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Chapter 10: Problem 20

In general, two functions \(f(x)\) and \(g(x)\) are inverses of each other if and only if \(f(g(x))=x\) and \(g(f(x))=x .\) Verify that the pairs of functions are inverses of each other. a) \(f(x)=5 x+10\) and \(g(x)=\frac{1}{5} x-2\) b) \(f(x)=\frac{x-1}{2}\) and \(g(x)=2 x+1\) c) \(f(x)=\sqrt[3]{x+1}\) and \(g(x)=x^{3}-1\) d) \(f(x)=5^{x}\) and \(g(x)=\log _{5} x\)

Short Answer

Expert verified
All given pairs of functions are inverses because they satisfy \(f(g(x)) = x\) and \(g(f(x)) = x\).

Step by step solution

01

- Verify for the pair (a)

First, substitute the function g(x) into f(x):\[f(g(x)) = f\left( \frac{1}{5}x - 2 \right)\]\[= 5\left(\frac{1}{5}x - 2\right) + 10 \]\[= x - 10 + 10 \]\[= x\]Now, substitute the function f(x) into g(x):\[g(f(x)) = g(5x + 10)\]\[= \frac{1}{5}(5x + 10) - 2\]\[= x + 2 - 2\]\[= x\]Since both conditions are satisfied, \(f(x) = 5x + 10\) and \(g(x) = \frac{1}{5}x - 2\) are inverses.
02

- Verify for the pair (b)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(2x + 1)\]\[= \frac{2x + 1 - 1}{2} \]\[= \frac{2x}{2} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g\left( \frac{x - 1}{2} \right) \]\[= 2\left( \frac{x - 1}{2} \right) + 1 \]\[= x - 1 + 1 \]\[= x\]Since both conditions are satisfied, \(f(x) = \frac{x-1}{2}\) and \(g(x) = 2x + 1\) are inverses.
03

- Verify for the pair (c)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(x^3 - 1)\]\[= \sqrt[3]{(x^3 - 1) + 1} \]\[= \sqrt[3]{x^3} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g(\sqrt[3]{x+1})\]\[= (\sqrt[3]{x+1})^3 - 1 \]\[= x + 1 - 1 \]\[= x\]Since both conditions are satisfied, \(f(x) = \sqrt[3]{x+1}\) and \(g(x) = x^3 - 1\) are inverses.
04

- Verify for the pair (d)

First, substitute the function g(x) into f(x):\[f(g(x)) = f(\log_5 x)\]\[= 5^{\log_5 x} \]\[= x\]Next, substitute the function f(x) into g(x):\[g(f(x)) = g(5^x)\]\[= \log_5(5^x)\]\[= x\]Since both conditions are satisfied, \(f(x) = 5^x\) and \(g(x) = \log_5 x\) are inverses.

