/*! 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 42 Determine whether the two functi... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The functions are inverses because \( w(z(x)) = x \) and \( z(w(x)) = x \).

Step by step solution

01

Understand Inverse Functions

Two functions are inverses if the composition of one with the other results in the identity function for both compositions. This means that if one function is substituted into the other, the result should be the original input variable.
02

Composition of Functions

First, check if composing \(w(x)\) with \(z(x)\) yields the identity function. Start by substituting \(z(x)\) into \(w(x)\): \[ w(z(x)) = w\bigg(\frac{6-2x}{x}\bigg) \] Now, substitute \( z(x) \): \[ w\bigg(\frac{6-2x}{x}\bigg) = \frac{6}{\frac{6-2x}{x} + 2} \]
03

Simplify the Composition

Simplify the result from the previous step: \[ w\bigg(\frac{6-2x}{x}\bigg) = \frac{6}{\frac{6-2x+2x}{x}} = \frac{6}{\frac{6}{x}} = 6 \cdot \frac{x}{6} = x \] So, \( w(z(x)) = x \), which is the identity function.
04

Reverse Composition Check

Next, check if composing \(z(x)\) with \(w(x)\) also gives the identity function. Substitute \(w(x)\) into \(z(x)\): \[ z(w(x)) = z\bigg(\frac{6}{x+2}\bigg) \] Now, substitute \( w(x) \): \[ z\bigg(\frac{6}{x+2}\bigg) = \frac{6-2\cdot \frac{6}{x+2}}{\frac{6}{x+2}} \]
05

Simplify the Reverse Composition

Simplify result from previous step: \[ z\bigg(\frac{6}{x+2}\bigg) = \frac{6 - 2 \cdot \frac{6}{x+2}}{\frac{6}{x+2}} = \frac{6 - \frac{12}{x+2}}{\frac{6}{x+2}} = \frac{\frac{6(x+2) - 12}{x+2}}{\frac{6}{x+2}} \]\[ = \frac{\frac{6x + 12 - 12}{x+2}}{\frac{6}{x+2}} = \frac{\frac{6x}{x+2}}{\frac{6}{x+2}} = x \] So, \( z(w(x)) = x \), which is the identity function.
06

Conclusion

Since both \(w(x)\) and \(z(x)\) compositions give the identity function, they are indeed inverses of each other.

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Key Concepts

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

Composition of Functions
The composition of functions involves applying one function to the results of another. For two functions to be inverses, their compositions must yield the identity function. This process involves substituting one function into the other. For example, when we composed the functions \( w(x) \) and \( z(x) \), we sought to determine if \( w(z(x)) \) would simplify back to \( x \). We started by substituting \( z(x) \) into \( w(x) \) and then performed algebraic manipulations to yield \( x \). If \( F(G(x)) = x \) and \( G(F(x)) = x \), then \( F \) and \( G \) are inverses of each other.

The core idea in composition is making sure we plug one function's output into another, checking to see if the operations cancel out to yield the original input variable, \( x \).
Identity Function
An identity function is a special function in mathematics represented as \( I(x) = x \). When you apply an identity function to any value, you simply get that value back. In our problem, we sought to verify if composing the functions \( w(x) \) and \( z(x) \) yields \( x \), which indicates an identity function.

This is a critical check because if the result of composing one function with another is the input variable itself, it demonstrates that the operations of the functions completely negate each other’s effects, leaving the input unchanged. This is the defining characteristic of inverse functions and an essential property when verifying inverse relationships.
Simplification of Expressions
Simplification is key to solving compositions and checking for identity functions. This involves breaking down complicated expressions to their simplest form. In our example, to simplify \( w(z(x)) \), we substituted \( z(x) = \frac{6-2x}{x} \) into \( w(x) = \frac{6}{x+2} \) and simplified step by step.

Through rigorous simplification, terms can be combined or cancelled out, helping reveal the underlying structure of the expression. We'll often convert fractions, combine like terms, or factor expressions as part of the process. Simplification makes it easier to see if the composed function simplifies to the identity function, \( x \).

Take it step by step: Identify the key parts of the expression. Perform arithmetic simplifications. Combine like terms. Factor if necessary. Check your work by ensuring you haven’t violated any mathematical rules.
Algebraic Manipulation
Algebraic manipulation involves a variety of techniques to transform expressions and equations. This includes operations like addition, subtraction, multiplication, and division, as well as more complex processes like factoring and expanding expressions. In verifying inverse functions, algebraic manipulation was essential to simplify the compositions \( w(z(x)) \) and \( z(w(x)) \) to check if they equal \( x \).

For example, to simplify \( w\left( \frac{6-2x}{x} \right) \), we had to handle complex fractions and combine terms within the numerators and denominators. Key techniques in algebraic manipulation include:
  • Distributing and factoring to simplify expressions
  • Combining like terms
  • Canceling out terms
Using these techniques correctly leads to clear and straightforward results. Mastery of algebraic manipulation is vital for solving more advanced mathematical problems confidently.

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Most popular questions from this chapter

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{2} 0.3 $$

Technetium- \(99\left({ }^{99 \mathrm{~m}} \mathrm{Tc}\right)\) is a radionuclide used widely in nuclear medicine. \({ }^{99 \mathrm{~m}} \mathrm{Tc}\) is combined with another substance that is readily absorbed by a targeted body organ. Then, special cameras sensitive to the gamma rays emitted by the technetium are used to record pictures of the organ. Suppose that a technician prepares a sample of \(^{99 \mathrm{~m}}\) Tc-pyrophosphate to image the heart of a patient suspected of having had a mild heart attack. a. At noon, the patient is given \(10 \mathrm{mCi}\) (millicuries) of \({ }^{99 \mathrm{~m}} \mathrm{Tc}\). If the half-life of \({ }^{99 \mathrm{~m}} \mathrm{Tc}\) is \(6 \mathrm{hr}\), write a function of the form \(Q(t)=Q_{0} e^{-k t}\) to model the radioactivity level \(Q(t)\) after \(t\) hours. b. At what time will the level of radioactivity reach \(3 \mathrm{mCi} ?\) Round to 1 decimal place.

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(10^{5+8 x}+4200=84,000\)

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(80=320 e^{-0.5 t}\)

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log _{8}\left(\frac{1}{w}\right)=-\log _{8} w $$

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