Precalculus with Limits
3rd Edition
ISBN: 9781133947202
Author: Ron Larson
Publisher: Cengage Learning
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Question
Chapter 1.9, Problem 99E
To determine
To prove 
if 
and 
are one to one functions.
Expert Solution & Answer
Explanation of Solution
Let
and
besets and let 
and
be functions.
If 
and 
are both invertible then 
is invertible.
Let
Then
Using the property
Where
is called Identity Function, a function from set 
to 
or 
Also,
Using the property
Where
is called Identity Function, a function from set 
to 
The function 
is therefore the inverse of 
Hence 
To prove 
Let 
and 
be sets and let 
and 
be functions.
Let 
Then
Chapter 1 Solutions
Precalculus with Limits
Ch. 1.1 - Prob. 1ECh. 1.1 - Prob. 2ECh. 1.1 - Prob. 3ECh. 1.1 - Prob. 4ECh. 1.1 - Prob. 5ECh. 1.1 - Prob. 6ECh. 1.1 - Prob. 7ECh. 1.1 - Prob. 8ECh. 1.1 - Prob. 9ECh. 1.1 - Prob. 10E
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Prob. 63ECh. 1.8 - Prob. 64ECh. 1.8 - Prob. 65ECh. 1.8 - Prob. 66ECh. 1.8 - Prob. 67ECh. 1.8 - Prob. 68ECh. 1.8 - Prob. 69ECh. 1.8 - Prob. 70ECh. 1.9 - Prob. 1ECh. 1.9 - Prob. 2ECh. 1.9 - Prob. 3ECh. 1.9 - Prob. 4ECh. 1.9 - Prob. 5ECh. 1.9 - Prob. 6ECh. 1.9 - Prob. 7ECh. 1.9 - Prob. 8ECh. 1.9 - Prob. 9ECh. 1.9 - Prob. 10ECh. 1.9 - Prob. 11ECh. 1.9 - Prob. 12ECh. 1.9 - Prob. 13ECh. 1.9 - Prob. 14ECh. 1.9 - Prob. 15ECh. 1.9 - Prob. 16ECh. 1.9 - Prob. 17ECh. 1.9 - Prob. 18ECh. 1.9 - Prob. 19ECh. 1.9 - Prob. 20ECh. 1.9 - Prob. 21ECh. 1.9 - Prob. 22ECh. 1.9 - Prob. 23ECh. 1.9 - Prob. 24ECh. 1.9 - Prob. 25ECh. 1.9 - Prob. 26ECh. 1.9 - Prob. 27ECh. 1.9 - Prob. 28ECh. 1.9 - Prob. 29ECh. 1.9 - Prob. 30ECh. 1.9 - Prob. 31ECh. 1.9 - Prob. 32ECh. 1.9 - Prob. 33ECh. 1.9 - Prob. 34ECh. 1.9 - Prob. 35ECh. 1.9 - Prob. 36ECh. 1.9 - Prob. 37ECh. 1.9 - Prob. 38ECh. 1.9 - Prob. 39ECh. 1.9 - Prob. 40ECh. 1.9 - Prob. 41ECh. 1.9 - Prob. 42ECh. 1.9 - Prob. 43ECh. 1.9 - Prob. 44ECh. 1.9 - Prob. 45ECh. 1.9 - Prob. 46ECh. 1.9 - Prob. 47ECh. 1.9 - Prob. 48ECh. 1.9 - Prob. 49ECh. 1.9 - Prob. 50ECh. 1.9 - Prob. 51ECh. 1.9 - Prob. 52ECh. 1.9 - Prob. 53ECh. 1.9 - Prob. 54ECh. 1.9 - Prob. 55ECh. 1.9 - Prob. 56ECh. 1.9 - Prob. 57ECh. 1.9 - Prob. 58ECh. 1.9 - Prob. 59ECh. 1.9 - Prob. 60ECh. 1.9 - Prob. 61ECh. 1.9 - Prob. 62ECh. 1.9 - Prob. 63ECh. 1.9 - Prob. 64ECh. 1.9 - Prob. 65ECh. 1.9 - Prob. 66ECh. 1.9 - Prob. 67ECh. 1.9 - Prob. 68ECh. 1.9 - Prob. 69ECh. 1.9 - Prob. 70ECh. 1.9 - Prob. 71ECh. 1.9 - Prob. 72ECh. 1.9 - Prob. 73ECh. 1.9 - Prob. 74ECh. 1.9 - Prob. 75ECh. 1.9 - Prob. 76ECh. 1.9 - Prob. 77ECh. 1.9 - Prob. 78ECh. 1.9 - Prob. 79ECh. 1.9 - Prob. 80ECh. 1.9 - Prob. 81ECh. 1.9 - Prob. 82ECh. 1.9 - Prob. 83ECh. 1.9 - Prob. 84ECh. 1.9 - Prob. 85ECh. 1.9 - Prob. 86ECh. 1.9 - Prob. 87ECh. 1.9 - Prob. 88ECh. 1.9 - Prob. 89ECh. 1.9 - Prob. 90ECh. 1.9 - Prob. 91ECh. 1.9 - Prob. 92ECh. 1.9 - Prob. 93ECh. 1.9 - Prob. 94ECh. 1.9 - Prob. 95ECh. 1.9 - Prob. 96ECh. 1.9 - Prob. 97ECh. 1.9 - Prob. 98ECh. 1.9 - Prob. 99ECh. 1.9 - Prob. 100ECh. 1.9 - Prob. 101ECh. 1.9 - Prob. 102ECh. 1.9 - Prob. 103ECh. 1.9 - Prob. 104ECh. 1.9 - Prob. 105ECh. 1.9 - Prob. 106ECh. 1.10 - Prob. 1ECh. 1.10 - Prob. 2ECh. 1.10 - Prob. 3ECh. 1.10 - Prob. 4ECh. 1.10 - Prob. 5ECh. 1.10 - Prob. 6ECh. 1.10 - Prob. 7ECh. 1.10 - Prob. 8ECh. 1.10 - Prob. 9ECh. 1.10 - Prob. 10ECh. 1.10 - Prob. 11ECh. 1.10 - Prob. 12ECh. 1.10 - Prob. 13ECh. 1.10 - Prob. 14ECh. 1.10 - Prob. 15ECh. 1.10 - Prob. 16ECh. 1.10 - Prob. 17ECh. 1.10 - Prob. 18ECh. 1.10 - Prob. 19ECh. 1.10 - Prob. 20ECh. 1.10 - Prob. 21ECh. 1.10 - Prob. 22ECh. 1.10 - Prob. 23ECh. 1.10 - Prob. 24ECh. 1.10 - Prob. 25ECh. 1.10 - Prob. 26ECh. 1.10 - Prob. 27ECh. 1.10 - Prob. 28ECh. 1.10 - Prob. 29ECh. 1.10 - Prob. 30ECh. 1.10 - Prob. 31ECh. 1.10 - Prob. 32ECh. 1.10 - Prob. 33ECh. 1.10 - Prob. 34ECh. 1.10 - Prob. 35ECh. 1.10 - Prob. 36ECh. 1.10 - Prob. 37ECh. 1.10 - Prob. 38ECh. 1.10 - Prob. 39ECh. 1.10 - Prob. 40ECh. 1.10 - Prob. 41ECh. 1.10 - Prob. 42ECh. 1.10 - Prob. 43ECh. 1.10 - Prob. 44ECh. 1.10 - Prob. 45ECh. 1.10 - Prob. 46ECh. 1.10 - Prob. 47ECh. 1.10 - Prob. 48ECh. 1.10 - Prob. 49ECh. 1.10 - Prob. 50ECh. 1.10 - Prob. 51ECh. 1.10 - Prob. 52ECh. 1.10 - Prob. 53ECh. 1.10 - Prob. 54ECh. 1.10 - Prob. 55ECh. 1.10 - Prob. 56ECh. 1.10 - Prob. 57ECh. 1.10 - Prob. 58ECh. 1.10 - Prob. 59ECh. 1.10 - Prob. 60ECh. 1.10 - Prob. 61ECh. 1.10 - Prob. 62ECh. 1.10 - Prob. 63ECh. 1.10 - Prob. 64ECh. 1.10 - Prob. 65ECh. 1.10 - Prob. 66ECh. 1.10 - Prob. 67ECh. 1.10 - Prob. 68ECh. 1.10 - Prob. 69ECh. 1.10 - Prob. 70ECh. 1.10 - Prob. 71ECh. 1.10 - Prob. 72ECh. 1.10 - Prob. 73ECh. 1.10 - Prob. 74ECh. 1.10 - Prob. 75ECh. 1.10 - Prob. 76ECh. 1.10 - Prob. 77ECh. 1.10 - Prob. 78ECh. 1.10 - Prob. 79ECh. 1.10 - Prob. 80ECh. 1 - Prob. 1RECh. 1 - Prob. 2RECh. 1 - Prob. 3RECh. 1 - Prob. 4RECh. 1 - Prob. 5RECh. 1 - Prob. 6RECh. 1 - Prob. 7RECh. 1 - Prob. 8RECh. 1 - Prob. 9RECh. 1 - Prob. 10RECh. 1 - Prob. 11RECh. 1 - Prob. 12RECh. 1 - Prob. 13RECh. 1 - Prob. 14RECh. 1 - Prob. 15RECh. 1 - Prob. 16RECh. 1 - Prob. 17RECh. 1 - Prob. 18RECh. 1 - Prob. 19RECh. 1 - Prob. 20RECh. 1 - Prob. 21RECh. 1 - Prob. 22RECh. 1 - Prob. 23RECh. 1 - Prob. 24RECh. 1 - Prob. 25RECh. 1 - Prob. 26RECh. 1 - Prob. 27RECh. 1 - Prob. 28RECh. 1 - Prob. 29RECh. 1 - Prob. 30RECh. 1 - Prob. 31RECh. 1 - Prob. 32RECh. 1 - Prob. 