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Key Concepts

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

Function Composition
To understand inverse functions, it's essential to know about function composition. Function composition involves applying one function to the results of another function. If you have two functions, say \(f(x)\) and \(g(x)\), you can create a new function by composing them in two ways: \(f(g(x))\) and \(g(f(x))\). Essentially, you substitute one function into the other.
For example, consider \(f(x) = 2x + 3\) and \(g(x) = x - 1\). To find \(f(g(x))\), substitute \(g(x)\) into \(f(x)\):
  • Step 1: Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = f(x - 1)\)
  • Step 2: Calculate the result: \(f(x - 1) = 2(x - 1) + 3 = 2x - 2 + 3 = 2x + 1\)
Similarly, for \(g(f(x))\):
  • Step 1: Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = g(2x + 3)\)
  • Step 2: Calculate the result: \(g(2x + 3) = (2x + 3) - 1 = 2x + 2\)
Function composition helps verify whether two functions are inverses.
Precalculus Verification
Precalculus verification is the process of confirming that two functions are indeed inverses. This involves checking two important conditions: \(f(g(x)) = x\) and \(g(f(x)) = x\). If both conditions hold true, the functions are inverses.
Let鈥檚 verify the pairs of functions provided in the exercise:
a) \(f(x) = 5x + 10\) and \(g(x) = \frac{1}{5}x - 2\)
  • Substitute \(g(x)\) into \(f(x)\): \(f(g(x)) = f\left( \frac{1}{5}x - 2 \right) = 5\left(\frac{1}{5}x - 2 \right) + 10 = x\)
  • Substitute \(f(x)\) into \(g(x)\): \(g(f(x)) = g(5x + 10) = \frac{1}{5}(5x + 10) - 2 = x\)
Since both conditions are satisfied, these functions are inverses.
We repeat similar verification steps for pairs b, c, and d as detailed in the original solution.
Inverse Properties
Inverse functions have unique properties that distinguish them. If \(f\) and \(g\) are inverses, applying one function after the other returns the starting value: \(f(g(x)) = x\) and \(g(f(x)) = x\). This cancellation effect is fundamental to the concept of inverse functions.
Consider a real-life analogy: If you lock a door (apply \(f\)), then unlock it (apply \(g\)), you return to the initial state of the door being unlocked.
This cancellation leads to several important implications:
  • One-to-One Functions: Only one-to-one functions (bijective functions) have inverses. These functions pass the horizontal line test.
  • Symmetry: If \(y = f(x)\), then \(x = g(y)\). The graph of \(f\) and its inverse \(g\) will be symmetric with respect to the line \(y = x\).
  • Notational Clarity: The inverse of a function \(f\) is denoted \(f^{-1}\), but this notation does not imply exponentiation.
Understanding these properties provides deeper insights into how inverse functions operate and their significance in mathematics.

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Most popular questions from this chapter

Let \(f_{1}(x)=x, f_{2}(x)=\frac{1}{x}, f_{3}(x)=1-x\), \(f_{4}(x)=\frac{x}{x-1}, f_{5}(x)=\frac{1}{1-x},\) and \(f_{6}(x)=\frac{x-1}{x}\). a) Determine the following. i) \(f_{2}\left(f_{3}(x)\right)\) ii) \(\left(f_{3} \circ f_{5}\right)(x)\) iii) \(f_{1}\left(f_{2}(x)\right)\) iv) \(f_{2}\left(f_{1}(x)\right)\) b) \(f_{6}^{-1}(x)\) is the same as which function listed in part a)?

Tobias is shopping at a local sports store that is having a \(25 \%\) -off sale on apparel. Where he lives, the federal tax adds \(5 \%\) to the selling price. a) Write the function, \(s(p),\) that relates the regular price, \(p,\) to the sale price, \(s,\) both in dollars. b) Write the function, \(t(s),\) that relates the sale price, \(s,\) to the total cost including taxes, \(t,\) both in dollars. c) Write a composite function that expresses the total cost in terms of the regular price. How much did Tobias pay for a jacket with a regular price of \$89.99?

The graph of \(y=f(x) g(x),\) where \(g(x)\) is a sinusoidal function, will oscillate between the graphs of \(f(x)\) and \(-f(x) .\) When the amplitude of the wave is reduced, this is referred to as damping. a) Given the functions \(f(x)=\frac{2}{x^{2}+1}\) and \(g(x)=\sin (6 x-1),\) show that the above scenario occurs. b) Does the above scenario occur for \(f(x)=\cos x\) and \(g(x)=\sin (6 x-1) ?\)

Given \(f(x)=\sqrt{x}\) and \(g(x)=x-1,\) sketch the graph of each composite function. Then, determine the domain and range of each composite function. a) \(y=f(g(x))\) b) \(y=g(f(x))\)

An even function satisfies the property \(f(-x)=f(x)\) for all \(x\) in the domain of \(f(x)\) An odd function satisfies the property \(f(-x)=-f(x)\) for all \(x\) in the domain of \(f(x)\). Devise and test an algebraic method to determine if the sum of two functions is even, odd, or neither. Show by example how your method works. Use at least three of the functions you have studied: absolute value, radical, polynomial, trigonometric, exponential, logarithmic, and rational.
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