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Chapter 2: Problem 39

The given function \(f\) is one-to-one. Find \(f^{-1}\). Proceed as in part (b) of Example 4 and find the domain and range of \(f^{-1}\). Then find the range of \(f\). $$ f(x)=\frac{1}{2 x-1} $$

Short Answer

Expert verified
The inverse function is \(f^{-1}(x) = \frac{1 + x}{2x}\), with domain \(x \neq 0\) and range \(y \neq \frac{1}{2}\). The range of \(f\) is \(y \neq 0\).

Step by step solution

01

Express the Function with y

To find the inverse function, start by replacing \(f(x)\) with \(y\). So the equation becomes \(y = \frac{1}{2x - 1}\).
02

Swap x and y

Change \(y\) to \(x\) and solve for \(y\). After swapping, the equation will be \(x = \frac{1}{2y - 1}\).
03

Solve for y

To find \(y\), manipulate the equation \(x = \frac{1}{2y - 1}\):1. Multiply both sides by \(2y - 1\): \(x(2y - 1) = 1\).2. Distribute \(x\): \(2xy - x = 1\).3. Add \(x\) to both sides: \(2xy = 1 + x\).4. Solve for \(y\): \(y = \frac{1 + x}{2x}\).Thus, the inverse function is \(f^{-1}(x) = \frac{1 + x}{2x}\).
04

Determine the Domain of \(f^{-1}\)

The domain of \(f^{-1}(x)\) is all real numbers except where the denominator is zero.For \(f^{-1}(x) = \frac{1 + x}{2x}\), the denominator is zero when \(x = 0\). Thus, the domain of \(f^{-1}\) is \(x eq 0\).
05

Determine the Range of \(f^{-1}\)

The range of \(f^{-1}\) is the same as the domain of the original function, \(f(x)\). For \(f(x) = \frac{1}{2x - 1}\), the denominator cannot be zero, so \(2x - 1 eq 0\). Solving gives \(x eq \frac{1}{2}\). Thus, the range of \(f^{-1}\) is \(y eq \frac{1}{2}\).
06

Determine the Range of f

Finally, the range of \(f\) is the same as the domain of \(f^{-1}\), which is all real numbers except zero. Therefore, the range of \(f\) is \(y eq 0\).

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Key Concepts

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

Domain and Range
When we talk about the **domain and range** of a function, it means we are discussing the input values the function can take (domain) and the output values it can produce (range). In this exercise, we are working with a rational function, which is a fraction where the numerator and denominator are both polynomials.

For the original function, \( f(x) = \frac{1}{2x - 1} \), let's discuss its domain:
  • The denominator \( 2x - 1 \) cannot be zero because division by zero is undefined.
  • Solving for when the denominator is zero: \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \).
  • Therefore, the domain of \( f(x) \) is all real numbers except \( x = \frac{1}{2} \).
Now, let's consider the range of \( f(x) \):
  • The range is the set of all possible \( y \) values that the function can output.
  • In this function, \( f(x) \) will never touch or return zero because it implies the denominator of the inverse function becomes undefined.
  • This means the range of \( f(x) \) is all real numbers except \( y = 0 \).
Finally, for the inverse function \( f^{-1}(x) = \frac{1+x}{2x} \):
  • The domain is all real numbers except zero, \( x eq 0 \), because zero makes the denominator undefined.
One-to-one Function
A **one-to-one function** is a special kind of function where each input (or \( x \)-value) is associated with exactly one output (or \( y \)-value), and vice versa. This is an essential property when finding inverse functions. If a function is not one-to-one, its inverse will not be a function unless the domain is restricted.

In the exercise problem, we have: \( f(x) = \frac{1}{2x - 1} \) is specified to be one-to-one. Here's how this works:
  • This means for every unique \( x \, (x eq \frac{1}{2}) \), there will be a unique \( y \).
  • Since no two different \( x \)-values can produce the same \( y \)-value, we can find an inverse.
To confirm a function is one-to-one, the horizontal line test is used:
  • Imagine drawing horizontal lines across the graph of the function. If any line touches the graph more than once, the function is not one-to-one.
  • For \( f(x) \), each horizontal line only touches in one place due to its nature as a rational function鈥攎aking it one-to-one.
Rational Functions
**Rational functions** are those that can be expressed as the quotient of two polynomials. They have the general form \( f(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.

They are characterized by the following:
  • They can have both vertical asymptotes and horizontal asymptotes.
  • The vertical asymptote occurs where the denominator is zero and the function is undefined.
  • The horizontal asymptote describes the behavior of the function as \( x \) approaches infinity.
In our exercise, the function \( f(x) = \frac{1}{2x - 1} \) is:
  • A simple rational function with a single vertical asymptote at \( x = \frac{1}{2} \), where \( 2x - 1 = 0 \).
  • There is no horizontal asymptote as \( x \) tends to infinity, because the degree of the numerator (0, constant 1) is less than the degree of the denominator (1).
Rational functions are versatile and can model many situations, but they require careful handling around their asymptotes and undefined points. Understanding their behavior around these areas is essential for graphing and solving equations involving them.

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Most popular questions from this chapter

Proceed as in Example 1 and translate the words into an appropriate function. Give the domain of the function. Express the distance from a point \((x, y)\) on the graph of \(x+y=1\) to the point (2,3) as a function of \(x\).

In Problems 35 and 36 , sketch the graph the given piece wise-defined function \(f\) for \(c=-2,-1,1,2\). $$ f(x)=\left\\{\begin{array}{ll} -(x-c)^{2}, & x \leq c \\ (x-c)^{2}, & x>c \end{array}\right. $$

Explain why the graph of a one-to-one function \(f\) can have at most one \(x\) -intercept.

Use the appropriate derivatives obtained in Problems 11-26. For the given function, find the point of tangency and slope of the tangent line at the indicated value of \(x .\) Find an equation of the tangent line at that point. $$ f(x)=x+\frac{1}{x}, \quad x=\frac{1}{2} $$

The given function \(f\) is one-to-one. Find \(f^{-1}\). Proceed as in part (b) of Example 4 and find the domain and range of \(f^{-1}\). Then find the range of \(f\). $$ f(x)=10-\frac{1}{5 x} $$
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