33RECh. 1 - Prob. 34RECh. 1 - Prob. 35RECh. 1 - Prob. 36RECh. 1 - Prob. 37RECh. 1 - Prob. 38RECh. 1 - Prob. 39RECh. 1 - Prob. 40RECh. 1 - Prob. 41RECh. 1 - Prob. 42RECh. 1 - Prob. 43RECh. 1 - Prob. 44RECh. 1 - Prob. 45RECh. 1 - Prob. 46RECh. 1 - Prob. 47RECh. 1 - Prob. 48RECh. 1 - Prob. 49RECh. 1 - Prob. 50RECh. 1 - Prob. 51RECh. 1 - Prob. 52RECh. 1 - Prob. 53RECh. 1 - Prob. 54RECh. 1 - Prob. 55RECh. 1 - Prob. 56RECh. 1 - Prob. 57RECh. 1 - Prob. 58RECh. 1 - Prob. 59RECh. 1 - Prob. 60RECh. 1 - Prob. 61RECh. 1 - Prob. 62RECh. 1 - Prob. 63RECh. 1 - Prob. 64RECh. 1 - Prob. 65RECh. 1 - Prob. 66RECh. 1 - Prob. 67RECh. 1 - Prob. 68RECh. 1 - Prob. 69RECh. 1 - Prob. 70RECh. 1 - Prob. 71RECh. 1 - Prob. 72RECh. 1 - Prob. 73RECh. 1 - Prob. 74RECh. 1 - Prob. 75RECh. 1 - Prob. 76RECh. 1 - Prob. 77RECh. 1 - Prob. 78RECh. 1 - Prob. 79RECh. 1 - Prob. 80RECh. 1 - Prob. 81RECh. 1 - Prob. 82RECh. 1 - Prob. 83RECh. 1 - Prob. 84RECh. 1 - Prob. 85RECh. 1 - Prob. 86RECh. 1 - Prob. 87RECh. 1 - Prob. 88RECh. 1 - Prob. 89RECh. 1 - Prob. 90RECh. 1 - Prob. 91RECh. 1 - Prob. 92RECh. 1 - Prob. 93RECh. 1 - Prob. 94RECh. 1 - Prob. 95RECh. 1 - Prob. 96RECh. 1 - Prob. 97RECh. 1 - Prob. 98RECh. 1 - Prob. 99RECh. 1 - Prob. 100RECh. 1 - Prob. 101RECh. 1 - Prob. 102RECh. 1 - Prob. 103RECh. 1 - Prob. 1CTCh. 1 - Prob. 2CTCh. 1 - Prob. 3CTCh. 1 - Prob. 4CTCh. 1 - Prob. 5CTCh. 1 - Prob. 6CTCh. 1 - Prob. 7CTCh. 1 - Prob. 8CTCh. 1 - Prob. 9CTCh. 1 - Prob. 10CTCh. 1 - Prob. 11CTCh. 1 - Prob. 12CTCh. 1 - Prob. 13CTCh. 1 - Prob. 14CTCh. 1 - Prob. 15CTCh. 1 - Prob. 16CTCh. 1 - Prob. 17CTCh. 1 - Prob. 18CTCh. 1 - Prob. 19CTCh. 1 - Prob. 20CTCh. 1 - Prob. 21CTCh. 1 - Prob. 22CTCh. 1 - Prob. 23CTCh. 1 - Prob. 24CTCh. 1 - Prob. 25CTCh. 1 - Prob. 26CTCh. 1 - Prob. 1PSCh. 1 - Prob. 2PSCh. 1 - Prob. 3PSCh. 1 - Prob. 4PSCh. 1 - Prob. 5PSCh. 1 - Prob. 6PSCh. 1 - Prob. 7PSCh. 1 - Prob. 8PSCh. 1 - Prob. 9PSCh. 1 - Prob. 10PSCh. 1 - Prob. 11PSCh. 1 - Prob. 12PSCh. 1 - Prob. 13PSCh. 1 - Prob. 14PSCh. 1 - Prob. 15PS
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- nd ave a ction and ave an 48. The domain of f y=f'(x) x 1 2 (= x<0 x<0 = f(x) possible. Group Activity In Exercises 49 and 50, do the following. (a) Find the absolute extrema of f and where they occur. (b) Find any points of inflection. (c) Sketch a possible graph of f. 49. f is continuous on [0,3] and satisfies the following. X 0 1 2 3 f 0 2 0 -2 f' 3 0 does not exist -3 f" 0 -1 does not exist 0 ve tes where X 0 < x <1 1< x <2 2arrow_forwardNumerically estimate the value of limx→2+x3−83x−9, rounded correctly to one decimal place. In the provided table below, you must enter your answers rounded