Chapter 7: Problem 46
Show that the graph of the inverse of \(f(x)=m x+b,\) where \(m\) and \(b\) are constants and \(m \neq 0,\) is a line with slope \(1 / m\) and \(y\) -intercept \(-b / m\).
Short Answer
Expert verified
The inverse graph has slope \(\frac{1}{m}\) and y-intercept \(-\frac{b}{m}\).
Step by step solution
01
Understanding the function
We have the function \(f(x) = mx + b\). Here, \(m\) is the slope and \(b\) is the y-intercept of the line.
02
Find the inverse function
To find the inverse, start by replacing \(f(x)\) with \(y\):\[y = mx + b\].We switch roles of \(x\) and \(y\) to get \(x = my + b\). Now solve for \(y\):\[x = my + b\] Subtract \(b\) from both sides:\[x - b = my\]Divide by \(m\):\[y = \frac{x-b}{m}\].Thus, the inverse function is \(f^{-1}(x) = \frac{x-b}{m}\).
03
Identify slope and y-intercept of the inverse
We rewrite \(f^{-1}(x) = \frac{x-b}{m}\) as \(f^{-1}(x) = \frac{1}{m}x - \frac{b}{m}\).Compare this with the slope-intercept form \(y = mx + b\). Here, the slope \((m')\) is \(\frac{1}{m}\) and the y-intercept \((b')\) is \(-\frac{b}{m}\).
04
Verification by differentiation
For further verification, take the derivative of the original function \(f(x) = mx + b\) to confirm the slope is \(m\). Then compute the derivative of the inverse \(f^{-1}(x) = \frac{x-b}{m}\). It gives \(\frac{d}{dx}(f^{-1}(x)) = \frac{1}{m}\), confirming the inverse slope is \(\frac{1}{m}\).
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.
Linear Equations
Linear equations are the simplest form of equations that describe a straight line on a coordinate plane. They come in the form of \(y = mx + b\), where:
- \(m\) is the slope of the line, representing the angle or steepness.
- \(b\) is the y-intercept, the value where the line crosses the y-axis.
Slope
Slope is a key component of linear equations and measures the line's steepness. It is denoted by \(m\) in the equation \(y = mx + b\). The slope can be calculated using the formula: \[m = \frac{{\Delta y}}{{\Delta x}} = \frac{{y_2 - y_1}}{{x_2 - x_1}}\] This formula tells us how much the \(y\) value changes for a unit change in \(x\). If the slope \(m\) is:
- Positive, the line rises as it moves from left to right.
- Negative, the line falls from left to right.
- Zero, the line is horizontal.
- Undefined, the line is vertical.
Y-intercept
The y-intercept is the point where a line crosses the y-axis on a graph. In the linear equation \(y = mx + b\), the y-intercept is represented by \(b\). This value is important because it provides a starting point for the line on the graph. You can think of it as the output value when the input \(x\) is zero. For an inverse function, such as \(f^{-1}(x) = \frac{x-b}{m}\), the y-intercept becomes \(-\frac{b}{m}\). This change occurs because when swapping \(x\) and \(y\) to find the inverse function, the operations on the variables lead to a necessary adjustment not only in the slope but also in the y-intercept. Recognizing these changes in parameters is essential for accurately plotting inverse lines.
Differentiation
Differentiation is a core concept in calculus that helps analyze how functions change. In terms of linear equations, differentiating helps confirm the slope. The derivative of the function \(f(x) = mx + b\) is simply \(m\), indicating a constant rate of change. This matches intuition, as linear functions change at a uniform rate across all points. When you differentiate the inverse function, \(f^{-1}(x) = \frac{x-b}{m}\), you find that its derivative, \(\frac{d}{dx}\left( \frac{x-b}{m} \right)\), is \(\frac{1}{m}\). This result confirms that the inverse's slope is indeed \(\frac{1}{m}\), providing a mathematical verification of this property. Utilizing differentiation gives a robust method for understanding these relationships and ensuring consistency between original and inverse functions.