/*! 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 Find \(f^{-1}(x) .\) State any r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find \(f^{-1}(x) .\) State any restrictions on the domain of \(f^{-1}(x)\) $$f(x)=3 x-7$$

Short Answer

Expert verified
The inverse function is \(f^{-1}(x) = (x + 7)/3\) with no domain restrictions.

Step by step solution

01

Swap \(x\) and \(y\)

To begin, replace \(f(x)\) with \(y\). The equation will be: \(y = 3x - 7\). Then, switch the roles of \(y\) and \(x\). That gives: \(x = 3y - 7\).
02

Solve for \(y\)

After swapping \(x\) and \(y\), we should solve for \(y\). To achieve this, add 7 to both sides: \(x + 7 = 3y\). Then, to isolate \(y\), divide both sides by 3. The solution is: \(y = (x + 7)/3\)
03

Finding the inverse function

Now we can write the inverse function \(f^{-1}(x)\). Therefore, \(f^{-1}(x) = (x + 7)/3\)
04

Domain Restriction

As for the domain of this function, since \(f^{-1}(x)\) is a linear function, there are no restrictions on the domain, i.e., \(x\) could be any real number.

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 91Ó°ÊÓ!

Key Concepts

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

Domain Restrictions
When delving into the concept of inverse functions, understanding domain restrictions is key. Essentially, the domain of a function encompasses all the possible input values it can accept. In some cases, functions might have certain restrictions on their domain due to factors such as division by zero or square roots of negative numbers, which aren't defined for real numbers.

However, the situation is simpler when dealing with linear functions, like the one in the exercise. The function given, f(x) = 3x - 7, is a linear function and, by nature, doesn't have any inherent restrictions. So when we find its inverse, f-1(x) = (x + 7) / 3, it maintains the lack of domain restrictions. Linear functions stretch infinitely in both the positive and negative directions, and there's nothing in the formula of the inverse that would limit the input values. Consequently, the domain for both f(x) and its inverse f-1(x) is all real numbers, symbolized by the set (-∞, ∞).
Solving for y
The process of 'solving for y' is a critical step when finding an inverse function. This generally translates to expressing the variable 'y' in terms of 'x'. The goal is to isolate 'y' on one side of the equation so that the function can be inverted. In our example, we start with y being defined implicitly by the equation y = 3x - 7.

Following the initial step of swapping 'x' and 'y', we work to rearrange the equation to solve for the new 'y'. By adding 7 to both sides and then dividing by 3, we isolate 'y' on one side, resulting in y = (x + 7) / 3. This algebraic manipulation reveals the inverse function, which is the rule that determines the output for any input 'x'. The ability to solve for 'y' is not just important for finding inverses; it's a fundamental skill used throughout precalculus and calculus to manipulate and understand functions.
Precalculus
Precalculus is an educational gateway to understanding functions, equations, and the concept of mathematical rigor that is crucial in calculus. It embodies a comprehensive set of topics that equip students with the necessary tools to handle calculus concepts effectively. One of those foundational topics is the concept of inverse functions, which underscores the importance of functions that 'undo' each other.

Precalculus also emphasizes the mastery of manipulating various types of functions — whether they are polynomials, exponentials, logarithmic, or trigonometric — and understanding their properties, such as domains, ranges, and behaviors. The example problem falls under the umbrella of linear functions, which are a straightforward, yet important kind of function explored in precalculus. This course of study ultimately combines algebraic and geometric thinking, setting the stage for the advance into the world of limits, derivatives, and integrals that define calculus.

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

Use a calculator to evaluate the exponential function for the given \(x\) -value. Round to the nearest hundredth. $$f(x)=3^{x}, x=-1.5$$

Evaluate the exponential function for the given \(x\) -values. $$h(x)=\left(\frac{3}{2}\right)^{x} ; x=2 \text { and } x=-3$$

Starting on the left side of a standard 88 -key piano, the frequency, in vibrations per second, of the \(n\) th note is given by \(f(n)=(27.5) 2^{(n-1) / 12}\) a. Using this formula, determine the frequency, to the nearest hundredth of a vibration per second, of middle C, key number 40 on an 88 -key piano. b. Is the difference in frequency between middle C (key number 40 ) and \(D\) (key number 42 ) the same as the difference in frequency between \(\mathrm{D}\) (key number 42 ) and \(\mathrm{E}\) (key number 44 )? Explain.

A cup of coffee is heated to \(180^{\circ} \mathrm{F}\) and placed in a room that maintains a temperature of \(65^{\circ} \mathrm{F}\). The temperature of the coffee after \(t\) minutes is given by \(T(t)=65+115 e^{-0.042 t}\) a. Find the temperature, to the nearest degree, of the coffee 10 minutes after it is placed in the room. b. Use a graphing utility to determine when, to the nearest tenth of a minute, the temperature of the coffee will reach \(100^{\circ} \mathrm{F}\)

A model for how long our aluminum resources will last is given by $$T=\frac{\ln (20,500 r+1)}{\ln (r+1)}$$ where \(r\) is the percent increase in consumption from current levels of use and \(T\) is the time (in years) before the resource is depleted. a. Use a graphing utility to graph this equation. b. If our consumption of aluminum increases by \(5 \%\) per year, in how many years (to the nearest year) will we deplete our aluminum resources? c. What percent increase in consumption of aluminum will deplete the resource in 100 years? Round to the nearest tenth of a percent.

See all solutions

Recommended explanations on Math Textbooks

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.