/*! 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 24 Determine whether the function i... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the function is one-to-one. $$f(x)=\frac{1}{x}$$

Short Answer

Expert verified
The function \(f(x) = \frac{1}{x}\) is one-to-one.

Step by step solution

01

Understand the Definition of One-to-One Function

A function \(f(x)\) is considered one-to-one if, for every \(a\) and \(b\) in its domain, \(f(a) = f(b)\) implies that \(a = b\). In other words, different inputs (\(a\) and \(b\)) must yield different outputs for the function to be one-to-one.
02

Write the Equation for Function Values

To determine if \(f(x) = \frac{1}{x}\) is one-to-one, assume \(f(a) = f(b)\). Thus, we have:\[\frac{1}{a} = \frac{1}{b}\] and we need to see if this implies \(a = b\).
03

Solve the Equation for Equality

Starting from the equation \(\frac{1}{a} = \frac{1}{b}\), multiply both sides by \(a \cdot b\) to eliminate the fractions:\[a \cdot \frac{1}{a} = b \cdot \frac{1}{b}\]This simplifies to \(b = a\). Thus, if \(f(a) = f(b)\), then it must follow that \(a = b\).
04

Check the Domain for Valid Inputs

Consider the domain of \(f(x) = \frac{1}{x}\), which is all real numbers except \(x = 0\) because \(x\) cannot be zero as it will make the denominator undefined. Since in all valid domains different inputs indeed produce different outputs, the condition for the function being one-to-one is satisfied.

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Key Concepts

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

Understanding the Function Domain
The domain of a function refers to the set of all possible input values (usually represented as "x") that the function can accept without leading to any undefined scenarios. When we talk about a rational function like \( f(x) = \frac{1}{x} \), we encounter a classic example of a domain restriction.

Because division by zero is undefined, the domain for this function includes all real numbers except zero. In mathematical notation, this is expressed as:
  • Domain of \( f(x) = \frac{1}{x} \): all real numbers \( x \), except \( x eq 0 \).
This means you can input any number into \( f(x) = \frac{1}{x} \) except zero, and you'll get a valid output. In any function problem, always start by identifying its domain, as it sets the stage for understanding more complex properties like one-to-one behavior.
Exploring Injective (One-to-One) Functions
An injective function, also known as a one-to-one function, is special because each element in its domain maps to a unique output in its range. For a function \( f(x) \) to be injective, if you take any two different inputs, having the same output must necessarily imply that the inputs were identical to begin with.

Mathematically, for any distinct \( a \) and \( b \) in the domain:
  • If \( f(a) = f(b) \), then it must be true that \( a = b \).

In our specific example with \( f(x) = \frac{1}{x} \), checking this injective property involves setting \( f(a) = f(b) \) and showing that this leads to \( a = b \). Multiplying through by \( a \cdot b \) to clear the fractions leads us straight to this conclusion. Once this is confirmed for all values in the domain, we can confidently say \( f(x) = \frac{1}{x} \) is an injective function.
Rational Functions Explained
Rational functions are functions defined by the ratio of two polynomials. The standard form is \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.

The function \( f(x) = \frac{1}{x} \) is a simple example of a rational function where our \( P(x) = 1 \) and \( Q(x) = x \). In rational functions, the denominator \( Q(x) \) must not be zero, as this would make the function undefined. This requirement is also what determines the domain of rational functions, excluding any x-values that make the denominator zero.
The fascinating aspect of rational functions is their behavior in terms of asymptotes: lines that the graph approaches but never quite touches.
  • Vertical asymptotes occur at points where the function is undefined.
  • Horizontal or slant asymptotes often depend on the degrees of the polynomials.
Understanding and analyzing these features is crucial for graphing and fully grasping rational functions.

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Most popular questions from this chapter

DISCUSS: Finding an Inverse "in Your Head" In the margin notes in this section we pointed out that the inverse of a function can be found by simply reversing the operations that make up the function. For instance, in Example 7 we saw that the inverse of $$f(x)=3 x-2 \quad \text { is } \quad f^{-1}(x)=\frac{x+2}{3}$$ because the "reverse" of "Multiply by 3 and subtract 2" is "Add 2 and divide by 3 ". Use the same procedure to find the inverse of the following functions. (a) \(f(x)=\frac{2 x+1}{5}\) (b) \(f(x)=3-\frac{1}{x}\) (c) \(f(x)=\sqrt{x^{3}+2}\) (d) \(f(x)=(2 x-5)^{3}\) Now consider another function: $$f(x)=x^{3}+2 x+6$$ Is it possible to use the same sort of simple reversal of operations to find the inverse of this function? If so, do it. If not, explain what is different about this function that makes this task difficult.

A linear function is given. (a) Find the average rate of change of the function between \(x=a\) and \(x=a+h .\) (b) Show that the average rate of change is the same as the slope of the line. $$f(x)=\frac{1}{2} x+3$$

A savings account earns \(5 \%\) interest compounded annually. If you invest \(x\) dollars in such an account, then the amount \(A(x)\) of the investment after one year is the initial investment plus \(5 \% ;\) that is, $$A(x)=x+0.05 x=1.05 x$$ Find $$\begin{aligned} &A \circ A\\\&A \circ A \circ A\\\&A \circ A \circ A \circ A\end{aligned}$$ What do these compositions represent? Find a formula for what you get when you compose \(n\) copies of \(A\).

A linear function is given. (a) Find the average rate of change of the function between \(x=a\) and \(x=a+h .\) (b) Show that the average rate of change is the same as the slope of the line. $$g(x)=-4 x+2$$

Find a function whose graph is the given curve. The line segment joining the points \((-3,-2)\) and \((6,3)\)

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