \( g(x) = \frac{x-1}{x+1} \).
02
Identify the Outer Function
Based on our choice of the inner function, the outer function should be what remains if we replace \( \frac{x-1}{x+1} \) with a single variable, typically \( u \), resulting in the operation \( \sqrt{u} \). Therefore, the outer function is \( f(u) = \sqrt{u} \).
03
Check the Composition
Verify that composing \( f \) and \( g \) will indeed reconstruct the original function. Compute \( f(g(x)) = f\left(\frac{x-1}{x+1}\right) = \sqrt{\frac{x-1}{x+1}} \), which matches the original function given in the exercise.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inner Function
When we talk about function composition, the **inner function** is the first operation applied to our input. Think of it as the "starting point" in a sequence of transformations. In this exercise, the **inner function** is the part of the expression that lies within the square root: \( \frac{x-1}{x+1} \). This can be written as \( g(x) = \frac{x-1}{x+1} \).
Understanding the inner function involves recognizing what operation the expression is primarily carrying out on the input \( x \). Here, it is subtracting 1 from \( x \) and then dividing the result by \( x+1 \). Each bit of this expression plays a part:
Understanding the inner function involves recognizing what operation the expression is primarily carrying out on the input \( x \). Here, it is subtracting 1 from \( x \) and then dividing the result by \( x+1 \). Each bit of this expression plays a part:
- **Numerator**: \( x-1 \) - reducing the value of \( x \) by 1.
- **Denominator**: \( x+1 \) - modifying the result with a division by \( x+1 \).
Outer Function
The **outer function** is the operation applied last, wrapping the final result of the inner function's calculation. If you imagine transforming \( x \) step-by-step, the outer function is the final flourish. In this case, it's the operation that takes the result of \( g(x) \) and processes it - the square root.
The outer function can be expressed as \( f(u) = \sqrt{u} \), where \( u \) represents the result from the inner function \( g(x) \). Here's why the outer function is important:
The outer function can be expressed as \( f(u) = \sqrt{u} \), where \( u \) represents the result from the inner function \( g(x) \). Here's why the outer function is important:
- **Encapsulation**: It effectively "wraps" the results of the inner function, defining the final form of the output.
- **Flexibility**: You can swap the inner function without affecting the structure of the outer function, proving its independence in the composition.
Square Root
The **square root** function is a core mathematical operation featured prominently in this exercise as the outer part of the function composition. It is denoted by \( \sqrt{} \), which signifies finding a number that, when multiplied by itself, gives the original number inside the root symbol.
For example, the square root of 9 is 3, since \( 3 \times 3 = 9 \). Understanding the square root is key in many areas of mathematics and science because:
For example, the square root of 9 is 3, since \( 3 \times 3 = 9 \). Understanding the square root is key in many areas of mathematics and science because:
- **Simplifies Expressions**: The square root operation can take complex numbers resulting from the inner function and reduce their complexity.
- **Graphical Representation**: Being able to visualize square roots on a graph helps in understanding shifts and transformations.