/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Express \(f\) as a composition o... [FREE SOLUTION] | 91Ó°ÊÓ

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Express \(f\) as a composition of two functions; that is,find \(g\) and \(h\) such that \(f=g \circ h .[\) Note: Each exercise has more than one solution. \(]\) $$ \text { (a) } f(x)=\frac{1}{1-x^{2}} \quad \text { (b) } f(x)=|5+2 x| $$

Short Answer

Expert verified
For (a), let \(g(u) = \frac{1}{1-u}\) and \(h(x) = x^2\). For (b), let \(g(u) = |u|\) and \(h(x) = 5 + 2x\).

Step by step solution

01

Identify Inner Function h(x) for Part (a)

For function \(f(x) = \frac{1}{1-x^2}\), we start by determining \(h(x)\). A natural choice for \(h(x)\) is the expression inside the denominator. Let's set \(h(x) = x^2\), which represents the simpler operation occurring first.
02

Determine Outer Function g(x) for Part (a)

Now that we have \(h(x) = x^2\), we need a function \(g(u)\) where \(u = h(x)\). The function \(g\) should output the entire function \(f\) when given \(h(x)\). Thus, let \(g(u) = \frac{1}{1-u}\). This ensures \(g(x^2) = f(x)\).
03

Identify Inner Function h(x) for Part (b)

For function \(f(x) = |5 + 2x|\), we consider simplifying the content of the absolute value expression first. Choose \(h(x) = 5 + 2x\) as it represents the inner modification before taking the absolute value.
04

Determine Outer Function g(x) for Part (b)

Knowing \(h(x) = 5 + 2x\), we need \(g(u)\) such that it performs the absolute value operation. Set \(g(u) = |u|\). This makes sure that \(g(h(x)) = f(x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inner Function
An "inner function" is the first step in composing two functions. When you look at a function and identify it as a composite (a combination) of functions, the inner function is the one that is applied initially. This function holds the variable expressions that become the input for the next function, known as the outer function.

To illustrate, consider the function from Part (a):
  • Function: \[f(x) = \frac{1}{1-x^2} \]
Here, we set our inner function as \(h(x) = x^2\). This choice is made because \(x^2\) lies inside the denominator of the rational function. It's the first modification happening to the input \(x\).In Part (b), the function involves an absolute value:
  • Function: \[f(x) = |5 + 2x| \]
The inner function here is chosen to be \(h(x) = 5 + 2x\). The arithmetic inside the absolute value is the inner operation, simplifying it before applying the absolute value as the outer function.
Outer Function
Once the inner function has been dealt with, the "outer function" is the next in line to be applied. The function composition process involves taking the result of the inner function and inserting it into the outer function.For Part (a), after \(h(x) = x^2\) is determined, the outer function needs to take this output and complete the original function:
  • Outer Function: \[g(u) = \frac{1}{1-u} \]
This leads us to fulfil the requirement \(g(h(x)) = f(x)\), which effectively reconstructs the original function \(f(x)\) using \(g(u)\).In Part (b), we set \(g(u) = |u|\) as the outer function:
  • Outer Function: \[g(u) = |u|\]
By following through with \(g(h(x))\), it ensures that the absolute value operation is the final process applied to \(x\), completing the reconstruction of \(f(x)\). This concept allows for clear understanding and manipulation of composed functions.
Absolute Value Function
The "absolute value function" concerns itself with altering any given real number by removing its sign and considering only its magnitude. This turns any negative input into a positive counterpart, leaving positive inputs unchanged.In Part (b) of the exercise, the expression \(f(x) = |5 + 2x|\) uses an absolute value function to highlight how it impacts the expression it contains. Here, the function \(g(u) = |u|\) is reserved for this purpose. Once we determine that \(h(x) = 5 + 2x\), applying \(g(u)\) finalizes the operation by taking every outcome of \(h(x)\) and converting it into its absolute value, effectively calculating \(f(x)\).Key features of the absolute value function include:
  • Non-negative results: The function does not allow negative values in its output.
  • Simplicity in computation: Easy application across both algebraic and graphical interpretations.
  • Used frequently in equations and inequalities: Offers a strategic technique to simplify expressions involving negative terms.
Understanding the absolute value function's role in compositions is crucial as it sets a defining condition for many mathematical solutions.

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Most popular questions from this chapter

Sketch the graph of the equation by making appropriate transformations to the graph of a basic power function. If you have a graphing utility, use it to check your work. $$ \begin{array}{ll}{\text { (a) } y=1-\sqrt{x+2}} & {\text { (b) } y=1-\sqrt[3]{x+2}} \\ {\text { (c) } y=\frac{5}{(1-x)^{3}}} & {\text { (d) } y=\frac{2}{(4+x)^{4}}}\end{array} $$

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