exactly to the correct number of decimals, based on the Numerical Conventions for MATH1044 (see lecture notes 1.3 Actions page 3). If there are more rows provided in the table than you need, enter NA for those output values in the table that should not be used. x→2+ x3−83x−9 2.1 2.01 2.001 2.0001 2.00001 2.000001arrow_forwardFind the general solution of the given differential equation. (1+x)dy/dx - xy = x +x2arrow_forwardEstimate the instantaneous rate of change of the function f(x) = 2x² - 3x − 4 at x = -2 using the average rate of change over successively smaller intervals.arrow_forwardGiven the graph of f(x) below. Determine the average rate of change of f(x) from x = 1 to x = 6. Give your answer as a simplified fraction if necessary. For example, if you found that msec = 1, you would enter 1. 3' −2] 3 -5 -6 2 3 4 5 6 7 Ꮖarrow_forwardGiven the graph of f(x) below. Determine the average rate of change of f(x) from x = -2 to x = 2. Give your answer as a simplified fraction if necessary. For example, if you found that msec = , you would enter 3 2 2 3 X 23arrow_forwardA function is defined on the interval (-π/2,π/2) by this multipart rule: if -π/2 < x < 0 f(x) = a if x=0 31-tan x +31-cot x if 0 < x < π/2 Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0. a= b= 3arrow_forwardUse the definition of continuity and the properties of limits to show that the function is continuous at the given number a. f(x) = (x + 4x4) 5, a = -1 lim f(x) X--1 = lim x+4x X--1 lim X-1 4 x+4x 5 ))" 5 )) by the power law by the sum law lim (x) + lim X--1 4 4x X-1 -(0,00+( Find f(-1). f(-1)=243 lim (x) + -1 +4 35 4 ([ ) lim (x4) 5 x-1 Thus, by the definition of continuity, f is continuous at a = -1. by the multiple constant law by the direct substitution propertyarrow_forward1. Compute Lo F⚫dr, where and C is defined by F(x, y) = (x² + y)i + (y − x)j r(t) = (12t)i + (1 − 4t + 4t²)j from the point (1, 1) to the origin.arrow_forward2. Consider the vector force: F(x, y, z) = 2xye²i + (x²e² + y)j + (x²ye² — z)k. (A) [80%] Show that F satisfies the conditions for a conservative vector field, and find a potential function (x, y, z) for F. Remark: To find o, you must use the method explained in the lecture. (B) [20%] Use the Fundamental Theorem for Line Integrals to compute the work done by F on an object moves along any path from (0,1,2) to (2, 1, -8).arrow_forwardhelp pleasearrow_forwardIn each of Problems 1 through 4, draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at t = 0, describe the dependency.1. y′ = 3 − 2yarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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