Explanation Given information: f ( x ) = 1 x 2 + 1 Formula used: Domain is the range of input value in the given function. Calculation: For the function f to be defined x 2 + 1 ≥ 0 . The inequality x 2 + 1 ≥ 0 holds for all real x . Therefore, f ( x ) ≥ 0 for all x . So, the domain of definition of function f ( x ) is { x : − ∞ ≤ x ≤ ∞ } . Now, consider the graph of f ( x ) = 1 x 2 + 1 Suppose f ( x 1 ) = f ( x 2 ) 1 x 1 2 + 1 = 1 x 2 2 + 1 x 2 2 + 1 = x 1 2 + 1 x 2 2 + 1 = x 1 2 + 1 x 2 2 = x 1 2 If both x 1 and x 2 are non-negative then x 2 2 = x 1 2 implies x 1 = x 2 . Therefore, f ( x ) is one-one on the restricted domain { x : x ≥ 0 }

4th Edition

ISBN: 9781319309671

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 1.5, Problem 15E

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A domain on which f is one to one and a formula for the inverse of f restricted to this domain. Sketch the graphs of f and f−1.

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Chapter 1.5, Problem 14E

Chapter 1.5, Problem 16E

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A factorization A = PDP 1 is not unique. For A= 7 2 -4 1 1 1 5 0 2 1 one factorization is P = D= and P-1 30 = Use this information with D₁ = to find a matrix P₁ such that - -1 -2 0 3 1 - - 1 05 A-P,D,P P1 (Type an integer or simplified fraction for each matrix element.)

Matrix A is factored in the form PDP 1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace. 30 -1 - 1 0 -1 400 0 0 1 A= 3 4 3 0 1 3 040 3 1 3 0 0 4 1 0 0 003 -1 0 -1 Select the correct choice below and fill in the answer boxes to complete your choice. (Use a comma to separate vectors as needed.) A basis for the corresponding eigenspace is { A. There is one distinct eigenvalue, λ = B. In ascending order, the two distinct eigenvalues are λ₁ ... = and 2 = Bases for the corresponding eigenspaces are { and ( ), respectively. C. In ascending order, the three distinct eigenvalues are λ₁ = = 12/2 = and 3 = Bases for the corresponding eigenspaces are {}, }, and { respectively.

Chapter 1 Solutions

CALCULUS: EARLY TRANS 4TH ED W/ ACCESS

Ch. 1.1 - Prob. 1PQ Ch. 1.1 - Prob. 2PQ Ch. 1.1 - Prob. 3PQ Ch. 1.1 - Prob. 4PQ Ch. 1.1 - Prob. 5PQ Ch. 1.1 - Prob. 6PQ Ch. 1.1 - Prob. 7PQ Ch. 1.1 - Prob. 8PQ Ch. 1.1 - Prob. 9PQ Ch. 1.1 - Prob. 10PQ

Ch. 1.1 - Prob. 1E Ch. 1.1 - Prob. 2E Ch. 1.1 - Prob. 3E Ch. 1.1 - Prob. 4E Ch. 1.1 - Prob. 5E Ch. 1.1 - Prob. 6E Ch. 1.1 - Prob. 7E Ch. 1.1 - Prob. 8E Ch. 1.1 - Prob. 9E Ch. 1.1 - Prob. 10E Ch. 1.1 - Prob. 11E Ch. 1.1 - Prob. 12E Ch. 1.1 - Prob. 13E Ch. 1.1 - Prob. 14E Ch. 1.1 - Prob. 15E Ch. 1.1 - Prob. 16E Ch. 1.1 - Prob. 17E Ch. 1.1 - Prob. 18E Ch. 1.1 - Prob. 19E Ch. 1.1 - Prob. 20E Ch. 1.1 - Prob. 21E Ch. 1.1 - Prob. 22E Ch. 1.1 - Prob. 23E Ch. 1.1 - Prob. 24E Ch. 1.1 - Prob. 25E Ch. 1.1 - Prob. 26E Ch. 1.1 - Prob. 27E Ch. 1.1 - Prob. 28E Ch. 1.1 - Prob. 29E Ch. 1.1 - Prob. 30E Ch. 1.1 - Prob. 31E Ch. 1.1 - Prob. 32E Ch. 1.1 - Prob. 33E Ch. 1.1 - Prob. 34E Ch. 1.1 - Prob. 35E Ch. 1.1 - Prob. 36E Ch. 1.1 - Prob. 37E Ch. 1.1 - Prob. 38E Ch. 1.1 - Prob. 39E Ch. 1.1 - Prob. 40E Ch. 1.1 - Prob. 41E Ch. 1.1 - Prob. 42E Ch. 1.1 - Prob. 43E Ch. 1.1 - Prob. 44E Ch. 1.1 - Prob. 45E Ch. 1.1 - Prob. 46E Ch. 1.1 - Prob. 47E Ch. 1.1 - Prob. 48E Ch. 1.1 - Prob. 49E Ch. 1.1 - Prob. 50E Ch. 1.1 - 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A domain on which f is one to one and a formula for the inverse of f restricted to this domain. Sketch the graphs of f and f − 1 .

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A domain on which f is one to one and a formula for the inverse of f restricted to this domain. Sketch the graphs of f and f−1.

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Chapter 1 Solutions