/*! 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 39 Determine whether the two functi... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine whether the two functions are inverses. \(m(x)=\frac{-2+x}{6}\) and \(n(x)=6 x-2\)

Short Answer

Expert verified
The functions are not inverses because \(m(n(x)) eq x\) and \(n(m(x)) eq x\).

Step by step solution

01

- Find the Composition of Functions

To determine if two functions are inverses, compute the composition of the two functions, both ways: \(m(n(x))\) and \(n(m(x))\). If both compositions result in \(x\), then the functions are inverses of each other.
02

- Compute \(m(n(x))\)

First, find the composition \(m(n(x))\):\[ n(x) = 6x - 2 \]Substitute \(n(x)\) into \(m(x)\):\[ m(n(x)) = m(6x - 2) \]Now, plug \(6x - 2\) into the function \(m(x)\):\[ m(6x - 2) = \frac{-2 + (6x - 2)}{6} \]Simplify the expression:\[ m(6x - 2) = \frac{6x - 4}{6} = x - \frac{4}{6} = x - \frac{2}{3} \]Since \(x - \frac{2}{3} eq x\), \(m(n(x)) eq x\). Therefore, the functions are not inverses.
03

- Verification (Optional)

If desired, verify the other composition \(n(m(x))\) for thoroughness:\[ m(x) = \frac{-2 + x}{6} \]Substitute \(m(x)\) into \(n(x)\):\[ n(m(x)) = n\left(\frac{-2 + x}{6}\right) \]Now, plug \(\frac{-2 + x}{6}\) into the function \(n(x)\):\[ n\left(\frac{-2 + x}{6}\right) = 6 \left(\frac{-2 + x}{6}\right) - 2 \]Simplify the expression:\[ 6 \left(\frac{-2 + x}{6}\right) - 2 = -2 + x - 2 = x - 4 \]Since \(x - 4 eq x\), \(n(m(x)) eq x\). This confirms that the functions are not inverses.

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Key Concepts

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

function composition
Understanding function composition is essential when determining if two functions are inverses. Function composition involves combining two functions such that the output of one function becomes the input of another. Mathematically, if we have functions \(f\) and \(g\), their composition is written as \((f \circ g)(x)\). This means you first apply \(g\) to \(x\) and then apply \(f\) to the result of \(g(x)\).

To illustrate, let's take functions \(m(x) = \frac{-2 + x}{6}\) and \(n(x) = 6x - 2\). To find \(m(n(x))\), we take \(n(x)\) and substitute it into \(m(x)\). Similarly, to find \(n(m(x))\), we take \(m(x)\) and substitute it into \(n(x)\). For these functions to be inverses, both compositions must simplify to \(x\).

When we computed \(m(n(x))\) and \(n(m(x))\) in the exercise, neither resulted in \(x\), indicating that \(m\) and \(n\) are not inverse functions.
algebraic expressions
Algebraic expressions form the backbone of manipulating functions during function composition. They consist of variables, numbers, and operations like addition, subtraction, multiplication, and division. Understanding how to work with algebraic expressions is crucial for simplifying compositions of functions.

In our exercise, we needed to handle expressions like \(m(6x - 2)\) and simplify them. The process involves careful substitution and arithmetic operations. For example, substituting \(n(x) = 6x - 2\) into \(m(x)\), we started by replacing \(x\) in \(m(x)\) with the expression \(6x - 2\). This led to:
  • \(m(6x - 2) = \frac{-2 + (6x - 2)}{6}\)
  • Simplified to \(\frac{6x - 4}{6} = x - \frac{2}{3}\)
Understanding each algebraic step ensures clarity and accuracy in the conclusion. Mastering these techniques empowers students to tackle increasingly complex problems confidently.
function properties
Knowing the properties of functions helps in understanding their behavior and relationships. Some key properties include domain, range, and the behavior under specific operations like composition. Functions have been classified into various types, such as linear, quadratic, polynomial, and rational functions. Each has its own set of rules and characteristics.

For example, the functions \(m(x) = \frac{-2 + x}{6}\) and \(n(x) = 6x - 2\) from our exercise have different forms. Function \(m(x)\) is a linear function represented as a rational expression, whereas \(n(x)\) is a straightforward linear function. Function properties like injectivity (one-to-one) and surjectivity (onto) help determine if functions can have inverses.
  • An injective function has no two different inputs mapping to the same output.
  • A surjective function covers the entire range of possible outputs.
For two functions to be inverses, each must be both injective and surjective. Our verification in the step-by-step exercise didn't meet these criteria, reinforcing that \(m(x)\) and \(n(x)\) are not inverse functions.

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Most popular questions from this chapter

If \(k>0,\) the equation \(y=y_{0} e^{k t}\) is a model for exponential (growth/decay), whereas if \(k<0,\) the equation is a model for exponential (growth/decay).

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\ln x+\ln (x-4)=\ln (3 x-10)\)

Solve for the indicated variable. \(A=P(1+r)^{t}\) for \(t\) (used in finance)

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log _{7}(12-t)=\log _{7}(t+6)\)

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

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