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Chapter 5: Problem 1

In Exercises \(1-8\), show that \(f\) and \(g\) are inverse functions (a) analytically and (b) graphically. \(f(x)=5 x+1\), \(g(x)=(x-1) / 5\)

Short Answer

Expert verified
\(f(x)\) and \(g(x)\) are inverse functions of each other.

Step by step solution

01

Analytical Method - \(f(g(x))\) Calculation

To start with, calculate \(f(g(x))\). Substitute \(g(x)\) in place of \(x\) in \(f(x)\), which results in: \(f(g(x)) = f((x-1)/5) = 5((x-1)/5) + 1 = x - 1 + 1 = x\).
02

Analytical Method - \(g(f(x))\) Calculation

Next, calculate \(g(f(x))\). Substitute \(f(x)\) in place of \(x\) in \(g(x)\), which results in: \(g(f(x)) = g(5x+1) = ((5x+1)-1)/5 = 5x/5 = x\).
03

Graphical Method - Sketch the Graphs of \(f(x)\) and \(g(x)\)

Sketch the graphs of \(f(x) = 5x + 1\) and \(g(x) = (x-1)/5\). The function \(f(x) = 5x + 1\) is a straight line with slope 5 and y-intercept 1. The function \(g(x) = (x-1)/5\) is a straight line with slope 1/5 and y-intercept -1/5.
04

Graphical Method - Compare the Graphs of \(f(x)\) and \(g(x)\)

Compare these two graphs. If \(g(x)\) is the reflection of \(f(x)\) in the line \(y = x\), then \(f\) and \(g\) are inverse of each other. For these two functions, this is indeed the case.
05

Conclusion

Since \(f(g(x)) = x\), \(g(f(x)) = x\), and the graph of \(g(x)\) is the reflection of \(f(x)\) in the line \(y=x\), it can be concluded that \(f\) and \(g\) are inverse functions of each other.

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Key Concepts

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

Analytical Verification
In order to demonstrate that two functions are inverses of each other analytically, we need to perform function composition and check if the original input is retrieved. This involves calculating both \( f(g(x)) \) and \( g(f(x)) \). If both of these compositions simplify to \( x \), then the functions are indeed inverses.

For the given functions \( f(x) = 5x + 1 \) and \( g(x) = \frac{x-1}{5} \):
  • Calculate \( f(g(x)) \): Substitute \( g(x) = \frac{x-1}{5} \) into \( f(x) \), which simplifies as follows: \( f(g(x)) = f\left(\frac{x-1}{5}\right) = 5\left(\frac{x-1}{5}\right) + 1 = x - 1 + 1 = x \).
  • Calculate \( g(f(x)) \): Substitute \( f(x) = 5x + 1 \) into \( g(x) \), which gives: \( g(f(x)) = g(5x+1) = \frac{(5x+1) - 1}{5} = \frac{5x}{5} = x \).
Each composition returns \( x \), thus confirming analytically that \( f \) and \( g \) are inverse functions.
Graphical Representation
Graphically proving that two functions are inverses involves looking at their reflections over the line \( y = x \). If one function's graph reflects perfectly onto the other across this line, they are inverses of each other.

For \( f(x) = 5x + 1 \) and \( g(x) = \frac{x-1}{5} \):
  • Graph \( f(x) \): This is a straight line with a slope of 5 and a y-intercept of 1.
  • Graph \( g(x) \): This line has a much gentler slope of \( \frac{1}{5} \) and a y-intercept of \(-\frac{1}{5}\).
To confirm the inverse relationship, observe these graphs. \( g(x) \) should appear as a mirror image of \( f(x) \) across \( y = x \). By doing so, you visually verify their inverse relationship.
Function Composition
Function composition is a crucial concept in mathematics to understand how two functions can effectively "undo" each other.

The composition of functions \( f(x) = 5x + 1 \) and \( g(x) = \frac{x-1}{5} \) helps illustrate how these two functions are designed to be inverses. Let's break it down:
  • When we compose \( f \) and \( g \) as \( f(g(x)) \), you substitute \( g(x) \) into \( f \), causing the operations of the function to cancel out the transformation of \( g \), returning \( x \).
  • Conversely, composing \( g \) and \( f \) as \( g(f(x)) \), you effectively reverse the transformation from \( f \) back to the original \( x \).
This type of function composition is pivotal in demonstrating that the two functions balance each other's transformations, confirming they are indeed inverse functions.

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Most popular questions from this chapter

Find the derivative of the function. $$ y=\tanh ^{-1}(\sin 2 x) $$

An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

Prove that if \(f\) has an inverse function, then \(\left(f^{-1}\right)^{-1}=f\).

The derivative of the function has the same sign for all \(x\) in its domain, but the function is not one-to-one. Explain. \(f(x)=\frac{x}{x^{2}-4}\)

Find any relative extrema of the function. Use a graphing utility to confirm your result. $$ g(x)=x \operatorname{sech} x $$
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