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Chapter 4: Problem 51

A one-to-one function is given. Write an equation for the inverse function. \(m(x)=4 x^{3}+2\)

Short Answer

Expert verified
The inverse function is \(m^{-1}(x) = \sqrt[3]{\frac{x - 2}{4}}\).

Step by step solution

01

Understand the function

The function given is a one-to-one function: \(m(x) = 4x^3 + 2\). To find the inverse, we need to find a function \(m^{-1}(x)\) that undoes what \(m(x)\) does.
02

Replace the function with y

Replace \(m(x)\) with \(y\) to make it easier to work with. Now we have \(y = 4x^3 + 2\).
03

Swap x and y

To find the inverse, switch the roles of \(x\) and \(y\). This gives us \(x = 4y^3 + 2\).
04

Solve for y

Next, solve the equation \(x = 4y^3 + 2\) for \(y\):1. Subtract 2 from both sides: \(x - 2 = 4y^3\).2. Divide both sides by 4: \(\frac{x - 2}{4} = y^3\).3. Take the cube root of both sides: \(y = \sqrt[3]{\frac{x - 2}{4}}\).
05

Write the inverse function

The inverse function can be written as \(m^{-1}(x) = \sqrt[3]{\frac{x - 2}{4}}\).

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Key Concepts

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

one-to-one functions
A one-to-one function is a fundamental concept in algebra. It's a function where each input has a unique output, and vice versa. This means that every x-value corresponds to one and only one y-value. If we imagine plotting a one-to-one function on a graph, it will pass the horizontal line test. This test involves drawing horizontal lines on the graph. If any horizontal line touches the graph more than once, the function is not one-to-one. Understanding this concept is crucial for finding inverse functions. When a function is one-to-one, we can be sure that its inverse, which reverses the operations of the original function, will also be a valid function.
inverse function
An inverse function reverses the operations of the original function. For a function \( m(x) \), its inverse is denoted as \( m^{-1}(x) \). To find the inverse of a function, follow these steps:
  • Replace the function notation \( m(x) \) with \( y \). This helps to simplify the equation.
  • Swap the roles of \( x \) and \( y \). This step essentially mirrors the function across the line \( y = x \).
  • Solve the resulting equation for \( y \). This will give you the equation for the inverse function.
As seen in the example given: 1. Start with \( y = 4x^3 + 2 \). 2. Swap \( x \) and \( y \) to get \( x = 4y^3 + 2 \). 3. Solve for \( y \):
\, Subtract 2 from both sides: \( x - 2 = 4y^3 \).
\, Divide by 4: \( \frac{x - 2}{4} = y^3 \).
\, Take the cube root: \( y = \sqrt[3]{\frac{x - 2}{4}} \).
So, the inverse function is \( m^{-1}(x) = \sqrt[3]{\frac{x - 2}{4}} \). This function undoes what the original function does.
solving equations
Solving equations is a core skill in algebra. It involves finding the value(s) of the variable that make the equation true. When solving for the inverse function, we need to handle each step carefully:
  • First, isolate the term containing the variable we need to solve for. This often involves addition or subtraction.
  • Next, divide or multiply to further isolate the variable.
  • Finally, if necessary, apply roots or exponents to solve for the variable completely.
All these steps rely on keeping the equation balanced. Whatever operation you perform on one side, you must do the same to the other side. In our example: 1. Subtract 2 from both sides to get \( x - 2 = 4y^3 \). 2. Divide both sides by 4 to get \( \frac{x - 2}{4} = y^3 \). 3. Take the cube root of both sides to find \( y = \sqrt[3]{\frac{x - 2}{4}} \). Paying careful attention to each step ensures you correctly find the inverse function.

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Most popular questions from this chapter

a. Graph \(f(x)=\ln x\) and \(g(x)=(x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\frac{(x-1)^{4}}{4}\) on the viewing window [-2,4,1] by [-5,2,1] . How do the graphs compare on the interval (0,2) ? b. Use function \(g\) to approximate \(\ln 1.5\). Round to 4 decimal places.

Graph the function. a. Graph \(Y_{1}=\log |x|\) and \(Y_{2}=\frac{1}{2} \log x^{2}\). How are the graphs related? b. Show algebraically that \(\frac{1}{2} \log x^{2}=\log |x|\).

Graph the following functions on the window [-3,3,1] by [-1,8,1] and comment on the behavior of the graphs near $$ \begin{array}{l} x=0 \\ \mathrm{Y}_{1}=e^{x} \\ \mathrm{Y}_{2}=1+x+\frac{x^{2}}{2} \\ \mathrm{Y}_{3}=1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6} \end{array} $$

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

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. \(11^{1-8 x}=9^{2 x+3}\)
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