2. In each item below, you are given sets A and B, and a formula for a (non?) function f: A → B. In each case, determine whether or not the formula gives a well-defined function. Fully justify your answers. (a) A = Q = B, f(r) = a root in B of the polynomial X²-r (b) A=C=B, f(2)= a root in B of the polynomial X² - z (c) A = P(R) - {0}¹, B = R, ƒ(E)= the smallest element of E (d) A = P(R), B = R, f(E)= the smallest element of E if E admits a smallest ele- ment, and f(E)= 0 otherwise A function f A → B is injective (or one-to-one) if it satisfies the following condition: for all a₁, a₂ € A, if f(a₁) = f(a₂), then a₁ = a2; f is surjective (or onto) if f(A) = B, or, more explicitly, if for each b E B there exists a € A such that f(a) = b; f is bijective if f is both injective and surjective. Here are some examples. • The squaring function f: RR given by f(x) = x² is not injective because f(-1) = 1 = f(1); f also fails to be surjective because -1 is not in the range of f. The function g: R→ [0,00) given by g(x) = x² is not injective for the same reason that the function f of the previous example is not injective, since g(-1) = 1 = g(1); but g is surjective, because every nonnegative real number admits a square root (this is a result in introductory real analysis). • The sine function sin: R R is very far from injective: it is 27-periodic, and assumes all of its values infinitely often. It is not surjective either, because its range is [-1,1]. • If we define S: [-π/2, π/2] → [-1, 1] by S(x) = sin(x), then methods from calculus show that S is both injective and surjective, i.e., S is bijective. Restricting the domain and codomain of sin in this way is how one obtains the arcsin function, which is the inverse of the function we have called S (see below for more on inverse functions). • The swapping function from Rx R to itself sending (x, y) to (y, z) is bijective. (Verify this!) • The exponential function exp: R→ (0, ∞) is bijective. (I write exp(r) for what is typically written e in calculus textbooks. This notation is pretty standard outside

Algebra & Trigonometry with Analytic Geometry
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Chapter4: Polynomial And Rational Functions
Section4.1: Polynomial Functions Of Degree Greater Than
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2. In each item below, you are given sets A and B, and a formula for a (non?)function
f: A → B. In each case, determine whether or not the formula gives a well-defined function.
Fully justify your answers.
(a) A = Q = B, f(r) = a root in B of the polynomial X² - r
(b) A = C = B, f(2)= a root in B of the polynomial X² – z
(c) A = P(R) - {0}¹, B = R₁ ƒ(E) = the smallest element of E
(d) A = P(R), B = R, ƒ(E) = the smallest element of E if E admits a smallest ele-
ment, and f(E)= 0 otherwise
A function f : A → B is injective (or one-to-one) if it satisfies the following condition:
for all a₁, a2 € A, if f(a₁) = f(a₂), then a₁ = a₂; f is surjective (or onto) if f(A) = B, or,
more explicitly, if for each b E B there exists a € A such that f(a) = b; f is bijective if f is
both injective and surjective. Here are some examples.
• The squaring function f : R → R given by f(x) = x² is not injective because
f(-1) = 1 = f(1); f also fails to be surjective because -1 is not in the range of f.
The function g: R→ [0, ∞) given by g(x) = x² is not injective for the same reason
that the function f of the previous example is not injective, since g(-1) = 1 = g(1);
but g is surjective, because every nonnegative real number admits a square root (this
is a result in introductory real analysis).
The sine function sin: R → R is very far from injective: it is 2-periodic, and
assumes all of its values infinitely often. It is not surjective either, because its range
is [−1,1].
If we define S: [-/2, π/2] → [1,1] by S(x) = sin(x), then methods from calculus
show that S is both injective and surjective, i.e., S is bijective. Restricting the domain
and codomain of sin in this way is how one obtains the arcsin function, which is the
inverse of the function we have called S (see below for more on inverse functions).
• The swapping function from RXR to itself sending (x, y) to (y, z) is bijective. (Verify
this!)
The exponential function exp: R→ (0, ∞) is bijective. (I write exp(r) for what is
typically written eª in calculus textbooks. This notation is pretty standard outside
the confines of first-year calculus.)
Transcribed Image Text:2. In each item below, you are given sets A and B, and a formula for a (non?)function f: A → B. In each case, determine whether or not the formula gives a well-defined function. Fully justify your answers. (a) A = Q = B, f(r) = a root in B of the polynomial X² - r (b) A = C = B, f(2)= a root in B of the polynomial X² – z (c) A = P(R) - {0}¹, B = R₁ ƒ(E) = the smallest element of E (d) A = P(R), B = R, ƒ(E) = the smallest element of E if E admits a smallest ele- ment, and f(E)= 0 otherwise A function f : A → B is injective (or one-to-one) if it satisfies the following condition: for all a₁, a2 € A, if f(a₁) = f(a₂), then a₁ = a₂; f is surjective (or onto) if f(A) = B, or, more explicitly, if for each b E B there exists a € A such that f(a) = b; f is bijective if f is both injective and surjective. Here are some examples. • The squaring function f : R → R given by f(x) = x² is not injective because f(-1) = 1 = f(1); f also fails to be surjective because -1 is not in the range of f. The function g: R→ [0, ∞) given by g(x) = x² is not injective for the same reason that the function f of the previous example is not injective, since g(-1) = 1 = g(1); but g is surjective, because every nonnegative real number admits a square root (this is a result in introductory real analysis). The sine function sin: R → R is very far from injective: it is 2-periodic, and assumes all of its values infinitely often. It is not surjective either, because its range is [−1,1]. If we define S: [-/2, π/2] → [1,1] by S(x) = sin(x), then methods from calculus show that S is both injective and surjective, i.e., S is bijective. Restricting the domain and codomain of sin in this way is how one obtains the arcsin function, which is the inverse of the function we have called S (see below for more on inverse functions). • The swapping function from RXR to itself sending (x, y) to (y, z) is bijective. (Verify this!) The exponential function exp: R→ (0, ∞) is bijective. (I write exp(r) for what is typically written eª in calculus textbooks. This notation is pretty standard outside the confines of first-year calculus.)
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