/*! 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 49 Let \(f(x)=\frac{1}{x}+x\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(f(x)=\frac{1}{x}+x\) and \(g(x)=\frac{2 x}{x^{2}+1} .\) Evaluate the following expressions. (a) \((f \circ f)(x)\) (b) \((f \circ f \circ f)(x)\)

Short Answer

Expert verified
(a) \((f \circ f)(x)\) simplifies to \(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\), and (b) \((f \circ f \circ f)(x)\) simplifies to \(\frac{x}{x^4 + 2x^2 + 1} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\)

Step by step solution

01

(a) Calculate \((f \circ f)(x)\)

First, let's find \((f \circ f)(x)\). This means that wherever we see an \(x\) in the function \(f(x)\), we need to instead put the whole function \(f(x)\). So,\[(f \circ f)(x) = f(f(x)) = f\left(\frac{1}{x}+x\right) = \frac{1}{(\frac{1}{x}+x)} + \frac{1}{x} + x.\]
02

Simplifying \((f \circ f)(x)\)

We now simplify this expression:\[\frac{1}{(\frac{1}{x}+x)} + \frac{1}{x} + x = \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]
03

(b) Calculate \((f \circ f \circ f)(x)\)

The term \((f \circ f \circ f)(x)\) means that we compose the function \(f(x)\) with itself twice. Now that we have \((f \circ f)(x)\), we can express this as \(f((f \circ f)(x))\), this means that wherever we see an \(x\) in the function \(f(x)\), we need to put the whole function \((f \circ f)(x)\):\[(f \circ f \circ f)(x) = f((f \circ f)(x)) = f\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right) = \frac{1}{\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right)} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]
04

Simplifying \((f \circ f \circ f)(x)\)

We now simplify this expression:\[\frac{1}{\left(\frac{x}{1 + x^{2}} + \frac{1}{x} + x\right)} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x = \frac{x}{x^4 + 2x^2 + 1} + \frac{x}{1 + x^{2}} + \frac{1}{x} + x.\]

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Key Concepts

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

Function Composition
Function composition is like stacking functions on top of one another. It involves inserting one function inside another. For example, when we talk about \(f \circ g\)(x), it means "apply function \(g\) to \(x\) first, and then apply function \(f\) to the result." In the context of the problem, we had functions \(f(x)\) and \(g(x)\). The exercise asked us to find \( (f \circ f)(x)\) and \( (f \circ f \circ f)(x)\). "(f ∘ f)(x)" means applying the function \(f\) to itself. More simply, it means taking the result of \(f(x)\) and plugging it back into \(f(x)\). So, you're essentially using the output of a function as the input for itself. This is a creative aspect of math that teaches us how functions can be reused in different ways.
To break it down further:
  • The process begins with applying the given function to an initial value or expression.
  • Then, the output of that first calculation becomes the new input for the next layer of the function application.
  • This can be layered multiple times to create more complex computations.
Understanding this concept helps one see how various mathematical operations interact with each other, which is crucial in advanced levels of math.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging and simplifying expressions to make calculations more manageable. When working with compositions like \( (f \circ f)(x)\), simplification is necessary to achieve a clearer form.
In the given solution, after using function composition, the result included several fractions. Algebraic manipulation allows us to simplify these into a neater form by following a few steps:
  • Combining like terms: Look for terms in the expression that can be combined or simplified.
  • Fraction simplification: When dealing with complex fractions, rewrite them as simpler expressions.
  • Canceling terms: If possible, cancel terms that appear in both the numerator and denominator to make things simpler.
Through these steps, rationalizing the initial composition leads to a more readable and usable form, which is crucial for further applications or additional operations.
Rational Functions
Rational functions are expressed as fractions where both the numerator and the denominator are polynomials. In this exercise, both given functions, \(f(x)\) and \(g(x)\), qualify as rational functions. To solve problems involving these functions, one must understand how to manage and manipulate their expressions.
Here are some key points about working with rational functions:
  • Identifying the function: Make sure to recognize that a function is rational if it can be written as a fraction of two polynomials.
  • Domain concerns: Rational functions can have restrictions on their domains, primarily where the denominator is zero. Always consider where the expression becomes undefined.
  • Simplification: Like with algebraic manipulation, simplifying rational functions can sometimes allow you to see solutions more clearly through reduced and manageable expressions.
These concepts are foundational in math, providing a thorough understanding of how functions behave and interact through compositions and manipulations.

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Most popular questions from this chapter

A photocopying shop has a xed cost of operation of $$\$ 4000$$ per month. In addition, it costs them $$\$ 0.01$$ per page they copy. They charge customers $$\$ 0.07$$ per page. (a) Write a formula for \(R(x)\), the shop s monthly revenue from making \(x\) copies. (b) Write a formula for \(C(x)\), the shop s monthly costs from making \(x\) copies. (c) Write a formula for \(P(x)\), the shop s monthly pro t (or loss if negative) from making \(x\) copies. Pro \(\mathrm{t}\) is computed by subtracting total costs from the total revenue. (d) How many copies must they make per month in order to break even? Breaking even means that the pro \(t\) is zero; the total costs and total revenue are equal. (e) Sketch \(C(x), R(x)\), and \(P(x)\) on the same set of axes and label the break-even point. (f) Find a formula for \(A(x)\), the shop s average cost per copy. (g) Make a table of \(A(x)\) for \(x=0,1,10,100,1000,10000\). (h) Sketch a graph of \(A(x)\).

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Most of the time, when a store provides coupons offering $$\$ 5$$ off any item in the store they include the clause except for sale items. Suppose that clause were omitted and you found an item you wanted on a \(30 \%\) off rack. There would be some ambiguity; should the $$\$ 5$$ be taken off the reduced price, or off the price before the \(30 \%\) discount? Let \(C\) be the function that models the effect of the coupon, \(S\) be the function that models the effect of the sale, and \(x\) be the original price of the item. (a) Which situation corresponds to \(C(S(x))\) ? (b) Which situation corresponds to \(S(C(x))\) ? (c) Which order of composition of the functions is in the buyer s favor?

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