/*! 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 29 Gives a formula for a function \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 29

Gives a formula for a function \(y=f(x) .\) In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1} .\) As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\). $$f(x)=1 / x^{2}, \quad x>0$$

Short Answer

Expert verified
The inverse is \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), with domain \( x > 0 \) and range \( y > 0 \).

Step by step solution

01

Understand the Original Function

The given function is \( f(x) = \frac{1}{x^2} \) with the restriction \( x > 0 \). This means that \( f(x) \) takes positive real numbers as input and outputs real numbers that are greater than zero.
02

Rearrange the Function for the Inverse

To find \( f^{-1}(x) \), start by expressing the function \( y = \frac{1}{x^2} \) in terms of \( x \). This needs to be rearranged to solve for \( x \) in terms of \( y \).
03

Swap Variables

From \( y = \frac{1}{x^2} \), swap \( x \) and \( y \) to start finding the inverse: \( x = \frac{1}{y^2} \). Solve this equation for \( y \) to find \( y = f^{-1}(x) \).
04

Solve for the Inverse Function

Taking reciprocals on both sides gives \( y = \frac{1}{\sqrt{x}} \). Thus, \( f^{-1}(x) = \frac{1}{\sqrt{x}} \). Since \( x > 0 \) in \( f(x) \), \( f^{-1}(x) \) must also take positive values for the domain.
05

Determine the Domain and Range of f^{-1}

The domain of \( f^{-1}(x) \) is \( x > 0 \), since \( \sqrt{x} \) is only defined for positive \( x \). The range of \( f^{-1}(x) \) is also \( y > 0 \), as it outputs positive numbers.
06

Verify f(f^{-1}(x))=x

Substitute \( f^{-1}(x) \) into \( f(x) \): \( f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\left(\frac{1}{\sqrt{x}}\right)^2} = x\).
07

Verify f^{-1}(f(x))=x

Substitute \( f(x) \) into \( f^{-1}(x) \): \( f^{-1}\left(\frac{1}{x^2}\right) = \frac{1}{\sqrt{\frac{1}{x^2}}} = x \).
08

Conclusion

The inverse function \( f^{-1}(x) = \frac{1}{\sqrt{x}} \) and it is verified that \( f(f^{-1}(x))=x \) and \( f^{-1}(f(x))=x \). The domain and range of \( f^{-1} \) are both \( x > 0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Domain and Range
When dealing with inverse functions, it's important to understand the concepts of domain and range as they lay the foundation for how functions behave and interact. The domain refers to the set of all possible input values for the function. For example, in our original function
  • The domain of f(x) is \( x > 0 \), meaning it only accepts positive real numbers.
The range, on the other hand, is the set of all possible output values. It's what the function can produce, or map to, from its inputs.
  • For our function \( f(x) = \frac{1}{x^2} \), since we only input positive x, it gives outputs greater than zero. Thus, its range is also \( y > 0 \).
When we consider the inverse function \( f^{-1}(x) = \frac{1}{\sqrt{x}} \), we switch roles:
  • The domain of the inverse function is the range of the original function, \( x > 0 \).
  • Similarly, the range of the inverse function becomes \( y > 0 \), as it maps positive outputs to inputs.
Function Composition
Function composition involves combining two functions such that the output of one function becomes the input of the other. This process allows us to explore the relationships between a function and its inverse. In our example,
  • We use function composition to show that performing a function and then its inverse returns us to our starting value: \( f(f^{-1}(x)) = x \).
  • This can be understood by looking at the operations: taking the inverse and then the original cancels out any transformation.
The same occurs in the reverse: applying the inverse after the original should also revert the output back to the initial input, stated as \( f^{-1}(f(x)) = x \).
  • This is because if you input \( x \) into the original function and then input the result into its inverse, you should logically end up with your initial input value, ensuring the properties of the inverse function.
This mathematical exploration form a crucial verification step, confirming the relationship between \( f \) and \( f^{-1} \).
Mathematical Verification
Verification is an essential step in confirming that we have correctly deduced the inverse. It involves checking through calculation to ensure the functions truly behave as expected. For verification:
  • We first substitute \( f^{-1}(x) \) back into \( f(x) \) and simplify. This should give us \( x \), the original input: \[ f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\left(\frac{1}{\sqrt{x}}\right)^2} = x. \]
  • Then, substitute \( f(x) \) into \( f^{-1}(x) \) yielding \( x \) again: \[ f^{-1}\left(\frac{1}{x^2}\right) = \frac{1}{\sqrt{\frac{1}{x^2}}} = x. \]
These checks, known as the inverse properties, confirm that \( f^{-1} \) is correctly derived. If these relationships hold true, we can be confident in our solution.
  • This ensures that the reverse process truly undoes the original functions, and validates the calculated inverse.
Through mathematical verification, we verify both the theoretical understanding and practical application of inverse functions in mathematics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To encourage buyers to place 100-unit orders, your firm's sales department applies a continuous discount that makes the unit price a function \(p(x)\) of the number of units \(x\) ordered. The discount decreases the price at the rate of \(\$ 0.01\) per unit ordered. The price per unit for a 100 -unit order is \(p(100)=\$ 20.09\) a. Find \(p(x)\) by solving the following initial value problem: $$\begin{aligned} &\text { Differential equation: } \quad \frac{d p}{d x}=-\frac{1}{100} p\\\ &\text { Initial condition: } \quad p(100)=20.09 \end{aligned}$$ b. Find the unit price \(p(10)\) for a 10 -unit order and the unit price \(p(90)\) for a 90 -unit order. c. The sales department has asked you to find out if it is discounting so much that the firm's revenue, \(r(x)=x \cdot p(x),\) will actually be less for a 100 -unit order than, say, for a 90 -unit order. Reassure them by showing that \(r\) has its maximum value at \(x=100\) d. Graph the revenue function \(r(x)=x p(x)\) for \(0 \leq x \leq 200\)

Which of the following functions grow faster than \(\ln x\) as \(x \rightarrow \infty ?\) Which grow at the same rate as \(\ln x ?\) Which grow slower? a. \(\log _{2}\left(x^{2}\right)\) b. \(\log _{10} 10 x\) c. \(1 / \sqrt{x}\) d. \(1 / x^{2}\) e. \(x-2 \ln x\) f. \(e^{-x}\) g. \(\ln (\ln x)\) h. \(\ln (2 x+5)\)

Use your graphing utility. Graph \(f(x)=\tan ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in relation to the signs and values of \(f^{\prime}\) and \(f^{\prime \prime}\).

L'H贸pital's Rule does not help with the limits. Try it-you just keep on cycling. Find the limits some other way. $$\lim _{x \rightarrow-\infty} \frac{2^{x}+4^{x}}{5^{x}-2^{x}}$$

a. Graph \(f(x)=(\sin x)^{x}\) on the interval \(0 \leq x \leq \pi .\) What value would you assign to \(f\) to make it continuous at \(x=0 ?\) b. Verify your conclusion in part (a) by finding \(\lim _{x \rightarrow 0^{+}} f(x)\) with l'H么pital's Rule. c. Returning to the graph, estimate the maximum value of \(f\) on \([0, \pi] .\) About where is max \(f\) taken on? d. Sharpen your estimate in part (c) by graphing \(f^{\prime}\) in the same window to see where its graph crosses the \(x\) -axis. To simplify your work, you might want to delete the exponential factor from the expression for \(f^{\prime}\) and graph just the factor that has a zero.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.