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Chapter 4: Problem 53

A one-to-one function is given. Write an equation for the inverse function. \(c(x)=\frac{5}{x+2}\)

Short Answer

Expert verified
The inverse function is \(c^{-1}(x) = \frac{5 - 2x}{x}\).

Step by step solution

01

- Write down the given function

The given function is \(c(x) = \frac{5}{x + 2}\).
02

- Replace the function notation

Replace \(c(x)\) with \(y\). This gives us \(y = \frac{5}{x + 2}\).
03

- Swap the variables

Interchange the roles of \(x\) and \(y\). This results in \(x = \frac{5}{y + 2}\).
04

- Solve for y

We need to solve for \(y\) in terms of \(x\).Step 1: Multiply both sides by \(y + 2\) to get rid of the fraction: \[ x (y + 2) = 5 \]Step 2: Distribute \(x\): \[ xy + 2x = 5 \]Step 3: Isolate \(y\): \[ xy = 5 - 2x \]\[ y = \frac{5 - 2x}{x} \]
05

- Write the inverse function

Replace \(y\) with the notation for the inverse function: \(c^{-1}(x) = \frac{5 - 2x}{x}\).

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Key Concepts

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

one-to-one function
A one-to-one function is one for which every element of the range is paired with exactly one element of the domain. This means that no two different inputs produce the same output.

To determine if a function is one-to-one, you can use the Horizontal Line Test: if any horizontal line intersects the graph of the function at most once, the function is one-to-one.

Understanding whether a function is one-to-one is crucial for finding its inverse. Only one-to-one functions have inverses that are also functions.
function notation
Function notation is a way to write functions in the form of equations. It provides a clear way to denote the function and its variables.

For example, the function notation for our example is written as:

function: \(c(x) = \frac{5}{x + 2}\), where \(c\) is the name of the function and \(x\) is the input variable.

Function notation allows us to easily replace variables and manipulate the equations to find solutions, important especially when dealing with inverse functions.
solving for y
Solving for \(y\) means isolating the variable \(y\) on one side of the equation. This is an important step in finding the inverse of a function.

In our example:
1. Start with the equation: \(x = \frac{5}{y + 2}\)
2. Multiply both sides by \(y + 2\): \(x (y + 2) = 5\)
3. Distribute \(x\): \(xy + 2x = 5\)
4. Isolate \(y\): \(xy = 5 - 2x\)
5. Finally, solve for \(y\): \(y = \frac{5 - 2x}{x}\).

Taking these steps ensures we have an expression for \(y\) in terms of \(x\). This is critical for defining the inverse function.
inverse function notation
Inverse function notation tells us how to label the inverse of a function. The inverse is indicated by a superscript \(-1\) on the function's name.

For example, if our original function is \(c(x)\), the inverse function would be denoted as \(c^{-1}(x)\).

Once we solve for \(y\) in terms of \(x\) from our original function, we replace \(y\) with the inverse notation to get:
\(c^{-1}(x) = \frac{5 - 2x}{x}\).

This notation helps us distinguish the original function from its inverse, making it easier to understand and work with both in various problems.

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Most popular questions from this chapter

The half-life of radium-226 is 1620 yr. Given a sample of \(1 \mathrm{~g}\) of radium \(-226,\) the quantity left \(Q(t)\) (in g) after \(t\) years is given by \(Q(t)=\left(\frac{1}{2}\right)^{t / 1620}\) a. Convert this to an exponential function using base \(e\). b. Verify that the original function and the result from (a) yield the same result for \(Q(0), Q(1620),\) and part \(Q(3240) .\) (Note: There may be round-off error.)

Write an equation for the inverse of the function. $$ f(x)=e^{x-2} $$

Explain why the product property of logarithms does not apply to the following statement. $$ \begin{array}{l} \log _{5}(-5)+\log _{5}(-25) \\ \quad=\log _{5}[(-5)(-25)] \\ \quad=\log _{5} 125=3 \end{array} $$

A function of the form \(P(t)=a b^{t}\) represents the population of the given country \(t\) years after January 1,2000 . a. Write an equivalent function using base \(e\); that is, write a function of the form \(P(t)=P_{0} e^{k t} .\) Also, determine the population of each country for the year 2000 . $$\begin{array}{|l|c|c|c|} \hline \text { Country } & P(t)=a b^{t} & P(t)=P_{0} e^{k t} & \begin{array}{c} \text { Population } \\ \text { in } 2000 \end{array} \\ \hline \text { Haiti } & P(t)=8.5(1.0158)^{t} & & \\ \hline \text { Sweden } & P(t)=9.0(1.0048)^{t} & & \\ \hline \end{array}$$ b. The population of the two given countries is very close for the year 2000 , but their growth rates are different. Determine the year during which the population of each country will reach 10.5 million. c. Haiti had fewer people in the year 2000 than Sweden. Why did Haiti reach a population of 10.5 million sooner?

Explain how to use the change-of-base formula and explain why it is important.
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