Plzz hurry :<
2 answers:
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Answer: First, injective means that for every y, there is only one x such f(x) = y, and surjective means that if f is a function that goes from the set {x} to the set {y}, for every element y in the codomain there is at least one element x in the domain such f(x) = y
(a) f:R----> + R given by f(x) = x2 (i guess you written )
The function is not injective, because f(-2) = f(2), and is surjective, because the codomain is +R, and is always a positive real number.
(b) f:N----> + N given by f(n) = n2 (i guess you written )
As the naturals have no negative numbers, in this case the function is injective, but isn't surjective, because there is no number that when squared is equal to 5, for example.
(c) f: Zx Z → Z given by f(n, k) = n +k:
here the domain is of the form (n,k) and the function is n +k, so the numbers (z,w) and (w,z) return the same value when evaluated in f, then f is not injective. And is easy to see that f is surjective, because in the sum you can reach al the integers on the codomain.
Answer:
B. No, the remainder is -50.
General Formulas and Concepts:
<u>Algebra I</u>
- Roots are when the polynomial are equal to 0
<u>Algebra II</u>
- Synthetic Division
Step-by-step explanation:
<u>Step 1: Define</u>
Function f(x) = x³ - 10x² + 27x - 12
Divisor/Root (x + 1)
<u>Step 2: Synthetic Division</u>
<em>See Attachment.</em>
To determine whether a given root is an actual root, the remainder must equal 0. Since we have a remainder of -50, the given root is not a factor of the polynomial.
<em>Please excuse the bad handwriting. Hope this helped!</em>
