Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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DEFINITION 3 (INJECTIVE FUNCTION). Let A, B C R be two subsets of the real numbers, and let f : A → B
be a function. We say that f is injective (we might alternatively say that f is 1-to-1) if for all a₁, a2 € A such
that f(a₁) = f(a₂) we have that a₁ = a2.
DEFINITION 4 (SURJECTIVE FUNCTION). Let A, B C R be two subsets of the real numbers, and let f : A → B
be a function. We say that f is surjective (we might alternatively say that f is onto) if for every b = B we can
find a € A such that f(a) = b.
Exercise 4. Let p: R → R be a polynomial function, that is, assume that p is given by an expression of the
form
p(x) = x² +an-1-¹+...+ao.
a) Assume that n is an odd number, that is, assume that n is not divisible by 2. Show that p is surjective.
b) Find an example of a polynomial p of degree three such that p is not injective.
c) Assume that n is an even number, that is, assume that n is divisible by 2. Show that p is not surjective
nor injective.
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Transcribed Image Text:DEFINITION 3 (INJECTIVE FUNCTION). Let A, B C R be two subsets of the real numbers, and let f : A → B be a function. We say that f is injective (we might alternatively say that f is 1-to-1) if for all a₁, a2 € A such that f(a₁) = f(a₂) we have that a₁ = a2. DEFINITION 4 (SURJECTIVE FUNCTION). Let A, B C R be two subsets of the real numbers, and let f : A → B be a function. We say that f is surjective (we might alternatively say that f is onto) if for every b = B we can find a € A such that f(a) = b. Exercise 4. Let p: R → R be a polynomial function, that is, assume that p is given by an expression of the form p(x) = x² +an-1-¹+...+ao. a) Assume that n is an odd number, that is, assume that n is not divisible by 2. Show that p is surjective. b) Find an example of a polynomial p of degree three such that p is not injective. c) Assume that n is an even number, that is, assume that n is divisible by 2. Show that p is not surjective nor injective.
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